}{O}{A}\psaxes[yAxis=false,labels=none,
arrowscale=3,arrows={->}](0,0)(0,0)(11.5,0)
\uput[d](0,0){0}\uput[d](3,0){3}
$$
\vspace{.5cm}
\bigskip Now, starting at $3$ move two units
left.
\vspace{.5cm}
$$\psset{unit=.5} \pstGeonode[PointName=none](0,0){O}(3,0){A}(1,0){B}\ncbar[angleA=90,nodesep=3pt,linewidth=2pt,linecolor=red]{->}{A}{B}\psaxes[yAxis=false,labels=none,
arrowscale=3,arrows={->}](0,0)(0,0)(11.5,0)
\uput[d](0,0){0}\uput[d](3,0){3}\uput[d](1,0){1}
$$\vspace{.5cm}
\bigskip
Since we landed at $1$, we conclude that $3-2=1$. \end{solu}
We raise the same objections that we raised against the geometric
method for addition: it becomes cumbersome for large values and it
is prone to error if we miscount. Let us now discuss an algorithm
for subtraction.
Suppose again that we have two arbitrary decimal integers, say
$$ \underline{d_nd_{n-1}\ldots d_2d_1d_0},\qquad \underline{e_pe_{p-1}\ldots e_2e_1e_0}, $$
with $n\leq p$ and $\underline{d_nd_{n-1}\ldots d_2d_1d_0}\leq
\underline{e_pe_{p-1}\ldots e_2e_1e_0}$. To subtract the integers,
line them up at the units:
$$ \begin{array}{cccccccccc}
& e_p & e_{p-1} & \ldots & e_n & e_{n-1} & \ldots & e_2 & e_1 & e_0\\
-& & & & d_n & d_{n-1} & \ldots & d_2 & d_1 & d_0\\
\hline
\end{array}$$
The algorithm is then as follows:
\begin{enumerate}
\item If it follows that $d_k\leq e_k$ for all $k\leq n$, then there is no regrouping and we simply put
the result of $e_k-d_k$ in the $k$th column.
\item If the previous item is not true, consider a string $i \leq j \leq n$ for which $d_i>e_i$,
$d_{i+1}\geq e_{i+1}$, \ldots , $d_j\geq e_j$ and $d_{j+1}<
e_{j+1}$. Starting at the index $i+1$, successively regroup so that
now in the $i$-th column you have $10+e_i-d_i$, on the $i+1$th
column you have $10+e_{i+1}-1-d_{i+1}$, and so on, until the $j$th
column where you will have $10+e_j-1-d_j$ and on the $j+1$th column
you will have $e_{j+1}-d_{j+1}$. Now all these subtractions can be
accomplished. Put the result below in the respective column. Repeat
again after the $j+1$th column if the situation arises again.
\end{enumerate}
\begin{exa}
Use the above algorithm to perform the subtraction
$9876666789-1234554321$.
\end{exa}
\begin{solu}
Since $9876666789>1234554321$, the subtraction is permissible.
Notice that each digit of $1234554321$ is smaller than each
corresponding digit of $9876666789$, and so there is no carrying.
The result is displayed thus:
$$
\opsub[voperator=bottom,carrysub=true]{9876666789}{1234554321}
$$
\end{solu}
\begin{exa}
Use the above algorithm to perform the subtraction
$98723112389-72430014467$.
\end{exa}
\begin{solu}Since $98723112389>72430014467$, the subtraction is
permissible. The subtraction on the units and the tens is effected
without complications, since both digits in the subtrahend are
smaller than the corresponding digits in the minuend:
$$
\opsub[voperator=bottom,resultstyle.3=\hole,resultstyle.4=\hole,resultstyle.5=\hole,resultstyle.6=\hole,resultstyle.7=\hole,resultstyle.8=\hole,resultstyle.9=\hole,resultstyle.10=\hole,resultstyle.11=\hole,
resultstyle.12=\hole]{98723112389}{72430014467}
$$
When we turn to the hundreds, we find that $3<4$, and so, we must
regroup from the digits to the left. In the thousands we find $2<4$.
In the ten thousands we find $1\leq 1$. We stop at the hundred
thousands, because $1>0$. Thus we take $1$ from the hundred
thousands (where we now have $0$) to the ten thousands. In the ten
thousands now we have $10 + 1=11$, but we take $1$ to the thousands,
leaving $10$ in the ten thousands. We now have $10 + 2=12$ in the
thousands, but we take $1$ to the hundreds, leaving $11$ in the
thousands. Finally, we have $10+3=13$ in the hundreds. We subtract
until the hundred thousands getting,
$$
\opsub[voperator=bottom,resultstyle.7=\hole,resultstyle.8=\hole,resultstyle.9=\hole,resultstyle.10=\hole,resultstyle.11=\hole,
resultstyle.12=\hole]{98723112389}{72430014467}\oplput(-3,3){{\tiny
13}}\oplput(-4,3){{\tiny 11}} \oplput(-5,3){{\tiny 10}}
\oplput(-6,3){{\tiny 0}}
$$
We now turn to the millions. Since $3<0$, subtraction goes without
any complications:
$$
\opsub[voperator=bottom,resultstyle.8=\hole,resultstyle.9=\hole,resultstyle.10=\hole,resultstyle.11=\hole,
resultstyle.12=\hole]{98723112389}{72430014467}\oplput(-3,3){{\tiny
13}}\oplput(-4,3){{\tiny 11}} \oplput(-5,3){{\tiny 10}}
\oplput(-6,3){{\tiny 0}}
$$
In the ten millions we find that $2<3$, so we must regroup from the
digits to the left. In the hundred millions we see that $7>4$ so we
stop there. We take a $1$ from the hundred millions, and so we have
a $6$ in the hundred millions. In the ten millions we now have
$10+2=12$. We perform the subtractions:
$$
\opsub[voperator=bottom,resultstyle.10=\hole,resultstyle.11=\hole,
resultstyle.12=\hole]{98723112389}{72430014467}\oplput(-3,3){{\tiny
13}}\oplput(-4,3){{\tiny 11}} \oplput(-5,3){{\tiny 10}}
\oplput(-6,3){{\tiny 0}} \oplput(-8,3){{\tiny 12}}
\oplput(-9,3){{\tiny 6}}
$$
Finally, the subtraction in the billions and the ten billions occur
without major complications. We display the result:
$$
\opsub[voperator=bottom]{98723112389}{72430014467}\oplput(-3,3){{\tiny
13}}\oplput(-4,3){{\tiny 11}} \oplput(-5,3){{\tiny 10}}
\oplput(-6,3){{\tiny 0}} \oplput(-8,3){{\tiny 12}}
\oplput(-9,3){{\tiny 6}}
$$
\end{solu}
\begin{exa}
A box of raisins was bought for $a$ dollars, and a firkin of
butter for $b$ dollars. If both were sold for $c$ dollars, how much
was gained?
\end{exa}
\begin{solu} The amount of money spent on raisins and butter gives
an original cost of $a+b$ dollars. Since we sold it for $c$, we must
take away our original cost, and hence we made a profit of $c-(a+b)
$ dollars.
\end{solu}
\section*{Homework}
\begin{multicols}{2}\columnseprule 1pt \columnsep 25pt\multicoltolerance=900\small
\begin{pro}
{\em Without availing from a calculator}, perform the addition
$123456789+987654321$.
\begin{answer}
$$\opadd[voperator=bottom]{123456789}{987654321}$$
\end{answer}
\end{pro}
\begin{pro}
{\em Without availing from a calculator}, perform the subtraction
$987654321-123456789$.
\begin{answer}
$$\opsub[voperator=bottom]{987654321}{123456789}$$
\end{answer}
\end{pro}
\begin{pro}\label{pro:number-cross}
Suppose we arrange the five numbers $1, 2, 3, 4, 5$ in each of the
five squares so that the horizontal and vertical lines both add to
$8$. Which number must go in the middle square?
\end{pro}
\vspace{1cm}
\begin{figurehere}
\centering \psset{unit=1pc}
\pspolygon[linewidth=2pt](-1,3)(-1,-3)(1,-3)(1,3)
\pspolygon[linewidth=2pt](3,-1)(-3,-1)(-3,1)(3,1)
\meinecaption{1}{Problem
\ref{pro:number-cross}.}\label{fig:number-cross}
\end{figurehere}
\begin{answer} $2$ \end{answer}
\begin{pro}
An oak tree is $42$ feet high. The oak tree is $18$ feet taller
than the fir tree. How tall is the fir tree?
\begin{answer}
$24 $ ft.
\end{answer}
\end{pro}
\begin{pro} Fill in the numbers from $1$ up to
$5$ in the five little circles, in such a way that in each triangle
the given number is the sum of the numbers on the vertices. Which
number appears in the shaded circle? \vspace*{2cm}
\begin{figurehere}
\rput(-8,0){
\begin{graph}(4,2)
\graphnodesize{.5} \roundnode{n1}(-2,0)[\graphnodecolour{1}]
\roundnode{n2}(0,0)[\graphnodecolour{1}]
\roundnode{n3}(2,0)[\graphnodecolour(1,.3,0)]
\roundnode{n4}(-1,-1)[\graphnodecolour{1}]\roundnode{n5}(1,-1)[\graphnodecolour{1}]
\edge{n1}{n2} \edge{n2}{n3} \edge{n1}{n4} \edge{n2}{n4}\edge{n2}{n5}
\edge{n3}{n5} \edge{n4}{n5} \rput(-1,-.5){6} \rput(0,-.6){10}
\rput(1,-.5){12}
\end{graph}}
\end{figurehere}
\begin{answer}
The figure below shows the correct arrangement. \vspace*{2cm}
\begin{figure}[h]
\begin{graph}(4,2)
\graphnodesize{.5} \roundnode{n1}(-2,0)[\graphnodecolour{1}]
\roundnode{n2}(0,0)[\graphnodecolour{1}]
\roundnode{n3}(2,0)[\graphnodecolour(1,.3,0)]
\roundnode{n4}(-1,-1)[\graphnodecolour{1}]\roundnode{n5}(1,-1)[\graphnodecolour{1}]
\edge{n1}{n2} \edge{n2}{n3} \edge{n1}{n4} \edge{n2}{n4}\edge{n2}{n5}
\edge{n3}{n5} \edge{n4}{n5} \rput(-1,-.5){6} \rput(0,-.6){10}
\rput(1,-.5){12} \nodetext{n1}{$1$} \nodetext{n2}{$3$}
\nodetext{n3}{$4$} \nodetext{n4}{$2$} \nodetext{n5}{$5$}
\end{graph}
\end{figure}
\end{answer}
\end{pro}
\begin{pro}
Can we find five even integers whose sum is $25$? \begin{answer} No.
When we add an even integer to another even integer the result is an
even integer. Thus the sum of five even integers is even, but $25$
is odd. \end{answer}
\end{pro}
\begin{pro}A bottle of wine and its cork cost $\$1$.
The bottle of wine costs $80$\textcent\ more than than the cork.
What is the price of the cork, in cents?
\begin{answer}
$10$\textcent. The cork costs $10$\textcent\ and the bottle
$90$\textcent.
\end{answer}
\end{pro}
\begin{pro}
Ingrid made $686$ biscuits. She sold some of them. If
$298$ were left over, how many biscuits did she sell?
\begin{answer}
$686-298=388$
\end{answer}
\end{pro}
\begin{pro}
What is the value of $$1 + 2 + 3 + \cdots + 7 + 8 + 9 + 8 + 7 + \cdots + 3 + 2 + 1,$$ where all of the
integers from $1$ through $9$ and then back down to $1$ are added together?
\begin{answer}
$81$
\end{answer}
\end{pro}
\begin{pro}
How many numbers are there in the set
$$ \{2,4,6,\ldots , 2006\}, $$
that is, in the set of all strictly positive natural numbers between
$2$ and $2006$?
What is
$$(2+4+\cdots + 2006) -(1+3+\cdots + 2005), $$
that is, what is the difference between the sum of all the odd
positive integers up to $2005$ and the sum of all the even positive
integers up to (and including) $2006$?
\begin{answer}
Observe that
$$ 2=2\cdot 1, \quad 4=2\cdot 2, \quad 6 = 2\cdot 3, \ldots 2006 = 2\cdot 1003, $$
and so there are $1003$ numbers.
\bigskip
The desired quantity is
$$(2-1) + (4-3) + \cdots + (2006-2005) = 1003. $$
\end{answer}
\end{pro}
\begin{pro}
Luigi saved $\$184$. He saved $\$63$ more than Brian.
How much did Brian save?
\begin{answer}
$184-63=121$
\end{answer}
\end{pro}
\begin{pro}
Alice is $4$ inches taller than Betty. Caroline is $8$ inches
shorter than Alice. Betty is $69$ inches tall. How tall is Caroline,
in inches?
\begin{answer}
Alice is $69+4=73$ inches tall and Caroline is $73-8=65$ inches
tall.
\end{answer}
\end{pro}
\begin{pro}
A tree was planted $54$ years before $1961$. How old was that
tree in $2008$?
\begin{answer}
$ 101$ years.
\end{answer}
\end{pro}
\begin{pro}
Lilliam had $20$ pieces of candy. She gave two pieces to her
sister.
\begin{enumerate}\item How many did she have left?
\item If she gave away 2 pieces each to 4 more people, how many
pieces would she have left?
\end{enumerate}
\begin{answer}\noindent
\begin{enumerate}\item $18$
\item $10$
\end{enumerate}
\end{answer}
\end{pro}
\begin{pro}
Buses need to be rented for $27$ children going on a field trip. Each bus
can take $12$ children in addition to the driver. How many buses must be
rented?
\begin{answer}
$3$
\end{answer}
\end{pro}
\begin{pro}
Collect like terms:
\begin{enumerate}
\item $(2a+8b)+(a+3b)$
\item $2a+8b-a-3b$
\item $(x^2+x+1)+(3x^2+2x+1)$
\end{enumerate}
\begin{answer}
\noindent
\begin{enumerate}
\item $3a+11b$
\item $a+5b$
\item $4x^2+3x+2$
\end{enumerate}
\end{answer}
\end{pro}
\begin{pro}
Bobby was playing on the elevator of a very tall building. Starting
from the floor where he was, he went five floors up, four down,
three up, four up, and two down. If he is now in the $30$th floor,
what was the original floor from where he started?
\begin{answer}
Start from $30$ and go back:
$$ 30+2-4-3+4-5=24, $$so he started on the $24$th floor.
\end{answer}
\end{pro}
\begin{pro}
How many addition signs should be put between digits of the number $987654321$ and where should we put them to get a total of $99$?
\begin{answer}
Either $9+8+7+65+4+3+2+1 = 99 $ or $9+8+7+6+5+43+21 = 99 $.
\end{answer}
\end{pro}
\begin{pro}
You have a seven inch gold bar, that is already segmented into seven
equal pieces. You are allowed to make two cuts to it. How can you
pay an employee that demands to be paid one gold piece daily for
the seven days that he works for you?
\begin{answer}
Cut the bar making pieces of $1$, $2$, and $4$ segments.
\end{answer}
\end{pro}
\begin{pro} Fill in the blank using either of the symbols '$=$' ,
'$\neq$', '$\in$', '$\notin$' as appropriate:
\begin{enumerate}
\item $3 \rule{1cm}{2pt} 3+0$ \item $5+1 \rule{1cm}{2pt} \naturals $ \item
$1+3 \rule{1cm}{2pt} 2+2$\item $64 \rule{1cm}{2pt} 604$
\item $\maltese \rule{1cm}{2pt} \naturals $
\item $ 5+6 \rule{1cm}{2pt} 7+ 8$
\end{enumerate}
\begin{answer}\noindent\begin{enumerate}
\item $3= 3+0$ \item $5+1\in \naturals $ \item
$1+3= 2+2$\item $64\neq 604$ \item $\maltese \notin \naturals $
\item $ 5+6 \neq 7+8$
\end{enumerate}
\end{answer}
\end{pro}
\begin{pro}
How many different sums can be made when two non-necessarily
distinct numbers from the set $\{1,3,4,5,7\}$ are taken?
\begin{answer}
From the addition table
$$ \begin{array}{|l||l|l|l|l|l|} \hline + & 1 & 3 & 4 & 5 & 7 \\
\hline\hline 1 & 2 & 4 & 5 & 6 & 8 \\
\hline 3 & 4 & 6 & 7 & 8 & 10 \\
\hline 4 & 5 & 7 & 8 & 9 & 11 \\
\hline 5 & 6 & 8 & 9 & 10 & 12 \\
\hline 7 & 8 & 10 & 11 & 12 & 14 \\
\hline
\end{array} $$the different sums belong to the set
$\{2, 4, 5, 6, 7, 8,9,10,11,12,14\}$, and so there are eleven different sums.
\end{answer}
\end{pro}
\begin{pro}
A frog is in a $10$ ft well. At the beginning of each day, it leaps
$5$ ft up, but at the end of the day it slides $4$ ft down. After how many days, if at all, will the
frog escape the well? \begin{answer} The frog will escape after
seven days. At the end of the sixth day, the frog has leaped $6$
feet. Then at the beginning of the seventh day, the frog leaps $5$
more feet and is out of the well.
\end{answer}
\end{pro}
\begin{pro}
Judith was imprisoned by a band of mathematicians and sent to
Guant\'{a}namo for crimes against Mathematics. Through the mercy of
the Brahmin mathematician, she was given the choice of being
released after $10$ years or be given freedom if she climbed the
$100$ steps of a $100$-step staircase subject to the following
rules:
\begin{enumerate}
\item She climbs up or down only one step per day.
\item She climbs up on every day of January, March, May, July,
September, and November.
\item She goes down on every day of February, April, June, August,
October, and December.
\end{enumerate}
Being adept at climbing, she chose this later option. If Judith
started on January 1 2001, when will she gain her freedom?
\begin{answer}
Let us see what happens in a typical non-leap year, and in a typical
leap year.
\bigskip
In a non-leap year\\
\begin{tabular}{|l|l|}\hline
by the end of & she has climbed \\
\hline 31 January & $31$ steps \\
\hline 28 February & $31-28=3$ steps \\
\hline 31 March & $3+31=34$ steps \\
\hline 30 April & $34-30=4$ steps \\
\hline 31 May & $31+4=35$ steps \\
\hline 30 June & $35-30=5$ steps \\
\hline 31 July & $31+5=36$ steps \\
\hline 31 August & $36-31=5$ steps \\
\hline 30 September & $30+5=35$ steps \\
\hline 31 October & $35-31=4$ steps \\
\hline 30 November & $30+4=34$ steps \\
\hline 31 December & $34-31=3$ steps \\
\hline
\end{tabular}
\bigskip
In a leap year\\
\begin{tabular}{|l|l|}\hline
by the end of & she has climbed \\
\hline 31 January & $31$ steps \\
\hline 29 February & $31-29=2$ steps \\
\hline 31 March & $2+31=33$ steps \\
\hline 30 April & $33-30=3$ steps \\
\hline 31 May & $31+3=34$ steps \\
\hline 30 June & $34-30=4$ steps \\
\hline 31 July & $31+4=35$ steps \\
\hline 31 August & $35-31=4$ steps \\
\hline 30 September & $30+4=34$ steps \\
\hline 31 October & $34-31=3$ steps \\
\hline 30 November & $30+3=33$ steps \\
\hline 31 December & $33-31=2$ steps \\
\hline
\end{tabular}
\bigskip
Now, $100-36=64$. Let us see how many years it takes her to climb
$64$ steps. By the end of the four-year range $2001-2004$, she
climbs $3+3+3+2=11$ steps. By the end of the four-year range
$2005-2008$, she has climbed $22$ steps. By the end of the four-year
range $2009-20012$, she has climbed $33$ steps. By the end of the
four-year range $2013-2016$, she has climbed $44$ steps. By the end
of the four-year range $2017-2020$, she has climbed $55$ steps. By
the end of the four-year range $2021-2024$, she has climbed $66$
steps. In fact, by 31 December 2023 she has climbed $64$ steps, and
by 31 July 2024 she has climbed $64+35=99$ steps. This means that
she needs to go into 2025. By 31 March 2025 she has climbed
$66+34=100$ steps and she is now free.
\end{answer}
\end{pro}
\begin{pro}
Using all the digits $\{0,1,2,3,4,5,6,7,8,9\}$, form two $5$-digit
numbers so that their difference is as large as possible.
\begin{answer}
$98765-10234=88531$. Why didn't we take $01234$ instead of $10234$?
\end{answer}
\end{pro}
\begin{pro}
Using all the digits $\{0,1,2,3,4,5,6,7,8,9\}$, form two $5$-digit
numbers so that their difference is as small as possible.
\begin{answer}
$50123-49876=247$.
\end{answer}
\end{pro}
\begin{pro}
You and I play the following game. I tell you to write down three
$2$-digit integers between $10$ and $89$. Then I write down three
$2$-digit integers of my choice. The answer comes to $297$, no
matter which three integers you choose (my choice always depends on
yours). For example, suppose you choose $12, 23, 48$. Then I choose
$87, 76, 51$. You add
$$12 + 23 + 48 + 87 + 76 + 51 = 297.$$ Again, suppose you chose
$33, 56, 89$. I then choose 66, 43, 10. Observe that
$$33 + 56 + 89 + 66 + 43 + 10 = 297.$$
Explain how I choose my numbers so that the answer always comes up
to be $297$ (!!!). \begin{answer} Notice that I always choose my
number so that when I add it to your number I get $99$, therefore, I
end up adding $99$ three times and $3 \cdot 99 =
297$.\footnote{Using algebraic language, observe that if you choose
$x, y, z$, then I choose $(99 - x), (99 - y), (99 - z)$. This works
because
$$x + (99 - x) + y + (99 - y) + z + (99 - z) = 3(99) = 297.$$}
\end{answer}
\end{pro}
\begin{pro}
A merchant bought $a$ barrels of sugar and $p$ barrels of molasses.
How many barrels in all did he buy?
\begin{answer}
$ a+p $
\end{answer}
\end{pro}
\begin{pro}
Jill bought a silk dress for $m$ dollars, a muff for $l$ dollars, a
shawl for $v$ dollars, and a pair of gloves for $c$ dollars. What
was the entire cost?
\begin{answer}
$m + l +v +c $
\end{answer}
\end{pro}
\begin{pro}
You start the day with $q$ quarters and $d$ dimes. How much money do
you have? Answer in cents. If by the end of the day you have lost
$a$ quarters and $b$ dimes, how much money do you now have? Answer
in cents. Write a one or two line explanation for your
answers.\begin{answer} You have $25q+10d$ cents at the beginning of
the day, and then you lose $25a+10b$, which leaves you with
$$(25q+10d)-(25a+10b)$$ cents.
\end{answer}
\end{pro}
\begin{pro}
Brian tore out several successive pages from a book. The first page that he tore up was page number $143$. The last page that he tore up
is also a three-digit number written with the same digits $\{1, 4, 3\}$ but in a different order. How many pages did he tear up?
\begin{answer}
The number of the last page torn must have opposite parity to the first one torn, thus it must be even, hence it must be $314$ and hence
$314-143=171$ pages.
\end{answer}
\end{pro}
\begin{pro}
A man bought a hat for $h$ dollars. He then bought a jacket and a
pair of trousers. If the jacket is thrice as expensive as the hat
and the trousers are $8$ dollars cheaper than jacket, how much money
did he spend in total for the three items?
\begin{answer}
The jacket costs $3h$ and the trousers cost $3h-8$. Hence he paid in
total
$$ h + 3h + 3h-8 = 7h-8. $$
\end{answer}
\end{pro}
\begin{pro}
Peter is $x$ years old, Paul is $y$, and Mary is $z$ years. What is
the sum of their ages?
\begin{answer}
$ x+y+z $
\end{answer}
\end{pro}
\begin{pro}
A boy bought a pound of butter for $y$ cents, a pound of meat for
$z$ cents, and a bunch of lettuce for $s$ cents. How much did they
all cost?
\begin{answer}
$y+z+s $\textcent
\end{answer}
\end{pro}
\begin{pro}
A man sells a carriage for $m$ dollars and loses $x$ dollars. What
was the cost of the carriage?
\begin{answer}
$ m+x $
\end{answer}
\end{pro}
\begin{pro}
I paid $c$\textcent\ for a pound of butter, and $f$\textcent\ for a
lemon. How much more did the butter cost than the lemon?
\begin{answer}
$c-f$ \textcent
\end{answer}
\end{pro}
\begin{pro}
Sold a lot of wood for $b$ dollars, and received in payment a barrel
of flour worth $e$ dollars. How many dollars remain due?
\begin{answer}
$b-e$
\end{answer}
\end{pro}
\begin{pro}
Complete the following addition:
$$\opadd[voperator=bottom,operandstyle.1.1=\hole,
operandstyle.1.3=\hole, operandstyle.2.4=\hole, resultstyle.2=\hole,
carryadd=false] {3679}{5283}$$
\begin{answer}
$3679+5283=8962$
\end{answer}
\end{pro}
\begin{pro}
Find the missing digits:
$$\opadd[voperator=bottom,operandstyle.1.2=\hole,
operandstyle.2.3=\hole, operandstyle.2.1=\hole, resultstyle.3=\hole,
carryadd=false] {714}{189}$$
\begin{answer}
$714+189=903$.
\end{answer}
\end{pro}
\begin{pro}
Complete the following subtraction:
$$\opsub[voperator=bottom,operandstyle.1.3=\hole, operandstyle.2.1=\hole, resultstyle.2=\hole,
resultstyle.4=\hole,carrysub=false] {5671}{1920}$$
\begin{answer}
$5671-1920=3751$
\end{answer}
\end{pro}
\begin{pro} By inspection (without any calculation) determine whether the following are true or false. Mention which property makes
the true statements true.
\begin{enumerate}
\item $1298+7459=7459+1298$ \item $1298+7459=7460+1298$
\item $(547+1250)+3=547+(1250+3)$ \item $(547+1250)+3=
547+(3+1250)$
\end{enumerate}
\begin{answer}\noindent
\begin{enumerate}
\item True by commutativity of addition. \item False.
\item True by associativity of addition. \item True. First observe that $(547+1250)+3= 547+(1250+3)$ by associativity and that
$(1250+3)=(3+1250)$ by commutativity. Then it follows that
$(547+1250)+3= 547+(3+1250)$ because equals replaced by equals are
equal.
\end{enumerate}
\end{answer}
\end{pro}
\begin{pro} Each square represents a digit. Find the value of
each
missing digit. $$\begin{array}{lllllll} & \blacksquare & 7 & 5 & \blacksquare & 6 \\
- & & \blacksquare & 5 & 6 & \blacksquare \\
\hline & 2 & 4 & \blacksquare & 7 & 5 \\
\end{array}
$$
\begin{answer}
$\psframebox{2}75\psframebox{3}6-\psframebox{2}56\psframebox{1} =
24\psframebox{9}75$.
\end{answer}
\end{pro}
\begin{pro}
Is it possible to replace the letter $A$ in the square below so
that every row has the same sum of every column?
\vspace*{1cm}
\PuzzleSolution
\begin{Puzzle}{3}{3}%
|{1}|{2}|{5}|.
|{3}|{3}|{2}|.
|{a}|{3} |{1}|.
\end{Puzzle}
\begin{answer}
Observe that the first and second rows, and the second and third
columns add up to $8$. Thus $A=4$ works.
\end{answer}
\end{pro}
\begin{pro}
Fill each square with exactly one number from
$$\{1,2,3,4,5,6,7,8,9\}$$ so that the square becomes a magic
square, that is, a square where every row has the same sum as every
column, and as every diagonal.
\vspace*{1cm}
\begin{Puzzle}{3}{3}%
|{\empty }|{\empty }|{\empty }|.
|{\empty }|{\empty }|{\empty }|.
|{\empty }|{\empty } |{\empty }|.
\end{Puzzle}
Is there more than one solution?
\begin{answer}
There are multiple solutions. They can be obtained by permuting the
entries of one another. Here are two:
\vspace*{1cm}
\begin{minipage}{5cm}
\PuzzleSolution
\begin{Puzzle}{3}{3}%
|{8}|{1}|{6}|.
|{3}|{5}|{7}|.
|{4}|{9} |{2}|.
\end{Puzzle}
\end{minipage}
\begin{minipage}{5cm}
\PuzzleSolution
\begin{Puzzle}{3}{3}%
|{4}|{9} |{2}|.
|{3}|{5}|{7}|.
|{8}|{1}|{6}|.
\end{Puzzle}
\end{minipage}
\end{answer}
\end{pro}
\begin{pro}
Vintik and Shpuntik agreed to go to the fifth car of a train.
However, Vintik went to the fifth car from the beginning, but
Shpuntik went to the fifth car from the end. How many cars has the
train if the two friends got to one and the same car?
\begin{answer}
Nine.
\end{answer}
\end{pro}
\begin{pro}
The desks in a classroom are arranged in straight rows. Jos\'{e} is in the third
row from the front and the fourth row from the back. He is also third from
the left end of a row and fifth from the right.
How many desks are in the classroom?
\begin{answer}
$42$.
\end{answer}
\end{pro}
\begin{pro}
In a contest to guess the number of balloons in a bunch, Adam guessed $25$,
Bob guessed $31$, Carlos guessed $29$, Daniel guessed $23$ and Edward
guessed $27$. Two guesses were wrong by $2$, and two guesses were wrong
by $4$. The other guess was correct. How many balloons are there?
\begin{answer}
$27$.
\end{answer}
\end{pro}
\begin{pro}
In the following array, some entries are missing. Find the value of
$x$. $$\begin{array}{|l|l|l|l|l||l|}
\hline & & & & & {\rm Row\ Totals} \\
\hline
& ? & ? & 5 & 7 & 24 \\
\hline & 5 & ? & ? & 1 & 21 \\
\hline & 7 & 8 & ? & ? & 26 \\
\hline & 9 & 3 & 2 & x & ? \\
\hline\hline {\rm Column\ Totals} & 29 & 21 & 22 & 19 & \\
\hline
\end{array}$$
\begin{answer}$6$\end{answer}
\end{pro}
\begin{pro}
An urn contains $28$ blue marbles, $20$ red
marbles, $12$ white marbles, $10$ yellow marbles, and $8$ magenta
marbles. How many marbles must be drawn from the urn in order to
assure that there will be $15$ marbles of the same color?
\begin{answer} If all the magenta, all the
yellow, all the white, $14$ of the red and $14$ of the blue marbles
are drawn, then in among these $8 + 10 + 12 + 14 + 14 = 58$ there
are no $15$ marbles of the same color. Thus we need $59$ marbles in
order to insure that there will be $15$ marbles of the same color.
\end{answer}
\end{pro}
\begin{pro}
In a $4$ by $4$ magic square the sum of the four entries in each
row, in each column, and in each of the two main diagonals are all
the same. If the magic square shewn below is completed, find $A+B$.
\vspace*{1cm}
\PuzzleSolution
\begin{Puzzle}{4}{4}%
|{\empty }|{\empty }|{7}|{12}|.
|{A}|{4}|{9}|{\empty}|.
|{10}|{5}|{\empty}|{3}|.
|{8}|{ 11 }|{\empty}|{B}|.\end{Puzzle}
\begin{answer}
$28$
\end{answer}
\end{pro}
\begin{pro}
In each of the following overlapping rings the sum of the entries is
$55$. What is $A-B$? \vspace*{2cm}
$$\psset{unit=3pc} \cnodeput(1;30){A}{11} \cnodeput(1;90){B}{8}
\cnodeput(1;150){C}{A}
\cnodeput(1;210){D}{9} \cnodeput(1;270){E}{9} \cnodeput(1;330){F}{B}
\ncline{A}{B}\ncline{B}{C}\ncline{C}{D}\ncline{D}{E}\ncline{E}{F}
\ncline{F}{A} \rput(1.75,0){\cnodeput(1;30){A}{2}
\cnodeput(1;90){B}{14}
\cnodeput(1;150){C}{11}
\cnodeput(1;210){D}{B} \cnodeput(1;270){E}{7}
\cnodeput(1;330){F}{13}
\ncline{A}{B}\ncline{B}{C}\ncline{C}{D}\ncline{D}{E}\ncline{E}{F}
\ncline{F}{A}} $$ \vspace*{1cm}
\begin{answer}
$2$
\end{answer}
\end{pro}
\begin{pro}
Let $\mathbb{E}=\{0,2,4,6, \ldots \}$ be the set of even natural
numbers. Is this set closed under addition? Is this set closed under
multiplication.
\begin{answer}
Yes. Yes.
\end{answer}
\end{pro}
\begin{pro}
Let $\mathbb{O}=\{1,3,5,7, \ldots \}$ be the set of odd natural
numbers. Is this set closed under addition? Is this set closed under
multiplication.
\begin{answer}
No. Yes.
\end{answer}
\end{pro}
\begin{pro}
Define the operation $\oplus$ of ``sooper-dooper addition'' by
$a\oplus b = a+b+10$, where $a$ and $b$ are natural numbers. For
example $2\oplus 3 = 2+3+10 = 15$ and $4\oplus 10 = 4+10+10 = 24$.
Is sooper-dooper addition a commutative operation? Is sooper-dooper
addition an associative operation?
\begin{answer}
It is commutative, because for arbitrary $a$ and $b$ we have
$a\oplus b = a+b+10$ and $b\oplus a = b+a+10$. Since $a+ b+10 =
b+a+10$ by the commutativity of the natural numbers, $\oplus$ is
commutative. It is also associative. For
$$ a\oplus (b\oplus c) = a\oplus (b+c+10) = a+(b+c+10)+10=a+b+c+20. $$Also
$$ (a\oplus b)\oplus c = (a+ b+10)\oplus c+ = (a+ b+10)+c+10=a+b+c+20. $$Since $ a\oplus (b\oplus c) = (a\oplus b)\oplus c$, the operation is associative.
\end{answer}
\end{pro}
\begin{pro}\label{pro:domino}
Consider $2005$ hexagonal ``domino'' pieces with the numbers
$1,2,3,4,5,6$ written on the edges in clockwise fashion, as in
figure \ref{fig:domino}. A ``chain'' is formed so that domino rules
are observed. This means that two edges from different pieces are
joined only when the numbers on the edges agree. The numbers on the
edges joining two different domino pieces are now deleted and the
remaining numbers are now added. What is the maximum value of this
sum? \vspace*{1cm}
\begin{figurehere}
\rput(2,-.5){\begin{graph}(1,1) \roundnode{A}(1,0)
\roundnode{B}(.5,.866) \roundnode{C}(-.5,.866) \roundnode{D}(-1,0)
\roundnode{E}(-.5,-.866) \roundnode{F}(.5,-.866) \edge{A}{B}
\edge{B}{C} \edge{C}{D} \edge{D}{E} \edge{E}{F} \edge{F}{A}
\edgetext{A}{B}{1} \edgetext{B}{C}{6}
\edgetext{C}{D}{5} \edgetext{D}{E}{4} \edgetext{E}{F}{3} \edgetext{F}{A}{2}
\end{graph}}
\meinecaption{1}{Problem \ref{pro:domino}.}\label{fig:domino}
\end{figurehere}
\begin{answer}In order to obtain the maximal sum, we must have a chain
where every piece is glued to the smallest possible of sides using
the smallest numbers. The chain then must have $2003$ pieces which
are joined on two sides and the two end pieces are joined at only
one side. If the $1$ and the $2$ were shared, the first and second
piece must be joined at $1$ and the second and third at $2$. But
this is impossible because then the first and the third piece would
be joined at $3$ and $6$, violating domino rules. Hence only $3$'s
and $1$'s are shared. The middle $2003$ contribute $2+4+5+6=17$
each, one of the end pieces contributes $2+3+4+5+6=20$, and the
other end piece contributes $1+2+4+5+6=18$ hence the maximum sum is
$(2003)(17)+1(20) + 1(18)=34089.$
\end{answer}
\end{pro}
\end{multicols}
\chapter{Arithmetic Progressions}
\begin{df}
An arithmetic progression is a succession of numbers where you start
with a given number and form the next element by always adding the
same fix quantity, that is, there are numbers $a$ and $d$ such that
the succession of numbers takes the form:
$$ a, \quad a+d,\quad a+d+d=a+2d, \quad \quad a+d+d+d=a+3d, \ldots ,
$$Here $a$ is called the {\em first term}, $a+d$ is called the {\em second term}, etc., and $d$ is the {\em
common} difference.
\end{df}
\begin{exa}\label{exa:arit-pro-1}
The arithmetic progression $$ 3, 10, 17, 24, \ldots , $$has first
term $3$ and common difference $7$.
\end{exa}
\bigskip
Observe that in example \ref{exa:arit-pro-1}, the pattern of
additions is
$$3=3+7\cdot 0, \quad 10 = 3+7\cdot 1, \quad 17 = 3+7\cdot 2, \quad 24=3+7\cdot 3, \ldots,$$
Thus if we wanted to find the fifth term we would compute $3+7\cdot
4 =3+28=31$. If we wanted to compute the hundredth term we would
find $3+7\cdot 99 = 696$.
\begin{exa}
Consider the arithmetic progression
$$5, 11, 17, 23, \ldots . $$
\begin{enumerate}
\item Find its next term.
\item Find its hundredth term.
\item Find a formula for the term in position $n$.
\item Is $100$ a term in this progression? If so, which term is it?
\item Is $101$ a term in this progression? If so, which term is it?
\end{enumerate}
\end{exa}
\begin{solu}
The first term is $5$ and the common difference is $6$. The term
that follows $23$ is $23+6=29$. Notice that
$$5=5+6\cdot 0, \quad 11 = 5+6\cdot 1, \quad 17 = 5+6\cdot 2, \quad 23=5+6\cdot 3, \ldots,$$
and so the hundredth term is $5+6\cdot 99 =599$. From this pattern
we infer that the term in position $n$ is $5+6(n-1)$.
Each time we are adding a multiple of $6$ to $5$. Hence the numbers
in this progression leave remainder $5$ upon division by $6$. Since
$100=4+16\cdot 6$ leaves remainder $4$ upon division by $6$, the
number $100$ is not in this progression. However, $101=5+16\cdot 6$,
which means that $101$ is the $17$th term of this progression. We
have in fact,
$$5, 11, 17, 23, 29, 35, 41, 47, 53, 59, 65, 71, 77, 83, 89, 95, 101,\ldots . $$
so we explicitly see how to get to $101$.
\end{solu}
\begin{exa}
How many terms are in the arithmetic progression
$$ 7,\quad 15,\quad 23, \quad \ldots ,\quad 1207? $$
\end{exa}
\begin{solu}
Observe that the first term is $7$ and that the common difference is
$8$. The terms have the law of formation
$$7= 7 +8\cdot 0, \quad 15= 7 +8\cdot 1, \quad 23= 7 +8\cdot 2, \quad \ldots, $$
so each time we are adding $7$ to a multiple of $8$. Now, simply
divide $1207$ by $8$ to find
$$1207 = 7 + 8\cdot 150. $$This means that $1207$ is the $151$st
position, that is, that there are $151$ terms in this progression.
\end{solu}
The terms of an arithmetic progression display a symmetry that makes
summing consecutive terms easy. The following classic example is
reputed to have been solved in seconds by the then seven-year-old K.
F. Gauss (a XIX century mathematician).
\begin{exa}
Find the sum of all the natural numbers from $1$ to $100$,
inclusive.
\end{exa}
\begin{solu}
The trick is to consider the fifty pairs
$$ 100+1, \quad 99 + 2, \quad 98+3, \quad \ldots , \quad 51+50. $$
Each pair adds up to $101$ and there are $50$ of them, hence
$$ 1+2+3+\cdots + 100 = 101\cdot 50=5050. $$
\end{solu}
The series of multiples of a natural number $n$ forms an arithmetic
progression with common difference $n$. For example, the multiples
of $3$ are
$$0,3,6,9,12,15,\ldots , $$
an arithmetic progression with common difference $3$, and the
multiples of $4$ are
$$ 0,4,8,12,16, \ldots , $$
an arithmetic progression with common difference $4$. The {\em least
common multiple} $\lcm (3,4)$ is the smallest strictly positive
number that is common in both lists, in this case $\lcm (3,4) =12$.
\begin{exa}
Find $\lcm (12,45)$. \label{exa:lcm-12-45}
\end{exa}
\begin{solu}
We form the progressions of the multiples of both numbers and circle
the first nonzero repeat:
$$ 0,12,24,36,48,60,72,84,96,108,120, 132, 144,156,168, \psovalbox{180}, 192, \ldots, $$
$$ 0, 45, 90, 135, \pscirclebox{180}, 205,\ldots $$ whence $\lcm
(12,45)=180$.
\end{solu}
The following sort of arithmetic progressions will be useful when we
study the division algorithm. Every natural number is either even or
odd, so we write
$$ \naturals = \{0,2,4,6,\ldots \}\cup \{1,3,5,7,\ldots\}, $$
where the symbol $\cup$ (read {\em union}) means that both sets are
joined. Notice that both sets are arithmetic progressions with
common difference $2$. Again, if the divisor is $3$, then the
possible remainders when we divide by $3$ are $0$, $1$, and $2$, and
we find
$$ \naturals = \{0,3,6,9,\ldots \}\cup \{1,4,7,10,\ldots\}\cup \{2,5,8,11,\ldots\}. $$
We say that these sets {\em partition} the natural numbers.
\begin{exa}
Partition the natural numbers into five infinite arithmetic
progressions.
\end{exa}
\begin{solu}Upon division by $5$, every natural number leaves
remainder $0$, $1$, $2$, $3$, or $4$. Hence we may take
$$ \naturals = \{0,5,10,15,\ldots \}\cup \{1,6,11,16,\ldots\}\cup \{2,7,12,17,\ldots\}\cup \{3,8,16,21,\ldots\}\cup \{4,9,13,18,\ldots\}. $$
\end{solu}
\section*{Homework}
\begin{multicols}{2}\columnseprule 1pt \columnsep 25pt\multicoltolerance=900\small
\begin{pro}
Find the $20$th number on the list
$$8, 11, 14, \ldots $$ assuming that the pattern is preserved.
\begin{answer}
$8 + 3\cdot 19=65$.
\end{answer}
\end{pro}
\begin{pro}
Find the $1000$th number on the list
$$1, 10, 19, \ldots $$ assuming that the pattern is preserved.
\begin{answer}
$1 + 9\cdot 999=8992$.
\end{answer}
\end{pro}
\begin{pro}
How many terms are there in the arithmetic progression
$$8, 11, 14, \ldots 3005? $$What is their sum?
\begin{answer}
Observe that to $8$ we must add multiples of $3$. Dividing $3005$ by
$3$ we obtain
$$ 3005=3\cdot 1001+2 = 3\cdot 999 + 3\cdot 2 +2 =3\cdot 999 +8, $$
and hence there are $999+1=1000$ terms.
Adding the $500$ pairs
$$ 8+3005=3013, \quad 11+3002=3013, \quad 14+2999=3013, \quad \ldots , \quad 1505+1508=3013, $$
we see that the sum is
$$8+ 11+14+ \cdots + 3005=500\cdot 3013 =1506500. $$
\end{answer}
\end{pro}
\begin{pro}Find a formula for the $n$-th term of the arithmetic progression
$$ 2, 7, 12, 17, \ldots . $$
\begin{answer}
We start with $2$, and then keep adding $5$, thus
$$ 2= 2+5\cdot 0, \qquad 7= 2+5\cdot 1, \qquad 12= 2+5\cdot 2, \qquad 17= 2+5\cdot 3, \ldots. $$
The general $n$th term is therefore of the form $2+5(n-1)$, where
$n=1,2,3,\ldots$ is a natural number.
\end{answer}
\end{pro}
\begin{pro}
Find a general formula for the $n$-th term of the arithmetic
progression $$1, 7, 13, 19, 25, \ldots . $$
\begin{answer} $1+6(n-1)$ where $n=1,2,3,\ldots$.\end{answer}
\end{pro}
\begin{pro}
You start counting by $11$s, starting with $3$:\qquad\qquad $$
3,14,25,36, \ldots . $$ Which number appears in the $100$th
position? Give a formula for the number appearing in the $N$th
position.
\begin{answer}
The pattern is as
follows,
$$ \begin{array}{|l|lll|}
\hline
\mathrm{Position} & \multicolumn{3}{|c|}{Number} \\
\hline
1 & 3 & = & 3 + 11\cdot 0 \\
2 & 14 & = & 3 + 11\cdot 1 \\
3 & 25 & = & 3 + 11\cdot 2 \\
4 & 36 & = & 3 + 11\cdot 3 \\
5 & 47 & = & 3 + 11\cdot 4 \\
6 & 59 & = & 3 + 11\cdot 5 \\
7 & 60 & = & 3 + 11\cdot 6 \\
\vdots & \vdots & \vdots & \vdots \\
100 & 1092 & = & 3 + 11\cdot 99 \\
\hline \end{array}$$
A general formula is $3 + 11(N-1)$ for $N =1,2,3, \ldots$.
\end{answer}
\end{pro}
\begin{pro}
Find $\lcm (30,36)$. \begin{answer} $180$
\end{answer}
\end{pro}
\begin{pro}
Find $\lcm (24,36)$. \begin{answer} $72$
\end{answer}
\end{pro}
\begin{pro}
Find $\lcm (30,24)$. \begin{answer} $120$
\end{answer}
\end{pro}
\begin{pro}
Partition the natural numbers into the union of four infinite
arithmetic progressions.
\begin{answer}Upon division by $4$, every natural number leaves
remainder $0$, $1$, $2$, or $3$. Hence we may take
$$ \naturals = \{0,4,8,12,\ldots \}\cup \{1,5,9,13,\ldots\}\cup \{2,6,10,14,\ldots\}\cup \{3,7,11,15,\ldots\}. $$
\end{answer}
\end{pro}
\end{multicols}
\chapter{Multiplication}
\begin{df}
Given natural numbers $m$ and $n$, we define their {\em product}
$mn$ as
$$mn = \underbrace{n + n + \cdots + n}_{m\ {\rm times}}. $$
The operation between $m$ and $n$ is called {\em multiplication}.
Here $m$ is called the {\em multiplier}, $n$ the {\em multiplicand}.
Both $m$ and $n$ are called {\em factors}.
\end{df}
\begin{rem} Ordinarily in Algebra, we use {\em juxtaposition}, that
is, putting one letter next to another, to indicate multiplication.
If we were multiplying two numbers, we would use Leibniz's raised
dot, as in $2\cdot 3$. We do not use the cross symbol $ \times $ so as
to avoid confusion with the letter $x$.
\end{rem}
A geometrical interpretation of multiplication is the following.
Consider an array with identical squares of area $1$ of width $m$
squares and height $n$, as in figure \ref{fig:mult-nat2}.
The product $mn$ is the number of squares, which is also the area
of the rectangle obtained once the squares are packed together and
all the white space is eliminated.
\vspace{2cm}
\begin{figure}[h]
\centering \psset{unit=2pc}
\def\cuadradito{\pstGeonode[PointName=none](0,0){O}(.75,0){A}(.75,.75){B}(0,.75){C}\pscustom[fillcolor=gray,fillstyle=solid]{\pstLineAB{O}{A}\pstLineAB{A}{B}\pstLineAB{B}{C}\pstLineAB{C}{O}}}
\rput(-2.5,-2.5){\cuadradito} \rput(-1.5,-2.5){\cuadradito}
\rput(0,-2){$\bullet\bullet\bullet$}\rput(1.5,-2.5){\cuadradito}\rput(2.5,-2.5){\cuadradito}
\pcline[offset=-12pt]{|-|}(-2.5,-2.5)(3.5,-2.5)\ncput*{$m$ squares}
\rput(-2.5,-1.5){\cuadradito}\rput(2.5,-1.5){\cuadradito}
\rput(-2.5,-.5){\cuadradito}\rput(2.5,-.5){\cuadradito}
\rput(-2.5,1.5){\cuadradito}\rput(2.5,1.5){\cuadradito}
\rput(0,-1){$\bullet\bullet\bullet$}
\rput(0,0){$\bullet\bullet\bullet$} \rput(-2,1){$\bullet$}
\rput(-2,.75){$\bullet$}
\rput(2.75,1){$\bullet$}\rput(2.75,.5){$\bullet$}\rput(2.75,.75){$\bullet$}
\rput(-2,.5){$\bullet$}
\pcline[offset=-12pt]{|-|}(-3.5,-2.5)(-3.5,2.25)\ncput*[nrot=:U]{$n$
squares} \meinecaption{3}{Multiplication in $\naturals$.}
\label{fig:mult-nat2}
\end{figure}
Just like addition of natural numbers, multiplication satisfies the
following axioms.
\begin{axi}[Closure] The product of any two natural numbers is a natural number, that is, $\naturals$ is {\em closed under multiplication}: if $a\in
\naturals$ and $b\in \naturals$ then also $ab\in \naturals$.
\end{axi}
\begin{axi}[Multiplicative Identity] If $1$ is multiplied to any natural number, the result is unchanged, that is $1\in\naturals$ is the
{\em multiplicative identity} of $\naturals$. It has the property
that for all $x\in\naturals$ it follows that $$x=1\cdot x=x\cdot 1.
$$
\end{axi}
\begin{axi}[Commutativity]Let $a\in\naturals$ and $b\in \naturals$ be arbitrary. Then
\mbox{$ab=ba$}.
\end{axi}
\begin{axi}[Associativity]
Let $a, b, c$ be arbitrary natural numbers. Then the order of
parentheses when performing multiplication is irrelevant, that is,
$$ a(bc)=(ab)c = abc. $$
\end{axi}
Since multiplication is stenography for addition, it is to be
understood that in a computation that involves sums and
multiplications, multiplications must be carried out first, unless
some other computation is coerced by parentheses. For example, in
$$ 2+3\cdot 4 =2+12=14, $$
we perform the multiplication first, but in
$$ (2+3)\cdot 4=5\cdot 4=20, $$
the addition is coerced by the parentheses, and hence, we perform it
first.
\bigskip
The next property links multiplication and addition.
\begin{axi}[Distributive Law]
Let $a, b, c$ be natural numbers. Then
$$a(b+c) = ab+ac, $$
and
$$(a+b)c=ac+bc. $$
\end{axi}
For example,
$$3(4+3)=3\cdot 7=21, \qquad 3\cdot 4 + 3\cdot 3 = 12 +9 =21. $$
In the case when the factors have each two summands, the distribute
law takes the form \begin{equation}\label{eq:mult-binom}
(a+b)(c+d)=a(c+d)+b(c+d)=ac+ad+bc+bd.
\end{equation}
For example,
$$ (3+4)(5+6)=3\cdot 5+ 3\cdot 6+ 4\cdot 5+4\cdot 6=15+18+20+24=77. $$
\begin{rem}
Needless to say, we could perform in a faster manner
$(3+4)(5+6)=7\cdot 11 =77,$ but our desire was to illustrate the
distributive law.
\end{rem}
\bigskip
Recall now the operation of exponentiation, which we have already
defined as, for $a\in \naturals$ and $n\in \naturals$, $n>0$,
$$a ^n = \underbrace{a\cdot a\cdots a\cdot a}_{n\ a\mathrm{'s}}. $$
We will have the convention that $a^0=1$ when $a\neq 0$, that is,
any non-zero number raised to the zeroth power is $1$ and that
$a^1=a$, that is, if the exponent is $1$ we do not write the
exponent.
\bigskip
We now analyse how to deal with products of exponential expressions
with the same base. For example, observe that
$$2^2=4, \quad 2^3=8, \quad 2^5=32, \implies 2^2\cdot 2^3 =4\cdot 8 = 32 =2^5, $$
and we deduce that $2^{2}\cdot 2^3=2^{2+3}=2^5$, that is, when
multiplying two exponential expressions with the same base, we added
the exponents. In general we have the following result.
\begin{thm}[First Law of Exponents]Let $a$ be a real number and $m, n$ natural numbers. Then
$$a^ma^n = a^{m+n}. $$
\end{thm}
\begin{pf}
We have $$\begin{array}{lll}a^ma^n & = & \underbrace{a\cdot a
\cdots a}_{m\ a\mathrm{'s}}\cdot \underbrace{a\cdot a \cdots a}_{n\
a\mathrm{'s}} \\
& = & \underbrace{a\cdot a \cdots a}_{m+n\ a\mathrm{'s}}\\
& = & a^{m+n}.
\end{array}$$
\end{pf}
\begin{exa}
We have: $(5x)(2x)=10x^{1+1}=10x^2$.
\end{exa}
\begin{exa}
We have: $(3x^2)(6x^4)=18x^{2+4}=18x^6$.
\end{exa}
Using the first law of exponents and the distributive law, we will
now give an algorithm for multiplication. Before explaining the
multiplication algorithm in general, let us consider an example.
Suppose we wanted to multiply $34$ to $56$.
We go through
the digits of $56$, starting left to right. We multiplied every
digit of $34$ by $6$, giving us
$$ 6\cdot 34=6(30+4)=6\cdot 30 + 6\cdot 4 = 180 + 24 =204. $$
Now we move to the digit $5$ of $56$. But the value of this digit is
$50$, and so we multiply
$$ 50\cdot 34 = 50 (30+4)=50 \cdot 30 + 50\cdot 4 = 1500 +200 =1700. $$
We conclude that
$$ 34\cdot 56 = 34(50+6)=34\cdot 50+34\cdot 6 = 1700+204=1904. $$
It is customary to display the result in the following manner:
$$ \opmul{34}{56} $$
Notice that here we shifted the multiplication by $5$ one unit left,
and pretended that we were in fact multiplying by $5$ and not $50$.
The two approaches have the same effect. If we wanted to see more
gory detail, and perhaps have a glimpse of how this algorithm
generalises to Algebra, we would probably proceed as follows. Since
$$ 34=3\cdot 10 +4, \quad 56=5\cdot 10+6, $$ using the distributive
law (\ref{eq:mult-binom}) we obtain,
$$ (3\cdot 10 +4)(5\cdot 10+6) = 3\cdot 10\cdot 5 \cdot 10 +3\cdot 10 \cdot 6 + 4\cdot 5 \cdot 10 + 4\cdot 6 =
1500+ 180+200+24= 1904. $$ The above ``vertical multiplication''
algorithm generalises to algebraic expressions. Suppose now that we
wanted to multiply $(3x+4)(5x+6)$. We first multiply every term of
$3x+4$ by $5x$:
$$\psmatrix & & 3x & + & 4 \\
\cdot & & 5x & + & 6 \\
\hline & \red{15x^2} & + & \red{20x} \\
\endpsmatrix \ncline[arrows=->,linecolor=red]{2,3}{1,3}\ncline[arrows=->,linecolor=red]{2,3}{1,5}$$
We now multiply every term of $3x+4$ by $6$, lining up the terms of
the same exponent of $x$ (in our case, exponent $1$ and exponent
$0$) with those of the first partial product:
$$\psmatrix & & 3x & + & 4 \\
\cdot & & 5x & + & 6 \\
\hline & 15x^2 & + & 20x \\
& & + & \red{18x} & + & \red{24} \\
\endpsmatrix \ncline[arrows=->,linecolor=red]{2,5}{1,3}\ncline[arrows=->,linecolor=red]{2,5}{1,5}$$
\bigskip
Finally, we add (collect) like terms:
$$\begin{array}{llllll} & & 3x & + & 4 \\
\cdot & & 5x & + & 6 \\
\hline & 15x^2 & + & 20x \\
& & + & 18x & + & 24 \\
\hline
& \red{15x^2} & + & \red{38x} & + & \red{24} \\
\end{array}$$
This multiplication can also be displayed horizontally, in a more
obvious display of (\ref{eq:mult-binom}):
$$ (3x+4)(5x+6)=(3x)(5x)+ (3x)(6)+4(5x)+4(6)=15x^2+18x+20x+24 =15x^2+38x+24. $$
Observe that if we substitute $x=10$ in the above we obtain,
$$15\cdot 10^2 +38\cdot 10 +24 =1500 +380 + 24=1904. $$
\bigskip
It is now easy to describe a multiplication algorithm akin to the
one used for addition. Let
$$ \underline{d_nd_{n-1}\ldots d_2d_1d_0},\qquad \underline{e_pe_{p-1}\ldots e_2e_1e_0}, $$
be two arbitrary decimal integers. To multiply them:
\begin{enumerate}
\item Multiply every digit of $\underline{d_nd_{n-1}\ldots
d_2d_1d_0}$ by $e_0$, taking into account all the possible carries.
\item Multiply every digit of $\underline{d_nd_{n-1}\ldots
d_2d_1d_0}$ by $e_1$, taking into account all the possible carries,
and write the result below the partial result for $e_0$, but shift
this partial result one unit to the left.
\item Keep multiplying by the $e_k$, taking care of all carries, and
each time shifting to the left each partial products.
\item Add all these partial products.
\end{enumerate}
\begin{exa}
Multiply $123$ by $456$, and display your result using the
``vertical'' multiplication algorithm. Then multiply $x^2+2x+3$ by
$4x^2+5x+6$ ``horizontally.''
\end{exa}
\begin{solu}
We have, $$ \opmul{123}{456} $$
and $$\begin{array}{lll} (x^2+2x+3)(4x^2+5x+6) & = &
x^2(4x^2+5x+6)+2x(4x^2+5x+6)+3(4x^2+5x+6)\\
& = & 4x^4+5x^3+6x^2+8x^3+10x^2+12x+12x^2+15x+18 \\
& = & 4x^4+13x^3+28x^2+27x+18.
\end{array}$$
Observe that if we substitute $x$ for $10$ we obtain
$$ 4\cdot 10^4 + 13\cdot 10^3 +28\cdot 10^2 + 27\cdot 10 + 18 =40000+13000+2800+270+18 = 56088.$$
\end{solu}
To conclude this section we will introduce the symbol $!$.
\begin{df}If $n\in \naturals$ we define $n!$ ($n$ {\em factorial}) as follows:
$$ 0! =1 , \quad 1! = 1, \quad 2! = 1\cdot 2=2, \quad 3! = 1\cdot 2 \cdot 3 = 6, \quad 4! = 1\cdot 2 \cdot 3\cdot 4 = 24, \quad n! = 1 \cdot 2 \cdots n. $$
\end{df}
\section*{Homework}
\begin{multicols}{2}\columnseprule 1pt \columnsep 25pt\multicoltolerance=900\small
\begin{pro}
If the figure shewn is folded to form a cube, then three faces meet
at every vertex. If for each vertex we take the product of the
numbers on the three faces that meet there, what is the largest
product we get?\vspace{1in}
$$
\psset{unit=2pc} \psline(-1, 0)(3,0)(3,1)(-1,1)(-1,0)
\psline(0,2)(0,-1)(1,-1)(1,2)(0,2) \psline(2,0)(2,1)
\rput(-.5,.5){4} \rput(.5, .5){2} \rput(1.5, .5){5} \rput(2.5,
.5){6} \rput(.5, 1.7){1} \rput(.5, -.7){3}
$$
\vspace{.25in}
\begin{answer}
Observe that $4$ and $5$ are on opposite sides, so they never meet.
The largest product is $3 \cdot 5 \cdot 6 = 90.$
\end{answer}
\end{pro}
\begin{pro}
Pencils are $8$ \textcent each. How much would $7$ pencils cost?
\begin{answer}
$56$ \textcent
\end{answer}
\end{pro}
\begin{pro}
Mr. Schremmer's class was doing a science experiment. There
were $7$ groups in the class. Each group got $4$ test tubes.
How many test tubes did the class use?
\begin{answer}
$28$
\end{answer}
\end{pro}
\begin{pro}
In some years an extra second is added on $31$ December to keep the
clocks in time with the rotation of the earth. How many seconds does
$31$ December have when this happens?
\begin{answer}
$24\cdot 60\cdot 60 +1=86401$
\end{answer}
\end{pro}
\begin{pro}
Sound travels at approximately $330$ meters per second. The sound of
an explosion took $28$ seconds to reach a person. Estimate how far
away was the person from the explosion.
\begin{answer}
$9240$ meters.
\end{answer}
\end{pro}
\begin{pro}
According to experts the first $4$ moves in a chess game can be played in $197299$
totally different ways. If it takes $30$ seconds to make one move,
what time, in seconds, would it take one player to try every
possible set of $4$ moves?
\begin{answer}
$30 \cdot 197299 = 5918970$ seconds.
\end{answer}
\end{pro}
\begin{pro}
When writing all the natural numbers from $1$ to $999$ in a row, as
follows,
$$123456789101112\ldots 998999 $$how many digits have I used?
\begin{answer}
I have written $9$ $1$-digit integers ($1$ through $9$), $90$
$2$-digit integers (from $10$ to $99$) and $900$ $3$-digit integers
(from $100$ to $999$). Altogether I have used
$$1\cdot 9 + 2\cdot 90 + 3 \cdot 900=2889 $$ digits.
\end{answer}
\end{pro}
\begin{pro}
The integers from $1$ to $1000$ are written in succession. Find the
sum of all the digits.
\begin{answer}
Pair up the integers from $0$ to $999$ as
$$ (0, 999), \ (1, 998), \ (2, 997), \ (3, 996), \ \ldots , (499, 500).
$$Each pair has sum of digits $27$ and there are $500$ such pairs.
Adding $1$ for the sum of digits of $1000$, the required total is
$$27\cdot 500 + 1 = 13501. $$
\end{answer}
\end{pro}
\begin{pro}
All the natural numbers---starting with $1$---are listed
consecutively
$$123456789101112131415161718192021\ldots$$Which digit occupies the $1002$nd place?
\begin{answer} When the number $99$ is written down, we have used
$$1\cdot 9 + 2\cdot 90 = 189$$ digits. If we were able to write
999, we would have used
$$1\cdot 9 + 2\cdot 90 + 3\cdot 900 = 2889$$digits, which is more than $1002$ digits. The $1002$nd
digit must be among the three-digit positive integers. We have $1002
- 189 = 813$ digits at our disposal, from which we can make
$\lfloor\dfrac{813}{3}\rfloor = 271$ three-digit integers, from 100
to $370$. When the $0$ in $370$ is written, we have used $189 +
3\cdot 271 = 1002$ digits. The $1002$nd digit is the $0$ in $370$.
\end{answer}
\end{pro}
\begin{pro}
Compute the following:
$$1^2, \quad 11^2, \quad 111^2, \quad 1111^2, \quad 11111^2, \quad 111111^2. $$
Do you notice a pattern?
\begin{answer}We have,
$$1^2=1, \quad 11^2=121, \quad 111^2=12321, \quad 1111^2=1234321, \quad 11111^2=123454321, \quad 111111^2=12345654321. $$
\end{answer}
\end{pro}
\begin{pro}
Multiply the numbers $12$ and $45$. Then multiply $x+2$ and $4x+5$.
\begin{answer}
We have,
$$ \opmul{12}{45} $$
Multiplying,
$$ \begin{array}{ccccc} & & x & + & 2 \\
\cdot & & 4x & + & 5 \\
\hline 4x^2 & + & 8x & & \\
& & 5x & + & 10 \\
\hline
4x^2 & + & 13x & + & 10.
\end{array}$$
\end{answer}
\end{pro}
\begin{pro}
Compute $5! - 4!$.
\begin{answer}
$120-24 = 96$
\end{answer}
\end{pro}
\begin{pro}
Peter is $x$ years old, Paul is $y$, and Mary is $z$ years. What is
the product of their ages?
\begin{answer}
$ xyz $
\end{answer}
\end{pro}
\begin{pro}
A farmer has $7$ ducks. He has $5$ times as many
chickens as ducks. How many more chickens
than ducks does he have?
\begin{answer}
$7\cdot 5 = 35 $
\end{answer}
\end{pro}
\begin{pro}
The sum of three consecutive integers is $2007$. Find their product.
\begin{answer}
$668\cdot 669\cdot 670=299417640$.
\end{answer}
\end{pro}
\begin{pro}
Multiply: $(2x^2)(4x^3)$.
\begin{answer}
$8x^5$.
\end{answer}
\end{pro}
\begin{pro}
Find the missing digits in the following product.
$$ \begin{array}{llllll} & & & & 2 & \blacksquare\\
\times & & & & \blacksquare & 9\\
\hline
& & & \blacksquare& \blacksquare & \blacksquare\\
& & & \blacksquare& 5 & \\
\hline
& & & 4 & \blacksquare & 5\\
\end{array} $$
\begin{answer}
$ 25\times 19 = 475. $
\end{answer}
\end{pro}
\begin{pro}
Find the numerical value of\quad $3\cdot 4 + 4^2$.
\begin{answer}
$28$.
\end{answer}
\end{pro}
\begin{pro}
Find the numerical value of\quad $1^12^23^3$.
\begin{answer}
$108$.
\end{answer}
\end{pro}
\begin{pro}
Prove that
$$ (3 + 2)^2 =25. $$
\end{pro}
\begin{pro}
Find the
last two digits of the integer $$ 1! + 2! + 3! + \cdots + 100!.
$$
\begin{answer} For $k \geq 10$, $k!$ ends in two or more $0$'s. Thus the last
two digits of the desired sum are the last two digits of
$$1! + 2! + 3! + 4! + 5! + 6! + 7! + 8! + 9! = 409113,
$$thus the desired sum ends in $13$.
\end{answer}
\end{pro}
\begin{pro}
Prove that
$$(2)(4) + 3^3 = 35. $$
\end{pro}
\begin{pro}
Prove that
$$3^2 + 2^3 = 17. $$
\end{pro}
\begin{pro}
Prove that
$$(3^2)(4)(5) = 180. $$
\end{pro}
\begin{pro}
Prove that
$$ (5 + (3 + 2(4))^2)^3 = 2000376.$$
\end{pro}
\begin{pro}
Calculate: $$(123^2\cdot 456^2 - 789 )^0 + 3\cdot 2^3.$$
\begin{answer}
$(123^2\cdot 456^2 - 789 )^0 + 3\cdot 2^3=1+3\cdot 2^3=1+3\cdot 8 =
1+24=25$.
\end{answer}
\end{pro}
\begin{pro}
Dale should have divided a number by $4$, but instead he
subtracted $4$. He got the answer $48$. What should his answer have
been?
\begin{answer} We work backwards. He obtained $48$ from $48 + 4 = 52$. This means
that he should have performed $52 \div 4 = 13$.
\end{answer}
\end{pro}
\begin{pro} When a number is multiplied by $3$ and
then increased by $16$, the result obtained is $37$. What is the
original number?
\begin{answer} We work backwards as follows. We obtained $37$ by adding $16$ to
$37 - 16 = 21$. We obtained this $21$ by multiplying by $3$ the
number $21 \div 3 = 7$. Thus the original number was a $7$.
\end{answer}
\end{pro}
\begin{pro}
A crossnumber is like a crossword except that the answers are
numbers with one digit in each square. Complete the following
crossnumber.
\vspace*{1cm}
\begin{minipage}{3cm}
\begin{Puzzle}{3}{3}%
|* |[1]{\empty}|[2]{\empty}|.
|[3]{\empty}|*|{\empty}|.
|[4]{\empty}|{\empty} |{\empty}|.
\end{Puzzle}
\end{minipage}
\begin{minipage}{5cm}
\begin{PuzzleClues}{\textbf{ACROSS}}\\
\Clue{1}{}{\mbox{See 3D.}}\\ \Clue{3}{}{Cube.}\\ \Clue{4}{}{\mbox{5 times 3D.}}\\
\end{PuzzleClues}
\begin{PuzzleClues}{\textbf{DOWN}}\\
\Clue{2}{}{Square} \\ \Clue{3}{}{\mbox{4 times 1A.}}\\
\end{PuzzleClues}
\end{minipage}
\end{pro}
\begin{pro}
Which of the digits $\{3,4,6,8,9\}$ cannot appear as one of the
missing digits in the product
$$\opmul[voperator=bottom,operandstyle.1.1=\hole, operandstyle.1.2=\hole,operandstyle.2.1=\hole, operandstyle.2.2=\hole, resultstyle.2=\hole,
resultstyle.4=\hole,carrysub=false] {5671}{1920}$$
\begin{answer}
$5671-1920=3751$
\end{answer}
\end{pro}
\begin{pro}
A quiz has $25$ questions with four points awarded for each correct
answer and one point deducted for each incorrect answer, with zero
for each question omitted. Anna scores $77$ points. How many
questions did she omit?
\begin{answer}
Anna answered $20$ questions correctly (She could not answer less
than $20$, because then her score would have been less than
$19 \cdot 4 = 76 < 77$; She could not answer more than $20$, because
her score would have been at least $21 \cdot 4 - 3=81>77$). To get
exactly $77$ points Anna had to answer exactly $3$ questions wrong,
which means she omitted $2$ questions.
\end{answer}
\end{pro}
\begin{pro}Doing only {\em one} multiplication,
prove that
$$\begin{array}{lcl}(666)(222) +(1)(333) +(333)(222) & & \\
+(666)(333) + (1)(445) + (333)(333) & & \\\qquad + (666)(445) + (333)(445) +
(1)(222) & = & 1000000. \end{array}$$
\begin{answer}
From the distributive law,
$$\begin{array}{lcl}
(666)(222) +(1)(333) +(333)(222) & & \\
\quad +(666)(333) + (1)(445) + (333)(333) & & \\
\quad + (666)(445) + (333)(445) + (1)(222) & = & (666 + 333 +
1)(445 + 333 + 222) \\
& = & (1000)(1000) \\
& = & 1000000.
\end{array}$$
\end{answer}
\end{pro}
\begin{pro}
A certain calculator gives as the result of the product
$$ 987654 \cdot 745321 $$ the number $7.36119E11$, which means
$736, 119, 000, 000$. Explain how to find the last six missing
digits.
\begin{answer}
The trick is to use a technique analogous to the one for
multiplying, but this time three-digits at a time:
$$321 \cdot 654=209934, $$
$$321 \cdot 987 = 316827, $$
$$745 \cdot 654 = 487230, $$
$$745 \cdot 987 = 735315. $$
Thus
$$\begin{array}{llll}
& & \psframebox{987} &\psframebox{654}\\
\cdot & & \psframebox{745} &\psframebox{321}\\
\hline & & 209 & 934 \\
& 316 & 827 & \\
& 487 & 230 & \\
735 & 315 & & \\
\hline
736& 119 & 266 & 934 \\
\end{array} $$
\end{answer}
\end{pro}
\begin{pro}
How many digits
does $4^{16}5^{25}$ have?
\begin{answer} There are $28$ digits, since
$$4^{16}5^{25} = 2^{32}5^{25} = 2^{7}2^{25}5^{25} = 128 \cdot
10^{25},
$$which is the $3$ digits of $128$ followed by $25$ zeroes.
\end{answer}
\end{pro}
\begin{pro}Each element of the set $$\{10, 11, 12, \ldots , 19, 20\}$$ is multiplied by each element of
the set $$\{21, 22, 23, \ldots , 29, 30\}.$$ If all these products
are added, what is the resulting sum? \begin{answer} This is asking
for the product $(10 + 11 + \cdots + 20)(21 + 22 + \cdots + 30)$
after all the terms are multiplied. But $10 + 11 + \cdots + 20 =
165$ and $21 + 22 + \cdots + 30 = 255.$ Therefore we want
$(165)(255) = 42075$.
\end{answer}
\end{pro}
\end{multicols}
\chapter{Primes and Factorisation}
We have already explored the multiplication of natural numbers. We
will now consider the reverse procedure, that is, given a number, we
would like to obtain two (or more) numbers whose product is the
given number. \begin{df}To {\em factor} a natural number greater
than $1$ means to find two or more natural numbers whose product is
equal to the original number. The multiplication itself is called a
{\em factorisation} of the number. The numbers in the product are
called {\em factors} of the original number.\end{df}
\begin{exa}One factorisation of $20$ is $20=4 \cdot 5$. Here $4$ and
$5$ are the factors. Other factorizations of $20$ are $20=1 \cdot
20$, or $20=2 \cdot 10$, or $20=2 \cdot 2 \cdot 5$.\end{exa}
\begin{df}A {\em prime number} is a natural number greater than $1$
with exactly two factors: $1$ and the number itself. If the natural
number $n>1$ is not prime, we say that it is {\em composite}.
Lastly, we say that $1$ is a {\em unit}.\end{df} Notice that the
list of the primes starts as $2,3,5,7,11,13,17,19,23,\ldots$.
\begin{rem}Every natural number is either a unit, prime, or composite.
\end{rem}
\begin{df}
We write $a|b$, if the natural number $a\neq 0$ divides the natural
number $b$. We say that $a$ is a {\em divisor} of $b$. Notice that
$1$ and $b$ (if $b\neq 0$) are always divisors of $b$. A divisor of
$b$ is {\em proper}, if it is smaller than $b$.
\end{df}
\bigskip
A procedure, called {\em Eratosthenes' Sieve} allows us to find out
the small primes. For example, let us find all the prime numbers
smaller than $25$. We first display these numbers:
$$\begin{array}{llllll} 2 & 3 & 4 & 5 & 6 & 7 \\
8 & 9 & 10 & 11 & 12 & 13 \\
14 & 15 & 16 & 17 & 18 & 19 \\
20 & 21 & 22 & 23 & 24 & 25. \end{array} $$
We leave $2$ uncrossed, and cross out every other multiple of
$2$:
$$\begin{array}{llllll} 2 & 3 & \cancel{4} & 5 & \cancel{6} & 7 \\
\cancel{8} & 9 & \cancel{10} & 11 & \cancel{12} & 13 \\
\cancel{14} & 15 & \cancel{16} & 17 & \cancel{18} & 19 \\
\cancel{20} & 21 & \cancel{22} & 23 & \cancel{24} & 25. \end{array} $$
The next uncrossed number after $2$ is $3$. Hence, we leave $3$
uncrossed and cross out every other multiple of $3$ (if a multiple
of $3$ is already crossed, it does not matter):
$$\begin{array}{llllll} 2 & 3 & \cancel{4} & 5 & \cancel{6} & 7 \\
\cancel{8} & \cancel{9} & \cancel{10} & 11 & \cancel{12} & 13 \\
\cancel{14} & \cancel{15} & \cancel{16} & 17 & \cancel{18} & 19 \\
\cancel{20} & \cancel{21} & \cancel{22} & 23 & \cancel{24} & 25. \end{array} $$
The next uncrossed number after $3$ is $5$. Hence, we leave $3$
uncrossed and cross out every other multiple of $5$ (if a multiple
of $5$ is already crossed, it does not matter):
$$\begin{array}{llllll} 2 & 3 & \cancel{4} & 5 & \cancel{6} & 7 \\
\cancel{8} & \cancel{9} & \cancel{10} & 11 & \cancel{12} & 13 \\
\cancel{14} & \cancel{15} & \cancel{16} & 17 & \cancel{18} & 19 \\
\cancel{20} & \cancel{21} & \cancel{22} & 23 & \cancel{24} & \cancel{25}. \end{array} $$
We continue this procedure with the next uncrossed numbers. We see
that after $13$ we are already out of range. The numbers that remain
uncrossed are the primes. Hence the primes smaller than $25$ are It
$$ 2, 3, 5, 7, 11, 13, 17, 19, 23, \ldots . $$
Table \ref{tab:primes} lists all primes under $1014$.
\begin{table}[!h]
\centering {\tiny\begin{tabular}{|rrrrrrrrrr|} \hline
2 & 3 & 5 & 7 & 11 & 13 & 17 & 19 & 23 & 29\\
31& 37 & 41 & 43 & 47& 53 & 59 & 61& 67& 71\\
73 & 79 & 83& 89 & 97 & 101 & 103 & 107 & 109 & 113\\
127 & 131& 137 & 139 & 149 &151 &157& 163 &167 &173\\
179 & 181& 191 & 193 & 197 &199& 211 & 223 & 227 &229\\
233 & 239& 241 & 251 & 257 &263& 269 & 271 & 277 &281\\
283 & 293& 307 & 311 & 313 &317& 331 & 337 & 347 &349\\
353 & 359& 367 & 373 & 379 &383& 389 & 397 & 401 &409\\
419 & 421& 431 & 433 & 439 &443& 449 & 457 & 461 &463\\
467 & 479& 487 & 491 & 499 &503& 509 & 521 & 523 &541\\
547 & 557& 563 & 569 & 571 &577& 587 & 593 & 599 &601\\
607 & 613& 617 & 619 & 631 &641& 643 & 647 & 653 &659\\
661 & 673& 677 & 683 & 691 &701& 709 & 719 & 727 &733\\
739 & 743& 751 & 757 & 761 &769& 773 & 787 & 797 &809\\
811 & 821& 823 & 827 & 829 &839& 853 & 857 & 859 &863\\
877 & 881& 883 & 887 & 907 &911& 919 & 929 & 937 &941\\
947 & 953& 967 & 971 & 977 &983& 991 & 997 & 1009 &1013\\
\hline
\end{tabular}
} \meinecaption{1}{Some Prime Numbers}\label{tab:primes}
\end{table}
Euclid proved (around 300 BC) that there are infinitely many prime
numbers, hence we can find arbitrarily large prime numbers, which
means that the table \ref{tab:primes} can be made as long as we
wish. \begin{rem}All prime numbers greater than $2$ are odd. But not
every odd number is prime. For example, $9$ is not prime, since it
has as three factors, $1$, $3$, and $9$.\end{rem} \begin{df}The
factorisation of a natural number greater than $1$ in which every
factor greater than $1$ is prime, is called the {\em prime
factorisation} of the number.\end{df} It can be proved that the
factorisation of a natural number into primes is unique, up to the
order of the factors. This means that if you and I start with the
same number and we factor it into primes, then we will end up with
the same factorisations, except perhaps in the order in which
display the factors.
\begin{exa}Find the prime
factorisations of $360$ and of $216$.\label{exa:fact-216}\end{exa}
\begin{solu}
A schematic procedure is shown in figures \ref{fig:factor-360}. The
idea is to start small and then substitute equals by equals. Then
choose all the prime factors in the circles. We conclude that
$$360=2 \cdot 2 \cdot 2 \cdot 3 \cdot 3 \cdot 5 = 2^3 \cdot 3^2 \cdot 5, $$
and that
$$ 216=2^33^3. $$
\end{solu}
\begin{figure}[!h]
\begin{minipage}{7cm}\centering \psset{unit=.25pc}
\pstree{\Toval{360}}%
{\Toval{2}%
\pstree{\Toval{180}}%
{\Toval{2}%
\pstree{\Toval{90}} %
{\Toval{2}%
\pstree{\Toval{45}} %
{
\Toval{5}%
\pstree{\Toval{9}} %
{
\Toval{3}%
\Toval{3}%
} } } } }\meinecaption{1}{Prime factorisation of
$360$.}\label{fig:factor-360}
\end{minipage}
\hfill
\begin{minipage}{7cm}\centering \psset{unit=.25pc}
\pstree{\Toval{216}}
{\Toval{2}%
\pstree{\Toval{108}}%
{\Toval{2}%
\pstree{\Toval{54}}%
{\Toval{2}%
\pstree{\Toval{27}} %
{\Toval{3}%
\pstree{\Toval{9}}%
{ \Toval{3} \Toval{3} } } } } } \meinecaption{1}{Prime
factorisation of $216$.}\label{fig:factor-216}
\end{minipage}
\end{figure}
\begin{exa}
How many divisors does $100$ have?
\end{exa}
\begin{solu}Clearly
$100=2^25^2$. Thus every divisor will be of the form $2^a5^b$, with
$a=0, 1,$ or $2$, and $b=0, 1$ or $2$. Taking $a=0,b=0$ we get
$2^05^0=1$. Taking $a=1,b=0$ we get $2^15^0=2$. Taking $a=2,b=0$ we
get $2^25^0=4$. Taking $a=0,b=1$ we get $2^05^1=5$. Taking $a=1,b=1$
we get $2^15^1=10$. Taking $a=2,b=1$ we get $2^25^1=20$. Taking
$a=0,b=2$ we get $2^05^2=25$. Taking $a=1,b=2$ we get $2^15^2=50$.
Taking $a=2,b=2$ we get $2^25^2=100$. The divisors of $100$ are thus
$$1, 2, 4, 8, 10, 20, 25, 50, 100, $$
$9$ in number.
\end{solu}
\begin{rem} In the preceding example there was no necessity of
listing all the divisors in order to know how many there are. Since
there are three choices for the exponent $a$, and three choices for
the exponent $b$, there is a total of $3\cdot 3 = 9$ divisors of a
$100.$
\end{rem}
\begin{df}
The {\em greatest common divisor} $\gcd (a, b)$ of two natural
numbers $a$ and $b$ is the largest natural number that divides both
$a$ and $b$.
\end{df}
For example, the divisors of $12$ are
$$ \{1,2,3,4,6,12\}, $$
and those of $30$ are
$$\{1,2,3,5,6,15,30\}, $$
and hence $\gcd (12,30)=6$, because $6$ is the largest number that
appears in both lists.
\bigskip
Using prime factorisations, we may find the greatest common divisor
of two numbers without having to list all their divisors. We look at
which primes are common to both numbers and then select the least
power appearing of the primes.
\begin{exa}
Find $\gcd (12,30)$ by finding the prime factorisation of $12$ and
$30$ first.
\end{exa}
\begin{solu}
First observe that $12=2^2\cdot 3$ and that $30 =2^1\cdot 3^1\cdot
5^1 $. The primes $2$ and $3$ appear in both numbers. The $2$
appears with least exponent $1$ and so does the $3$, and so,
therefore, $\gcd (12,30)=2\cdot 3=6$
\end{solu}
\begin{exa}
Find $\gcd (300,360)$.
\end{exa}
\begin{solu}
First observe that $300=2^2\cdot 3\cdot 5^2$ and that $360 =2^3\cdot
3^2\cdot 5$. The primes $2$, $3$, and $5$ appear in both numbers.
The $2$ appears with least exponent $2$, the $3$ appears with least
exponent $1$, and the $5$ appears with least exponent $1$.
Therefore, $\gcd (300,360)=2^2\cdot 3 \cdot 5 =60$.
\end{solu}
\begin{exa}
Find $\gcd (132,504)$.
\end{exa}
\begin{solu}
First observe that $132=2^2\cdot 3\cdot 11$ and that $504 =2^3\cdot
3^2\cdot 7$. Only the primes $2$ and $3$ appear in both numbers. The
$2$ appears with least exponent $2$, the $3$ appears with least
exponent $1$. Therefore, $\gcd (132,504)=2^2\cdot 3 = 12$.
\end{solu}
\begin{exa}
Find $\gcd (100,343)$.
\end{exa}
\begin{solu}Observe that $100=2^25^2$ and $343=7^3$. They have no
prime factor in common, their only common divisor is $1$.
Therefore, $\gcd (100,343)=1$.
\end{solu}
\begin{df}
Two natural numbers $a$ and $b$ are said to be {\em relatively
prime} if $\gcd (a,b)=1$.
\end{df}
A similar procedure for finding the least common multiple of two
numbers can now be outlined. We first find their prime
factorisations, take all the primes appearing in them, and choose
the powers with the maximum exponents for this primes.
\begin{exa}
Find $\lcm (12,45)$.
\end{exa}
\begin{solu}Notice that this is example \ref{exa:lcm-12-45}.
The prime factorisations are $12=2^2\cdot 3$ and $45=3^2\cdot 5$.
All primes occurring at least once are $2$, $3$, and $5$. The prime
$2$ appears with maximum exponent $2$, the primes $3$ appears with
maximum exponent $2$ and the prime $5$ appears with maximum exponent
$1$. Hence $\lcm (12,45)=2^23^25 = 180$.\end{solu}
\begin{exa}
Find $\lcm (300,360)$.
\end{exa}
\begin{solu}
First observe that $300=2^2\cdot 3\cdot 5^2$ and that $360 =2^3\cdot
3^2\cdot 5$. The primes $2$, $3$, and $5$ appear at least once. The
$2$ appears with maximum exponent $3$, the $3$ appears with maximum
exponent $2$, and the $5$ appears with maximum exponent $2$.
Therefore, $\lcm (300,360)=2^3\cdot 3^2 \cdot 5^2 =1800$.
\end{solu}
\begin{exa}
Find $\lcm (132,504)$.
\end{exa}
\begin{solu}
First observe that $132=2^2\cdot 3\cdot 11$ and that $504 =2^3\cdot
3^2\cdot 7$. The primes $2$, $3$, $7$, and $11$ appear at least
once. The $2$ appears with maximum exponent $3$, the $3$ appears
with maximum exponent $2$, the $7$ appears with maximum exponent
$1$, and the $11$ appears with maximum exponent $1$. Therefore,
$$\lcm (132,504)=2^3\cdot 3^2 \cdot 7^1\cdot 11 ^1 =5544.$$
\end{solu}
\section*{Homework}
\begin{multicols}{2}\columnseprule 1pt \columnsep 25pt\multicoltolerance=900\small
\begin{pro}
Give the prime factorisation of $24$.
\begin{answer}
$24=2^33$
\end{answer}
\end{pro}
\begin{pro}
Give the prime factorisation of $36$.
\begin{answer}
$36=2^23^2$
\end{answer}
\end{pro}
\begin{pro}
How many prime numbers less than $10000$ have digits adding up to
$2$?
\begin{answer}
Three. They are $2$, $11$, $101$.
\end{answer}
\end{pro}
\begin{pro}
Demonstrate that $$\gcd (24,36)\cdot \lcm (24,36)=24\cdot 36.$$
\begin{answer}
Since $24=2^33$ and $36=2^23^2$, we have $\gcd (24,36) = 2^2\cdot
3=12$ and $\lcm (24,36) = 2^3\cdot 3^2=72$, from where $$\gcd
(24,36)\cdot \lcm (24,36)= 12\cdot 72 = 12\cdot 2 \cdot 36= 24\cdot
36.$$
\end{answer}
\end{pro}
\begin{pro}
How many divisors does $36$ have?
\begin{answer} $36=2^23^2$, so it has $(2+1)(2+1)=9$ divisors.
\end{answer}
\end{pro}
\begin{pro}
How many divisors does $38$ have?
\begin{answer} $38=2^1\cdot 19^1$, so it has $(1+1)(1+1)=4$ divisors.
\end{answer}
\end{pro}
\begin{pro}
How many divisors does $40$ have?
\begin{answer} $40=2^35^1$, so it has $(3+1)(1+1)=8$ divisors.
\end{answer}
\end{pro}
\begin{pro}
An unsolved problem in Number Theory is the {\em Goldbach
Conjecture}, which asserts that every even natural number greater
than $6$ can be written as the sum of distinct primes. Verify
Goldbach's conjecture from every even integer between $8$ and $36$,
inclusive.
\begin{answer}In some cases there may be more than one answer. For
example, $20$ is $7+13$ and also $3+17$, etc.
$$ \begin{array}{lllll} 8 = 3 + 5 & 10 = 3 + 7 & 12 = 5 + 7 & 14 = 3 + 11 & 16 = 5 + 11 \\
18 = 7+11& 20 =3+17 & 22 = 3+19 & 24 = 7+17 & 26 =3+23 \\
28 = 11+17 & 30 =13+17 & 32 = 13+19 & 34 =11+23 & 36 =13+23 \\
\end{array} $$
\end{answer}
\end{pro}
\begin{pro} Given that there is a unique digit
$d\in\{0,1,2,3,4,5,6,7,8,9\}$ so that the nine-digit number
$19700019d$ is prime, find it.
\begin{answer}
$d$ cannot be even, as then the number would be divisible by $2$.
Now, $1+9+7+0+0+0+1+9+d=27+d$. If $d\in\{3,6,9\}$, then the sum of
the digital sum would be divisible by $3$, and so the number would
be divisible by $3$. If $d=5$, the number would be divisible by $5$.
If $d=7$, the number would be divisible by $197$. This leaves $d=1$
as the only possible digit, and so it must be this one.
\end{answer}
\end{pro}
\begin{pro}
A natural number is called {\em perfect} if it is the sum of all its
proper divisors. For example, the proper divisors of $6$ are $1, 2,
3$ and $6=1+2+3$. Also, the proper divisors of $28$ are $1, 2, 4, 7,
14$ and $28=1+2+4+7+14$, so $28$ is perfect. All perfect numbers
known to date are even, and it is a conjecture, still unsolved, that
there are no odd perfect numbers. Prove that $496$ is a perfect
number by finding all its proper divisors and adding them.
\begin{answer}
The proper divisors of $496$ are
$$ \{1,2,4,8,16,31,62,124,248\} $$ and we have
$$ 1+2+4+8+16+31+62+124+248=496. $$
\end{answer}
\end{pro}
\begin{pro}
How many divisors does a prime number have?
\begin{answer} If $p$ is a prime, then it has only $2$ divisors,
namely $1$ and $p$.
\end{answer}
\end{pro}
\begin{pro}
Factor $111111$ (six $1$'s) into primes.
\begin{answer}We have,$$\begin{array}{lll} 111111
& = & (111)(1001) \\
& = & (3)(37)(11)(91) \\
& = & (3)(37)(11)(7)(13) \\
& = & (3)(7)(11)(13)(37) \\
\end{array}
$$
\end{answer}
\end{pro}
\begin{pro}
Let $p$ be a prime number. List all the divisors of $p^2$.
\begin{answer}
$\{1,p,p^2\}$
\end{answer}
\end{pro}
\begin{pro}
Prove that among any $101$ integers taken from the set $\{1,2,\ldots
, 200\}$, there will always be $2$ which are relatively prime.
\begin{answer}
Notice that consecutive integers are always relatively primes. Any
$101$ integers is bound to have the pair of one the one hundred sets
$$ \{1,2\}, \{3,4\},\ldots , \{197,198\},\{199,200\}. $$
\end{answer}
\end{pro}
\begin{pro}
Which numbers have an odd number of divisors?
\begin{answer}
Only perfect squares.
\end{answer}
\end{pro}
\begin{pro}
Four comrades are racing side by side down a dusty staircase.
Frodo goes down two steps at a time, Gimli three, Legolas four, and
Aragorn five. If the only steps with all four's footprints are at
the top and the bottom, how many steps have just one footprint?
\begin{answer} Number the steps from $0$ to $N$, where $N$ is the last step, and hence there are
$N+1$ steps. Notice that the top and the bottom of the stairs are
counted as steps. Let us determine first the number of steps. Frodo
steps on steps $0$, $2$, $4$, \ldots , $N$; Gimli steps on steps
$0$, $3$, $6$, \ldots , $N$; Legolas steps on steps $0$, $4$, $8$,
\ldots , $N$; and Aragorn steps on steps $0$, $5$, $10$, \ldots ,
$N$. For this last step to be common, it must be a common multiple
of $2$, $3$, $4$, and $5$, and hence
$$N = \lcm (2,3,4,5) = 60. $$
Since every step that Legolas covers is also covered by Frodo, we
don't consider Legolas in our count. Frodo steps alone on the steps
which are multiples of $2$, but not multiples of $3$ or $5$. Hence
he steps on the $8$ steps
$$ 2, 14 , 22, 26, 34, 38, 46, 58. $$
Gimli steps alone on the $8$ steps that are multiples of $3$ but not
multiples of $2$ nor $5$:
$$ 3, 9, 21, 27, 33, 39, 51,
57. $$
Finally, Aragorn steps alone on the $4$ steps that are multiples of
$5$ but not multiples of $2$ or $3$:
$$5,25, 35, 55.$$
Our final count is thus $8+8+4=20$.
\end{answer}
\end{pro}
\begin{pro}[The Locker-room Problem] A locker room contains $100$ lockers, numbered $1$ through $100$.
Initially all doors are open. Person number $1$ enters and closes
all the doors. Person number $2$ enters and opens all the doors
whose numbers are multiples of $2$. Person number $3$ enters and if
a door whose number is a multiple of $3$ is open then he closes it;
otherwise he opens it. Person number $4$ enters and changes the
status (from open to closed and viceversa) of all doors whose
numbers are multiples of $4$, and so forth till person number $100$
enters and changes the status of door number $100$. Which doors are
locked?
\begin{answer}
Notice that if $d$ divides $n$ so does $\dfrac{n}{d}$. Thus we can
pair up every the different divisors of $n$, and have an even
number of divisors as long as we do not have $d=\dfrac{n}{d}$. This
means that the integers with an even number of divisors will have
all doors open, and those with an odd number of divisors will all
all doors closed. This last event happens when
$d=\dfrac{n}{d}\implies n=d^2$, that is, when $n$ is a square. Hence
only the ten doors $$ 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, $$are
closed.
\end{answer}
\end{pro}
\begin{pro}
Find an infinite set of positive integers such that the sum of any
finite number of distinct elements of the set is not a square.
\begin{answer}
Consider the set $$ \{2^n: n\in \naturals, n\ \mathrm{odd}\}. $$
\end{answer}
\end{pro}
\end{multicols}
\section*{References}
Much of the material of this section belongs to a branch of
Mathematics formerly called ``the Higher Arithmetic'', nowadays
called Number Theory. Some references follow.
\begin{description}
\bibitem[NiZuMo]{NiZuMo} I. Niven, H. Zuckermann, H. Montgomery ``Introduction to Number Theory''.
\end{description}
\chapter{Division}
\begin{df}[Definition of Division]
Let $m, n, x$ be natural numbers, with $n\neq 0$. Then the statement
$m\div n=x$ means that $m=xn$.
\end{df}
\begin{exa}
To compute $15\div 3$ we think of which number when multiplied $3$
gives $15$. Clearly then $15\div 3=5$ since $15=5\cdot 3$.
\end{exa}
\begin{rem}
Division is neither closed in $\naturals$, nor commutative, nor
associative. For example, $$4 \div 8 \notin \naturals, \quad 4\div 8
\neq 8 \div 4, \quad 8\div (4\div 2 )\neq (8\div 4)\div 2.
$$\end{rem}
Our links to division and reality are the ideas of {\em sharing} and
{\em containment}. The example above can be thought as the solution
to the following problem: $15$ customers are going to be handled
equally by $3$ waitresses, how many people will each waitress
handle? The result $5$ means that $5+5+5=15$, that is, that each of
the $3$ waitresses will handle or receive an {\em equal share} of
$5$ customers. This corresponds to the idea of division as repeated
subtraction:
$$\underbrace{15-3 =12}_{\mathrm{first\ instance}}\to \underbrace{12-3 =9}_{\mathrm{second\ instance}} \to
\underbrace{9-3=6}_{\mathrm{third\ instance}} \to
\underbrace{6-3=3}_{\mathrm{fourth\ instance}} \to
\underbrace{3-3=0}_{\mathrm{fifth\ instance}}
$$
Since the subtraction was able to be evenly carried out five times,
$15\div 3=5$.
One may also think of the problem in the following way. If there are
$15$ customers to be served, and each waitress receives $5$, how
many waitresses are there? In this case we are asserting that
$3+3+3+3+3=15$.
\bigskip
The above definition of division depends on the existence of a
natural number $x$ so that $m=xn$, which effectively means, that we
have division without remainder. What happens in general? Let us
first introduce some notation.
\begin{df}
By $\floor{x}$ (pronounced {\em floor of $x$}) we mean either $x$ if
$x$ is an integer, or the integer just to the left of $x$.
\end{df}
For example, $\floor{3}=3$, $\floor{3.1}=3$, $\floor{3.5}=3$,
$\floor{3.9}=3$, $\floor{4.1}=4$. Notice that this is not rounding.
\bigskip
Observe that the following inequality is satisfied:
\begin{equation}
\floor{x}\leq x < \floor{x}+1.
\end{equation}
Let us now consider the task of defining $a\div b$, by considering
first an example. Consider the problem of performing the division
$23\div 5$. We look for multiples of $5$ that straddle $23$. We
readily see that
$$ 20<23<25\implies 5\cdot 4 < 23 < 5\cdot 5. $$
We divide through by $5$ and assume that inequalities are preserved:
$$ 5\cdot 4 < 23 < 5\cdot 5
\implies (5\cdot 4)\div 5 < 23\div 5 < (5\cdot 5) \div 5 \implies 4
< 23\div 5 < 5.
$$This means that $$ \floor{23\div 5}=4. $$ In fact, we have
$$ 23=5\cdot 4 + 3 \implies 23=5\cdot\floor{23\div 5} +3. $$
The excess $3$ over the multiple of $5$ just to the left of $23$ is
the {\em remainder}.
We are ready to state the following result, which we will admit
without proof.
\begin{thm}[Division Algorithm]
Given two natural numbers $a$ and $b$ with $b\neq 0$, we may find
two unique integers $\floor{a\div b}$ and $r$ with $0 \leq r ~~&
\~~ & \& \

**$ {\em is strictly greater than}, \item[] $\leq$ {\em is less
than or equal to} and,
\item[] $\geq$ {\em is greater than or equal to}.
\end{enumerate}\end{center}
We also mention the symbols $=$, {\em equal to}, and $\neq$, {\em
not equal to}.
\end{df}
\begin{exa}
The farther to the right a number is on the number line, the larger
it is. Thus we have
$$1914<1939, \quad 666>100, \quad 1 \leq 1. $$
\end{exa}
\bigskip
We now introduce the operation of $+$ or {\em addition} in
$\naturals$. We can think of $+$ as {\em joining}, {\em
concatenating}, {\em appending}, that is, {\em and}. In fact, the
$+$ used today evolved from the fact that in Latin manuscripts
people used to write {\em et}---meaning {\em and}---for ``plus.''
This was shortened to just ``{\em t}'', and then this was slightly
altered to give the modern sign of plus. Let us introduce some fancy
terminology.
\begin{df}
In the expression $a+b=s$, $a$ is called the {\em augend}, $b$ is
called the {\em addend} and $s$ is called the {\em sum}. Both the
augend and the addend are called {\em summands}.
\end{df}
Thus in $1+4=5$, $1$ and $4$ are the summands and $5$ is their sum.
Now, joining one marble to four marbles gives five marbles, that
is, $1$ marble $+ 4$ marbles $= 5$ marbles. Concatenating a line
segment of length $b$ inches to one of length $a$ inches gives a
line segment of length $a+b$ inches. See figures \ref{fig:add-join}
and \ref{fig:add-conc}
\vspace*{3cm}
\begin{figure}[h]
\begin{minipage}{7cm}
\centering \psset{unit=1pc} \pscircle*[linecolor=red](-5,0){.7}
\pscircle*[linecolor=red](-3,0){.7}
\pscircle*[linecolor=red](-1,0){.7}
\pscircle*[linecolor=red](1,0){.7}
\pscircle*[linecolor=red](3,0){.7}
\pspolygon(-4.15,1)(-4.15,-1)(3.9,-1)(3.9,1)
\meinecaption{1}{Joining.} \label{fig:add-join}
\end{minipage}
\hfill
\begin{minipage}{7cm}
\centering \psset{unit=1pc}
\psline[linecolor=gray,linewidth=2pt](-1,4)(1,4)
\pcline[offset=12pt]{|-|}(-1,4)(1,4)
\ncput*[nrot=:U]{$a$}\rput(0,3){$+$}
\psline[linecolor=blue,linewidth=2pt](1,2)(5,2)
\pcline[offset=-12pt]{|-|}(1,2)(5,2)
\ncput*[nrot=:U]{$b$}\rput(0,0){$=$}
\psline[linecolor=green,linewidth=2pt](-1,-1)(5,-1)\pcline[offset=-12pt]{|-|}(-1,-1)(5,-1)
\ncput*[nrot=:U]{$a+b$}\meinecaption{1}{Concatenation.}
\label{fig:add-conc}
\end{minipage}
\end{figure}
Observe that in the examples above we added like quantities. Thus
marbles were added to marbles and inches to inches. We did not dare
to add marbles to inches! We call these similar entities the {\em
denominator} of the term. Hence in the sum $1$ marble $+ 4$ marbles
$= 5$ marbles, marbles are the denominator. In the sum $1$ inch $+
4$ inches $= 5$ inches, inches are the denominator. It should not be
a stretch of one's imagination to progressively reach the
following levels of abstraction:
\begin{itemize}
\item $1m + 4m=5m$
\item $1A + 4A=5A$
\item $\dfrac{1}{401} + \dfrac{4}{401}=\dfrac{5}{401}$
\item $1+4=5$
\end{itemize}
In the first sum, $m$ is the common denominator, we are adding two
like objects. Similarly, $A$ is the common denominator for the
second sum, and $\dfrac{\cdot}{401}$ is the one for the third sum.
The fourth sum is more abstract, in the sense that even though we do
not have a common denominator, we are asserting that if to a unit of
whatever we add four units of the same whatever, the result is five
of the whatever. It is important to realise that this last
abstraction came from our agreement that can only add terms with a
common denominator. Here are more challenging examples.
\begin{exa}
Ariel walked $x$ miles and rode $10$ miles. How far did he go?
\end{exa}
\begin{solu}
The total of miles he went was $ x+10 $.
\end{solu}
\begin{exa}
Collect like terms: $$ a+2b+3a+4b. $$
\end{exa}
\begin{solu}
We combine the $a$'s with the $a$'s and the $b$'s with the $b$'s:
$$ a+2b+3a+4b = 4a+6b. $$
\end{solu}
\begin{rem}
In the preceding example there is the temptation to further combine
the $4a$ with the $6b$. But this temptation has no basis: we do not
know what the $a$'s are and what the $b$'s are. They are different
denominators and we cannot combine them without further information.
So, if the $a$'s were {\em apples} and the $b$'s {\em bananas}, we
cannot combine them because they are different fruit! But you may
object and say: ``Well, $4$ apples and $6$ bananas do give $10$
fruit! There! I combined them! But you have cheated! You have
created a common denominator, {\em fruit}, for apples and bananas,
and that was not asked in the problem. You cannot deviate from what
is given to you!
\end{rem}
Again, consider the following way of adding $731$ and $695$. Since
$$731= 7\cdot 10^2 + 3\cdot 10 + 1, \qquad 6\cdot 10^2 + 9\cdot 10 + 5, $$
we could add in the following fashion, without worrying about
carrying:
$$(7\cdot 10^2 + 3\cdot 10 + 1) +(6\cdot 10^2 + 9\cdot 10 + 5)= 13\cdot 10^2+12\cdot 10 + 6=1300+120+6=1426. $$
The above reasoning can be extended to other than powers of ten.
\begin{exa}
Collect like terms:
$$ (7x^2+3x+1)+(6x^2+8x+5). $$
\end{exa}
\begin{solu}
We match the squares with the squares, the $x$ to the first power
with the $x$ to the first power, and the constant term with the
constant term to get
$$ (7x^2+3x+1)+(6x^2+8x+5)=13x^2+11x+6. $$
\end{solu}
As we begin to see more examples of addition of natural numbers, the
following properties emerge.
\begin{axi}[Closure] The sum of any two natural numbers is a natural number, that is, $\naturals$ is {\em closed under addition}: if $a\in
\naturals$ and $b\in \naturals$ then also $a+b\in \naturals$.
\end{axi}
\begin{rem}
Thus if we add two natural numbers, we don't get eggs, we don't get
pastrami, we get a natural number.
\end{rem}
\begin{axi}[Additive Identity] If $0$ is added to any natural number, the result is unchanged, that is $0\in\naturals$ is the
{\em additive identity} of $\naturals$. It has the property that for
all $x\in\naturals$ it follows that $$x=0+x=x+0. $$
\end{axi}
Again, it is easy to see that when two natural numbers are added,
the result does not depend on the order. This is encoded in the
following axiom.
\begin{axi}[Commutativity]Let $a\in\naturals$ and $b\in \naturals$ be arbitrary. Then
\mbox{$a+b=b+a$}.
\end{axi}The last axiom has to do with grouping.
\begin{axi}[Associativity]
Let $a, b, c$ be arbitrary natural numbers. Then the order of
parentheses when performing addition is irrelevant, that is,
$$ a+(b+c)=(a+b)+c = a+b+c. $$
\end{axi}
How do we actually {\em add} in decimal notation? Let us first
present a geometric method to add integers along the number line.
\begin{exa}Use the number line to
perform the addition $2+3$. \end{exa}
\begin{solu}First locate $2$ by starting from $0$ and moving two
units right.
\vspace{.5cm}
$$\psset{unit=.5} \pstGeonode[PointName=none](0,0){O}(2,0){A}\ncbar[angleA=90,nodesep=3pt,linewidth=2pt,linecolor=red]{->}{O}{A}\psaxes[yAxis=false,labels=none,
arrowscale=3,arrows={->}](0,0)(0,0)(11.5,0)
\uput[d](0,0){0}\uput[d](2,0){2}
$$
\vspace{.5cm}
\bigskip Now, starting at $2$ move three units
right.
\vspace{.5cm}
$$\psset{unit=.5} \pstGeonode[PointName=none](0,0){O}(2,0){A}(5,0){B}\ncbar[angleA=90,nodesep=3pt,linewidth=2pt,linecolor=red]{->}{A}{B}\psaxes[yAxis=false,labels=none,
arrowscale=3,arrows={->}](0,0)(0,0)(11.5,0)
\uput[d](0,0){0}\uput[d](2,0){2}\uput[d](5,0){5}
$$\vspace{.5cm}
\bigskip
Since we landed at $5$, we conclude that $2+3=5$. \end{solu}
The above method of addition can, in principle, accomplish all the
sums of natural numbers that we think of. It suffers from the
limitation of us having to locate numbers on the number line and
counting at least twice. If the numbers are really large, then we
might miscount. Countless cultures that did not know our decimal
positional notation were able to carry out these computations. But
the cumbersomeness of these calculations meant that the operations
could be carried out by expert accountants. Today, one hopes, even a
child can perform these operations.
In order to explain an algorithm for addition, let us first think of
how we can give the decimal representation of an arbitrary number.
Let us agree that the notation
$$ \underline{d_nd_{n-1}\ldots d_2d_1d_0}, $$where the $d$'s are
digits, means
$$ \underline{d_nd_{n-1}\ldots d_1d_0} = d_n\cdot 10^n+d_{n-1}\cdot 10^{n-1}+\cdots +d_2\cdot 10^2 + d_2\cdot 10 + d_0. $$
For example, if the numeral is $60637$, then this has $5$ digits,
and we have,\footnote{The arrow $\implies$ is read {\em implies}.}
$$\underline{d_4d_3d_2d_1d_0}=60637 \implies d_4=6, \quad d_3=0, \quad d_2 =6, \quad d_1 =3, \quad d_0 =7. $$
If the numeral is $1234567$, then this has $7$ digits, and we have
$$\underline{d_6d_5d_4d_3d_2d_1d_0}=1234567 \implies d_6=1, \quad d_5=2, \quad d_4=3, \quad d_3=4, \quad d_2 =5, \quad d_1 =6, \quad d_0 =7. $$
Suppose now that we have two arbitrary decimal integers, say
$$ \underline{d_nd_{n-1}\ldots d_2d_1d_0},\qquad \underline{e_pe_{p-1}\ldots e_2e_1e_0}. $$
Here it may be the case that $n=p$, that is, the integers have the
same number of digits, but we do not know that in advance. Let us
suppose for the sake of argument that $n****0$ is the divisor, then
the possible remainders are the $b$ numbers $0$, $1$, $2$, \ldots ,
$b-1$.
\begin{exa}
Take $b=2$ as the divisor. The possible remainders are $0$ and
$2-1=1$. Thus every natural number comes in one of the two flavours
$$ \{0,2,4,6,\ldots \}\quad \mathrm{or}\quad \{1,3,5,7,\ldots\}, $$
that is, as a number that leaves remainder $0$ upon division by $2$
or as a number that leaves remainder $1$ upon division by $2$. In
fact, we see that
$$ \naturals = \{0,2,4,6,\ldots \}\cup \{1,3,5,7,\ldots\}, $$
and so the division algorithm serves to partition the natural
numbers into two parts. We can also put this in algebraic terms, by
saying that every natural number is of the form $2a$ or $2a+1$ for
some natural number $a$.
\end{exa}
\begin{exa}
Take $b=3$ as the divisor. The possible remainders are $0$, $1$, and
$3-1=2$. Thus every natural number comes in one of the three
flavours
$$ \{0,3,6,9,\ldots \}\quad \mathrm{or}\quad \{1,4,7,10,\ldots\}\quad \mathrm{or}\quad \{2,5,8,11,\ldots\}. $$
that is, as a number that leaves remainder $0$ upon division by $3$,
as a number that leaves remainder $1$ upon division by $3$
or as a number that leaves remainder $2$ upon division by $2$. In
fact, we see that
$$ \naturals = \{0,3,6,9,\ldots \}\cup \{1,4,7,10,\ldots\}\cup \{2,5,8,11,\ldots\}. $$
and so the division algorithm serves to partition the natural
numbers into three parts. We can also put this in algebraic terms,
by saying that every natural number is of the form $3a$, $3a+1$ or
$3a+2$ for some natural number $a$.
\end{exa}
\begin{exa}
Prove that the sum of two even natural numbers is even. Can you find
five natural numbers whose sum is $25$?
\end{exa}
\begin{solu}
Let $2x$ and $2y$ be two even natural numbers. Then using the
distributive law in reverse,
$$ 2x+2y = 2(x+y). $$Since $x+y$ is a natural numbers, $2(x+y)$ is
twice a natural number and so even.
The answer to our second question is clearly NO, since $25$ is an
odd number and the sum of five even numbers would be even.
\end{solu}
\bigskip
We now discuss criteria for a given natural number to be (evenly)
divisible by another number. Granted, we will only consider a few
cases---just the criteria for divisibility by $2$, $3$, $5$, and
$9$---but these are important in their own merit.
\bigskip
The task we have at hand is the following. Suppose that we had a
very large number, let us say $1234509876.$ How do we know, {\em
without actually performing the division}, whether it is divisible
by $2$? By $3$? By $5$? It turns out that the decimal positional
notation makes answering these questions extremely easily.
\bigskip
For the case of divisibility by $2$, we want to investigate whether
$1234509876$ belongs to the set
$$ \{0,2,4,6,8,10,\ldots \} $$ of integers that leave no remainder
when divided by $2$. But we may write
$$ 1234509876 = 1234509870+6= 123450987\cdot 10+6=123450987\cdot 2\cdot 5 + 6.$$
Clearly, $123450987\cdot 2\cdot 5 $ is divisible by $2$. Hence the
divisibility of $1234509876$ hinges on whether its last digit, $6$
is divisible by $2$, which is clearly the case. We conclude that
thus $1234509876$ is divisible by $2$. Notice that we did not need
to perform any division. A more general argument can then establish
the following result.
\begin{thm}
A natural number is divisible by $2$ (even) if and only if its last
digit of its decimal expansion is even, that is, it ends in one of
$\{0,2,4,6,8\}$.
\end{thm}
For the case of divisibility by $5$, we want to investigate whether
$1234509876$ belongs to the set
$$ \{0,5,10,15,20,25,\ldots \} $$ of integers that leave no remainder
when divided by $5$. But we may write
$$ 1234509876 = 1234509870+6= 123450987\cdot 10+6=123450987\cdot 2\cdot 5 + 6.$$
Clearly, $123450987\cdot 2\cdot 5 $ is divisible by $5$. Hence the
divisibility of $1234509876$ hinges on whether its last digit, $6$
is divisible by $5$, which is clearly not the case. We conclude that
thus $1234509876$ is not divisible by $5$. Notice that we did not
need to perform any division. We state the general result.
\begin{thm}
A natural number is divisible by $5$ if and only if its last digit
of its decimal expansion ends in one of $\{0,5\}$.
\end{thm}
\begin{rem}
In fact, by writing $ 1234509876 = 1234509875+1$, since by the
preceding theorem $ 1234509875$ is evenly divisible by $5$, we see
that $ 1234509876$ leaves remainder $1$ upon division by $5$.
\end{rem}
Before we introduce a criterion for divisibility by $3$ or $9$, we
need the concept of {\em digital sum}. Consider any natural number,
say $1234509876$. Sum its digits successively, and then sum the
digits of each successive sum, until only a $1$-digit integer
remains:
$$1234509876\to 1+2+3+4+5+0+9+8+7+6=45\to 4+5=9, $$
whence the digital sum is $9$.
\bigskip
Our criterion for divisibility by $3$ or $9$ will be linked to the
digital sum of the number in question. For the example at hand,
observe that
$$\begin{array}{lll}1234509876 & = & 1\cdot 10^9 + 2\cdot 10^8 + 3\cdot 10^7+ 4\cdot 10^6+ 5\cdot 10^5
+ 0\cdot 10^4 + 9\cdot 10^3+ 8\cdot 10^2 + 7\cdot 10^1+6\\
& = & 1\cdot 999999999+2\cdot 99999999 + 3\cdot 9999999+4\cdot
999999 + 5\cdot 99999+ 0 \cdot 9999 + 9\cdot 999 + 8\cdot 99 + 7
\cdot 9\\
& & \qquad + (1+2+3+4+5+0+9+8+7+6),
\end{array}
$$
Thus whether $1234509876$ is divisible by $3$ or $9$ depends
exclusively on whether its digital sum $1+2+3+4+5+0+9+8+7+6=45$ is.
In turn $45$ is divisible by $3$ or $9$ if $4+5=9$ is divisible by
$3$ or $9$. The string of sums make us conclude that $1234509876$ is
divisible by $9$. In general, we may state the following result.
\begin{thm}
A natural number is divisible by $3$ (or $9$) if and only if its
digital sum is divisible by $3$ (or $9$).
\end{thm}
\section*{Homework}
\begin{multicols}{2}\columnseprule 1pt \columnsep 25pt\multicoltolerance=900\small
\begin{pro}
You put $54$ marbles into $6$ bags, ending up with the same number
of marbles in each bag. How many marbles would be in each bag
if there were $6$ bags?
\begin{answer}
$9$
\end{answer}
\end{pro}
\begin{pro}
Find the quotient and the remainder when $23$ is divided by $4$.
\begin{answer}
$23=4\cdot 5 + 3$ and so the quotient is $5$ and the remainder $3$.
\end{answer}
\end{pro}
\begin{pro}
Find the quotient and the remainder when $4$ is divided by $23$.
\begin{answer}
$4=0\cdot 23 + 4$ and so the quotient is $0$ and the remainder $4$.
\end{answer}
\end{pro}
\begin{pro}
If today is Thursday, what day will it be $100$ days from now?
\begin{answer}Since $100 = 7 \cdot 14 + 2$, $98$ days from today will be a
Thursday, and $100$ days from today will be a Saturday.
\end{answer}
\end{pro}
\begin{pro}
Bilbo and Frodo have just consumed a plateful of cherries. Each
repeats the rhyme `Tinker, tailor, soldier, sailor, \mbox{rich man,}
poor man, beggar man, thief' over and over again as he runs through
his own heap of cherry stones. Bilbo finishes on `sailor', whereas
Frodo finishes on `poor man'. What would they have finished on if
they had run through both heaps together?
\begin{answer}
This problem is from [Gard]. At `tailor.'
\end{answer}
\end{pro}
\begin{pro}
A runner ran $3000$ m in $8$ minutes. What is his average speed in
meters per second?
\begin{answer}
$6\tfrac{1}{4}$ meters per second.
\end{answer}
\end{pro}
\begin{pro}
Peter and Paul were selling magazines, of which Peter sold $60$ and
Paul $80$. Each magazine cost the same. If they received $\$700$
altogether for the magazines, how much money did Peter get?
\begin{answer}
$\$300$
\end{answer}
\end{pro}
\begin{pro}
Brian is drawing coloured camels in sequence. He draws a yellow
camel, a green camel, a blue camel, a red camel, a brown camel, an
orange camel, and again, a yellow camel, a green camel, a blue
camel, etc. Which colour is the $2008$-th camel?
\begin{answer}
Red.
\end{answer}
\end{pro}
\begin{pro}
What is the greatest number of Mondays that can occur in $45$
consecutive days?
\begin{answer}
If one puts Monday at the very beginning of the period one obtains
$\ceil{\dfrac{45}{7}} = 7$ Mondays.
\end{answer}
\end{pro}
\begin{pro}
A club has $86$ members, and there are $14$ more girls than boys in
the club. How many are there of each?
\begin{answer}
There are $36$ boys and $50$ girls.
\end{answer}
\end{pro}
\begin{pro}
When a number is doubled and the result increased by nine, one obtains fifteen. What was the original number?
\begin{answer}
$3$
\end{answer}
\end{pro}
\begin{pro}
When a number is doubled and the result diminished by nine, one obtains fifteen. What was the original number?
\begin{answer}
$12$
\end{answer}
\end{pro}
\begin{pro}
When a number is halved and the result increased by nine, one obtains fifteen. What was the original number?
\begin{answer}
$12$
\end{answer}
\end{pro}
\begin{pro}
When a number is halved and the result diminished by nine, one obtains fifteen. What was the original number?
\begin{answer}
$48$
\end{answer}
\end{pro}
\begin{pro}
There are $5,064$ marbles that need to be packed in boxes. There are
$6$ boxes. We want to put the same number of marbles in each box.
How many marbles will fit into each box?
\begin{answer}
$844 $
\end{answer}
\end{pro}
\begin{pro}
The four hundred thirty seven students of a school were called to assembly for a drug raid. If the police line up as many as possible
into twenty five equal rows, how many pupils are left out?
\begin{answer}
$12$
\end{answer}
\end{pro}
\begin{pro}
A book publisher sent $140$ copies of a book to a bookstore. The
publisher packed the books in two types of boxes. One type held $8$
copies of the book and the other held $12$. The boxes were all full
and there were an equal number of each type of box. How many boxes
of each type were sent to the bookstores?
\begin{answer}
$7$
\end{answer}
\end{pro}
\begin{pro}
With what possible values can the digit $a$ be replaced so that the
four-digit number $32a2$ be divisible by $9$?
\begin{answer}
The sum of the digits is $3+2+a+2=7+a$. We want $a$ to be one of
$\{0,1,2,3,4,5,6,7,8,9\}$ and also we want $7+a$ to be a multiple of
$9$. This happens only when $a=2$.
\end{answer}
\end{pro}
\begin{pro}
With what possible values can the digit $a$ be replaced so that the
five-digit number $32aa2$ be divisible by $9$?
\begin{answer}
The sum of the digits is $3+2+a+a+2=7+2a$. We want $a$ to be one of
$\{0,1,2,3,4,5,6,7,8,9\}$ and also we want $7+2a$ to be a multiple
of $9$. This happens only when $a=1$.
\end{answer}
\end{pro}
\begin{pro}
A book publisher must bind $4500$ books. One contractor can bind all
these books in $30$ days and another contractor in $45$ days. How
many days would be needed if both contractors are working
simultaneously?
\begin{answer}
The first contractor binds $4500\div 30 = 150$ books per day, and
the second, $4500 \div 45 = 100$ books per day. Therefore, if they
worked simultaneously, they could bind $150+100 = 250$ books per
day, and it would take them $4500 \div 250 = 18$ days to bind all
books.
\end{answer}
\end{pro}
\begin{pro}
For commencement exercises, the students of a school are arranged in
nine rows of twenty eight students per row. How many rows could be
made with thirty six students per row?
\begin{answer}
Seven.
\end{answer}
\end{pro}
\begin{pro}
For which values of the natural number $n$ is $36 \div n$ a natural
number?
\begin{answer}
If $36 \div n$ is a natural number, then $n$ must evenly divide
$36$, which means that $n$ is a divisor of $36$. Thus $n$ can be any
one of the values in $\{1,2,3,4,6,9,12,18,36\}$.
\end{answer}
\end{pro}
\begin{pro}
A two-digit natural number is divided by the sum of its digits. What
is the largest possible remainder?
\begin{answer}
$15$
\end{answer}
\end{pro}
\begin{pro}
As a publicity stunt, a camel merchant has decided to pose the
following problem: ``If one gathers all of my camels into groups of
$4$, $5$ or $6$, there will be no remainder. But if one gathers them
into groups of $7$ camels, there will be $1$ camel left in one
group.'' The number of camels is the smallest positive integer
satisfying these properties. How many camels are there?
\begin{answer}The least common multiple of $4, 5$ and $6$ is $60$, hence
we want the smallest positive multiple of $60$ leaving remainder $1$
upon division by $7$. This is easily seen to be $120$.
\end{answer}
\end{pro}
\begin{pro}
What is the least and the largest number of Friday the $13$th that
there can be in a given year?
\begin{answer}
Least: $1$. Largest: $3$.
\end{answer}
\end{pro}
\begin{pro}
A number of bacteria are placed in a glass. One second later each bacterium splits in two, the next second each of the now existing bacteria
splits in two, et cetera. After one minute the glass is full. When was the glass half full?
\begin{answer}
At $59$ seconds.
\end{answer}
\end{pro}
\begin{pro}
A lotus flower doubles its surface each day, taking $32$ days to cover the whole surface of a pond.
How long will it take two lotus flowers to cover the same
pond?
\begin{answer}
Thirty one days. Explanation: consider just one lotus flower. On the second day it covers as much as two and thirty one days remain for it to
cover the pond.
\end{answer}
\end{pro}
\begin{pro}
A boy and a girl collected $24$ nuts. The boy collected twice as
many nuts as the girl. How many did each collect?
\begin{answer}
We can represent the boy's and girl's amounts by boxes, each box
having an equal amount of nuts, and there being two boxes for the
boy and one for the girl:
$$ \psframebox[fillstyle=solid,fillcolor=red]{B} \qquad \psframebox[fillstyle=solid,fillcolor=red]{B} \qquad \psframebox[fillstyle=solid,fillcolor=green]{G} $$
Then clearly, each box must have $8$ nuts. The boy has $16$ and the
girl $8$. This problem was taken from [Toom].
\end{answer}
\end{pro}
\begin{pro}
Prove that the sum of two odd numbers is even.
\begin{answer}
If $2x+1$ and $2y+1$ are the odd numbers, their sum is
$$ 2x+1+2y+1=2x+2y+2 =2(x+y+1), $$twice an integer, and hence even.
\end{answer}
\end{pro}
\begin{pro}
Prove that the product of two odd numbers is odd.
\begin{answer}
If $2x+1$ and $2y+1$ are the odd numbers, their product is
$$ (2x+1)(2y+1)=(2x)(2y)+2x(1) + 1(2y)+1(1) = 2xy +2x+2y +1 =2(xy+x+y)+1, $$twice an integer plus one, and hence odd.
\end{answer}
\end{pro}
\begin{pro}
How many integers in the set $\{100,101,102, \ldots , 198, 199\}$ of
$100$ consecutive integers {\bf are not} the sum of four consecutive
integers?
\begin{answer}
If $n-1, n, n+1, n+2$ are four consecutive integers, then their sum
is $4n+2$, that is, the integers sought do not leave remainder $2$
upon division by $4$. Since three quarters of the integers leave
some other remainder, the answer is $\dfrac{3}{4}\cdot 100 = 75$.
\end{answer}
\end{pro}
\begin{pro} Prove that the square of any natural number either leaves
remainder $0$ when divided by $4$ or remainder $1$ when divided by
$4$. That is, prove that the square of any natural number is of the
form $4k$ or $4k + 1$.
\begin{answer} By the Division Algorithm, any integer comes in one of two
flavours: $2a$ or $2a + 1$. Squaring, $$ (2a)^2 = 4a^2,\,\,\, (2a +
1)^2 = 4(a^2 + a) + 1$$and so the assertion follows.
\end{answer}
\end{pro}
\begin{pro} Prove that no integer in the sequence $$ 11, 111, 1111, 11111, \ldots$$
is the square of an integer.
\begin{answer} The previous problem asserts that if a number leaves either
remainder $2$ or remainder $3$ when divided by $4$, then the number
cannot be a square.
Observe that
$$ 11=8+3, 111= 100 + 8 +3, 1111=1100 +8 +3, 11111=11100+8+3, \ldots$$
so every integer in the sequence leaves remainder $3$ upon division
by $4$, and by the preceding problem, they cannot be squares.
\end{answer}
\end{pro}
\begin{pro}
{\em One Price Shoes} sell all their shoes for the same price, which
is an integral number of dollars. Anacleta and Sinforosa go shoe
shopping to {\em One Price Shoes}. Anacleta has \$200 and she buys
as many pairs of shoes as possible, and there remain \$32. Sinforosa
has \$150 and she buys as many pairs of shoes as possible, and there
remain \$24. What is the price of a pair of shoes?
\begin{answer}
Let $A$ be the number of pairs of shoes Anacleta buys and let $S$ be
the number of pairs of shoes Sinforosa buys. Let $P$ be the price,
in dollars, of each pair of shoes. Observe that $P > 32.$ Then $200
= AP + 32, 150 = SP + 24,$ whence
$$AP = 168 = 2^3\cdot 3\cdot 7, SP = 126 = 2\cdot 3^2\cdot 7.$$ The
price of each pair of shoes is a common divisor of 168 and 126, but
it must exceed $32$. The only such divisor is $42$. Hence the shoes
cost $\$42$.
\end{answer}
\end{pro}
\begin{pro}
Find the greatest multiple of $8$ using all ten digits and with all
the digits different.
\begin{answer}
For a number to be divisible by $8$, the number formed by its last
three digits must the divisible by $8$. The largest number that can
be formed with all ten digits is $9876543210$, but this is not
divisible by $8$. A permutation of the last three digits to
$9876543120$ yields a number divisible by $8$, and the largest such.
\end{answer}
\end{pro}
\begin{pro}
Find the smallest strictly positive natural number $n$ such that
$\dis{\frac{n}{2}}$ is a (perfect) square, $\dis{\frac{n}{3}}$ is a
(perfect) cube, and $\dis{\frac{n}{5}}$ is a (perfect) fifth power.
\begin{answer}
The integer must be of the form $n = 2^a3^b5^c,$ with $a, b, c$
positive integers. The conditions imply that $a - 1$ is even and
that $a$ is divisible by 15; that $b - 1$ is divisible by 3 and that
$b$ is divisible by 10; and that $c - 1$ is divisible by 5 and $c$
divisible by 6. Clearly, the smallest positive integers satisfying
those conditions are $a = 15, b = 10, c = 6.$ Hence the integer
sought is $n = 2^{15}3^{10}5^6 = 30233088000000.$
\end{answer}
\end{pro}
\end{multicols}
\section*{References}
The excellent \cite{Aha} is a good guide for adults trying to teach
their children the basic of Arithmetic. I have borrowed some of his
ideas for these notes.
\begin{description}
\bibitem[Aha]{Aha} Ron AHARONI, ``Arithmetic for Parents''\end{description}
\chapter{Long Division}
Let us now discuss an algorithm for division. The way we display it
here is perhaps somewhat different from what you learned in school,
but the result will be the same. We will start with simple examples
in the hopes of elucidating the general procedure. Before we begin,
let us introduce some notation.
\begin{df}[Fraction Notation] Let $a$, $b$ be two natural numbers,
with $b\neq 0$. Then we define the {\em fraction} $\dfrac{a}{b}$ as
$$\dfrac{a}{b} = a\div b. $$Here $a$ is the {\em numerator} of the
fraction and $b$ the {\em denominator}.
\end{df}
There it is! A fraction is a division that we are too lazy to carry
out!
\bigskip
Now, recall that from the distributive law, we have the equality
$$ a(b+c) = ab+ac, $$
that is, we are able to ``multiply term by term.'' Suppose now that
instead of the factor $a$ we multiplied by $\dfrac{1}{a}$, with
$a\neq 0$. Then we would have
$$ \dfrac{1}{a}\left(b+c\right)=\dfrac{b+c}{a}=\dfrac{b}{a}+\dfrac{c}{a}, $$
which gives us a ``division term by term.''
\bigskip
For example, suppose we were considering the task of dividing $9639$
by $3$. Taking advantage of the decimal representation of the number
could proceed as follows:
$$9639\div 3 = \dfrac{9639}{3} =
\dfrac{9000+600+30+9}{3} =
\dfrac{9000}{3}+\dfrac{600}{3}+\dfrac{30}{3}+\dfrac{9}{3} =
3000+200+10+3=3213.
$$
Notice that we first isolated the thousands, divided them by $3$,
then the hundreds, divided them by $3$, etc. We could actually
display this ``vertically'' in the following manner.
$$ \opidiv[displayintermediary=all]{9639}{3} $$
In the above display, we first begin by the thousands. Divide $9$ by
$3$, obtain $3$, write this in the space for the quotient. Multiply
$3$ by $3$, get $9$ and write this below the $9$ of the dividend.
Subtract and obtain $0$. Bring down the next digit from the
dividend, which is $6$. Divide $6$ by $3$, obtain $2$, write this in
the space for the quotient. Multiply $2$ by $3$, get $6$ and write
this below the $6$ of the dividend. Subtract and obtain $0$. Bring
down the next digit from the dividend, which is $3$. Divide $3$ by
$3$, obtain $1$, write this in the space for the quotient. Multiply
$1$ by $3$, get $3$ and write this below the $3$ of the dividend.
Subtract and obtain $0$. Bring down the next digit from the
dividend, which is $9$. Divide $9$ by $3$, obtain $3$, write this in
the space for the quotient. Multiply $3$ by $3$, get $9$ and write
this below the $9$ of the dividend. Subtract and obtain $0$. As a
check: $$9639 = 3\cdot 3213, $$and so the quotient is $3213$ and
the remainder $0$.
\begin{rem}
You may prefer to display your division using a more Anglo Saxon
display: $$\longdiv{9639}{3} $$
\end{rem}
The general algorithm is thus:
\begin{enumerate}
\item Find the number of times the divisor is contained into as many
of the dividend figures (as read from left to right) an place this
number in the place for the quotient.
\item Multiply the divisor by the number found above and place the
product under the figures used in the part above from the dividend.
\item Subtract the product found above from the said figures of the
dividend. Bring down the next figure from the dividend to the right
of the difference just obtained.
\item Rinse and repeat with the remainder just increased by the next
figure in the step above.
\end{enumerate}
\begin{exa} Perform the division: $21416\div 2$
\end{exa}
\begin{solu}
We write
$$\dfrac{21416}{2} = \dfrac{20000}{2}+\dfrac{1000}{2}+ \dfrac{400}{2}+ \dfrac{10}{2}+\dfrac{6}{2} =
10000+500+200+5+3=10708. $$ Using the vertical algorithm for
division:
$$ \opidiv[displayintermediary=all]{21416}{2} $$
This can also be displayed as
$$\longdiv{21416}{2} $$
Verification: $21416=2\cdot 10708$, hence the quotient is $10708$
and the remainder $0$.
\end{solu}
\begin{exa}
Here we display $9634\div 3$:
$$ \opidiv[displayintermediary=all]{9634}{3} $$
This can also be displayed as
$$\longdiv{9634}{3} $$
Verification: $9634=3\cdot 3211+1$, hence the quotient is $3211$ and
the remainder $1$.
\end{exa}
\begin{exa}
Here we display $1326039\div 13$:
$$ \opidiv[displayintermediary=all]{1326039}{13} $$
This can also be displayed as
$$\longdiv{1326039}{13} $$
Verification: $1326039=13\cdot 102003$, hence the quotient is
$10708$ and the remainder $0$.
\end{exa}
\begin{exa}
Here we display $789\div 123$:
$$ \opidiv[displayintermediary=all]{789}{123} $$
This can also be displayed as
$$\longdiv{789}{123} $$
Verification: $789=123\cdot 6+51$, hence the quotient is $6$ and the
remainder $51$.
\end{exa}
\begin{exa}
Here we display $321123\div 123$:
$$ \opidiv[displayintermediary=all]{321123}{123} $$
This can also be displayed as
$$\longdiv{321123}{123} $$
Verification: $1326039=123\cdot 2610+93$, hence the quotient is
$2610$ and the remainder $93$.
\end{exa}
We now discuss algebraic division. Let us start with what is perhaps
the easiest case: powers of $10$. Observe that
$$ \dfrac{10^5}{10^2} = \dfrac{100000}{100}= \dfrac{1000\cancel{0}\cancel{0}}{1\cancel{0}\cancel{0}} = 1000 =10^3, $$
that is to perform the division, we cancelled as many zeroes as the
smaller power had. But notice what happened to the exponents. We in
fact had
$$\dfrac{10^5}{10^2} =10^{5-2}=10^3, $$
that is, to perform division of powers with the same base, we
subtracted the exponents. To further our evidence we again observe
that
$$ \dfrac{2^6}{2^2} = \dfrac{64}{4}=16=2^4 \implies \dfrac{2^6}{2^2} = 2^{6-2} =2^4. $$
Again,
$$\dfrac{a^5}{a^2} = \dfrac{aaaaa}{aa} = \dfrac{aaa\cancel{a}\cancel{a}}{\cancel{a}\cancel{a}} = a^3, $$
and, of course, $a^3=a^{5-2}$. This generalises as follows.
\begin{thm}[Second Law of Exponents]Let $a\neq 0$ be a real number and $m, n$ natural numbers, such that $m \geq n$. Then
$$\dfrac{a^m}{a^n} = a^{m-n}. $$
\end{thm}
\begin{pf}
We have $$\begin{array}{lll}\dfrac{a^m}{a^n} & = &
\dfrac{\underbrace{a\cdot a \cdots a}_{m\ a\mathrm{'s}}}{
\underbrace{a\cdot a \cdots a}_{n\
a\mathrm{'s}}} \\
& = & \underbrace{a\cdot a \cdots a}_{m-n\ a\mathrm{'s}}\\
& = & a^{m-n}.
\end{array}$$
\end{pf}
\begin{exa}
$$ \dfrac{2^9}{2^4} = 2^{9 - 4} = 2^5 = 32. $$
\end{exa}
\begin{exa}$\dfrac{15x^5y^7z^4}{5x^2y^2z^2} = 3x^3y^5z^2.$\end{exa}
\bigskip
In the case when the numerator possesses more than one term, we use
the distributive law. \begin{exa} $\dfrac{x^3+x^4}{x^2}=
\dfrac{x^3}{x^2}+\dfrac{x^4}{x^2}=x+x^2$.
\end{exa}
We will postpone the study when the denominator possesses more than
one term till an Algebra course.
\section*{Homework}
\begin{multicols}{2}\columnseprule 1pt \columnsep 25pt\multicoltolerance=900\small
\begin{pro}
Perform the division without a calculator: $100200300400\div 25$.
\begin{answer}
$\opidiv[displayintermediary=all]{100200300400}{25}$
\end{answer}
\end{pro}
\begin{pro}
Perform the division without a calculator: $123123\div 123$.
\begin{answer}
$\longdiv{123123}{123}$
\end{answer}
\end{pro}
\begin{pro}
Perform the division without a calculator: $123123\div 1001$.
\begin{answer}
$\longdiv{123123}{1001}$
\end{answer}
\end{pro}
\begin{pro}
Perform the division without a calculator: $1234567890\div 90$.
\begin{answer}
$\longdiv{1234567890}{90}$
\end{answer}
\end{pro}
\begin{pro}
Perform the division without a calculator: $10111213\div 9$.
\begin{answer}
$\longdiv{10111213}{9}$
\end{answer}
\end{pro}
\begin{pro}
Perform the division without a calculator: $10111213\div 14$.
\begin{answer}
$\longdiv{10111213}{14}$
\end{answer}
\end{pro}
\begin{pro}
Divide: $\dfrac{a^{12}}{a^6}$
\begin{answer} $a^6$
\end{answer}
\end{pro}
\begin{pro}
Divide: $\dfrac{a^{8}x^6}{a^6x}$
\begin{answer} $a^2x^5$
\end{answer}
\end{pro}
\begin{pro}
Divide term by term: $\dfrac{30x^6+9x^9}{3x^3}$
\begin{answer} $10x^3+3x^6$
\end{answer}
\end{pro}
\begin{pro}
Divide term by term: $\dfrac{x^2+x^3}{x^2}$
\begin{answer} $1+x$
\end{answer}
\end{pro}
\begin{pro}Find the exact numerical value of: \quad $\dfrac{20^6}{25^5}$.
\begin{answer} $\dfrac{4096}{625}$ \end{answer}
\end{pro}
\begin{pro}Calculate: $$\dfrac{100^3+100^3+100^3+100^3+100^3+100^3}{100^2+100^2+100^2}.$$
\begin{answer} $\dfrac{100^3+100^3+100^3+100^3+100^3+100^3}{100^2+100^2+100^2}=\dfrac{6\cdot 100^3}{3\cdot 100^2}=2\cdot 100=200$. \end{answer}
\end{pro}
\begin{pro}Find the exact numerical value of: $$\dfrac{25^2+25^2+25^2+25^2 }{25^4}.$$
\begin{answer} $\dfrac{4}{625}$ \end{answer}
\end{pro}
\begin{pro}
If
$$\left(\dfrac{4^5 + 4^5 + 4^5 + 4^5}{3^5 + 3^5 + 3^5}\right)\left(\dfrac{6^5 + 6^5 + 6^5 + 6^5 + 6^5 + 6^5}{2^5 + 2^5}\right) = 2^n,$$find $n$.
\begin{answer}
The product equals
$$\frac{4(4^5)}{3(3^5)}\cdot\frac{6(6^5)}{2(2^5)} = 4(4^5) =
2^{12},$$so $n=12$.
\end{answer}
\end{pro}
\begin{pro}
If
$$ 3^{2001} + 3^{2002} + 3^{2003} = a3^{2001},$$find $a$.
\begin{answer}
We have $$ 3^{2001} + 3^{2002} + 3^{2003} = 3^{2001}(1 +
3 + 3^2) = (13)3^{2001},$$whence $a = 13$.
\end{answer}
\end{pro}
\end{multicols}
\chapter{Roots and Geometric Progressions}
We introduce now the operation of extracting roots. Notice that we
will introduce this ``new'' operation by resorting to the
``reverse'' of an ``old'' operation. This is often the case in
Mathematics.\index{root}\index{root extraction}
\begin{df}[Roots] Let $m$ be a natural number greater than or equal
to $2$, and let $a$ and $b$ be any natural numbers. We write that
$\sqrt[m]{a}=b$ if $a=b^m$. In this case we say that $b$ is the {\em
$m$-th root} of $a$. The number $m$ is called the {\em index} of the
root.
\end{df}
\begin{rem}
In the special case when $m=2$, we do not write the index. Thus we
will write $\sqrt{a}$ rather than $\sqrt[2]{a}$. The number
$\sqrt{a}$ is called the {\em square root of $a$}. The number
$\sqrt[3]{a}$ is called the {\em cubic root of $a$}.
\end{rem}
\begin{exa}
We have\\
\centerline{\begin{tabular}{lll} $\sqrt{1}=1$ & because &
$1^2=1$,\\
$\sqrt{4}=2$ & because &
$2^2=4$,\\
$\sqrt{9}=3$ & because &
$3^2=9$,\\
$\sqrt{16}=4$ & because &
$4^2=16$,\\
$\sqrt{25}=5$ & because &
$5^2=25$,\\
$\sqrt{36}=6$ & because &
$6^2=36$.\\
\end{tabular}}
\end{exa}
\begin{exa}
We have\\
\centerline{\begin{tabular}{lll} $\sqrt[10]{1}=1$ & because &
$1^{10}=1$,\\
$\sqrt[5]{32}=2$ & because &
$2^5=32$,\\
$\sqrt[3]{27}=3$ & because &
$3^3=27$,\\
$\sqrt[3]{64}=4$ & because &
$4^3=64$,\\
$\sqrt[3]{125}=5$ & because &
$5^3=125$.\\
$\sqrt[10]{1024}=2$ & because &
$2^{10}=1024$.\\
\end{tabular}}
\end{exa}
\begin{df}
Let $a, r$ be non-zero natural numbers. A {\em geometric
progression} is a sequence of the form
$$ a, ar, ar^2, ar^3, \ldots, $$that is, a sequence where we
successively multiply by the same number. Here $a$ is the {\em first
term} of the progression, and $r$ is its {\em common ratio}.
\end{df}
\begin{exa}
What is the next sequence of the geometric progression
$$ 3, 6, 12, \ldots? $$
\end{exa}
\begin{solu}
Observe that each time we are multiplying by $2$, hence the next
term is $12\cdot 2 = 24$.
\end{solu}
\begin{exa}Find a formula for the $n$-th term of the geometric progression
$$ 6, 12, 24, 48, \ldots . $$ \end{exa}
\begin{solu}
We start with $6$, and then keep multiplying by $2$, thus
$$ 6= 6\cdot 2^0, \qquad 12=6\cdot 2^1, \qquad 24= 6\cdot 2^2, \qquad 48= 6\cdot 2^3, \ldots. $$
The general term is therefore of the form $3\cdot 2^{n-1}$, where
$n=1,2,3,\ldots$ is a natural number.
\end{solu}
\section*{Homework}
\begin{multicols}{2}\columnseprule 1pt \columnsep 25pt\multicoltolerance=900\small
\begin{pro}
Find $\sqrt{49}$. \begin{answer} $7$
\end{answer}
\end{pro}
\begin{pro}
Find $\sqrt{64}$. \begin{answer} $8$
\end{answer}
\end{pro}
\begin{pro}
Find $\sqrt[5]{243}$. \begin{answer} $3$
\end{answer}
\end{pro}
\begin{pro}
Find $\sqrt{81}+\sqrt{100}$. \begin{answer} $9+10=19$
\end{answer}
\end{pro}
\begin{pro}
Find $\sqrt[3]{343}$. \begin{answer} $7$
\end{answer}
\end{pro}
\begin{pro}Find
$$ \sqrt{1+2000\sqrt{1+2001\sqrt{1+2002\sqrt{1+2003\cdot 2005}}}}. $$
\begin{answer}
$2001$
\end{answer}
\end{pro}
\begin{pro}
What is the closest (integer) perfect square to $2003$?
\begin{answer} Observe that $\sqrt{2003}
> 44$. We see that $44^2 = 1936$ and $45^2 = 2025$. Thus the closest square is $2025$.
\end{answer}
\end{pro}
\begin{pro}
Find the next term of the geometric progression $$ 2, 6, 18, \ldots
.
$$
\begin{answer} $54$.
\end{answer}
\end{pro}
\begin{pro} The positive non-squares are ordered
in increasing order: $$ 2,\quad 3, \quad 5, \quad 6, \quad 7, \quad
8, \quad 10, \ldots .
$$Find the $100$-th term.
\begin{answer}
Since there are $10$ squares between $1$ and $110$ inclusive $110$
is the $100$-th term.
\end{answer}
\end{pro}
\begin{pro}
A crossnumber is like a crossword except that the answers are
numbers with one digit in each square. Complete the following
crossnumber. \vspace*{2cm}
\begin{minipage}{3cm}
\begin{Puzzle}{3}{3}%
|* |[1]{\empty}|[2]{\empty}|.
|[3]{\empty}|{\empty}|{\empty}|.
|{\empty}|* |{\empty}|.
\end{Puzzle}
\end{minipage}
\begin{minipage}{5.5cm}
\begin{PuzzleClues}{\textbf{ACROSS}}\\
\Clue{1}{}{\mbox{Prime Number}}\\ \Clue{3}{}{\mbox{Square of 3D.}}\\
\end{PuzzleClues}
\begin{PuzzleClues}{\textbf{DOWN}}\\
\Clue{1}{}{\mbox{Prime Number.}} \\ \Clue{2}{}{\mbox{Square of 1D.}}\\
\Clue{3}{}{\mbox{ Sqr. root of 3A.}} \\
\end{PuzzleClues}
\end{minipage}
\end{pro}
\begin{pro}
The product of three consecutive even integers is $87*****8$, where
the asterisks represent five missing digits. What is the sum of the
integers?
\begin{answer}Let the integers be $n-2, n, n+2$. Forcedly, they must end in $2, 4, 6$, since their product ends in
$8$. The product of the integers is $n^3-4n \approx n^3$. Now $n^3
\approx 87000008 \implies n \approx 443$. Inspection gives $n=444$,
and the sum is $442+444+446=1332$.
\end{answer}
\end{pro}
\end{multicols}
\part{Fractions}
\chapter{Fractions}
\QUOTEME{I continued to do arithmetic with my father, passing
proudly through fractions to decimals. I eventually arrived at the
point where so many cows ate so much grass, and tanks filled with
water in so many hours I found it quite enthralling.}{\mbox{Agatha
CHRISTIE}}
\bigskip
\dropping{2}{I}n this section we review some of the arithmetic
pertaining fractions. We have already defined fractions in the
lecture on long division, but here is the definition again.
\begin{df}
A {\em (positive numerical) fraction} is a number of the form $m\div
n=\dfrac{m}{n}$ where $m$ and $n$ are natural numbers and $n\neq 0$.
Here $m$ is the {\em numerator} or the fraction and $n$ is the {\em
denominator} of the
fraction.\index{fraction}\index{numerator}\index{denominator}
\end{df}
Given a natural number $n\neq 0$, we divide the interval between
consecutive natural numbers $k$ and $k+1$ into $n$ equal pieces.
Figures \ref{fig:den-2-frac} \ref{fig:den-3-frac}, and
\ref{fig:den-4-frac}, shew examples with $n=2$, $n=3$, and $n=4$,
respectively. Notice that the larger $n$ is, the finer the
partition. \vspace*{1cm}
\begin{figure}[h]
\psset{unit=6pc} \psaxes[yAxis=false,
subticks=2]{|->}(0,0)(0,0)(7.2,0) \multido{\nx=0+.5,
\n =0+1}{15}{\uput[d](\nx, 0){$\frac{\n}{2}$}}
\meinecaption{1}{Fractions with denominator $2$.}
\label{fig:den-2-frac}
\end{figure}
\vspace*{1cm}
\begin{figure}[h]
\psset{unit=6pc} \psaxes[yAxis=false,
subticks=3]{|->}(0,0)(0,0)(7.2,0)
\multido{\nx=0+.33333333,
\n =0+1}{21}{\uput[d](\nx, 0){$\frac{\n}{3}$}}
\meinecaption{1}{Fractions with denominator $3$.}
\label{fig:den-3-frac}
\end{figure}
\vspace*{1cm}
\begin{figure}[h]
\psset{unit=6pc} \psaxes[yAxis=false,
subticks=4]{|->}(0,0)(0,0)(7.2,0)
\multido{\nx=0+.25,
\n =0+1}{28}{\uput[d](\nx, 0){$\frac{\n}{4}$}}
\meinecaption{1}{Fractions with denominator $4$.}
\label{fig:den-4-frac}
\end{figure}
\begin{rem}You will notice that in many of the fractions written
above, the numerator is larger than the denominator. In primary
school vocabulary these fractions are commonly called {\em
improper}, and perhaps, you were taught to convert them to a {\em
mixed number}. For example, when we write $\dfrac{9}{2}$, your
primary school teacher might have preferred to write $4\frac{1}{2}$.
We will, however, prefer to use improper fractions as this is the
common usage in Algebra and in modern computer algebra
programmes.\index{mixed number}\index{fraction!improper}
\end{rem}
Note that some divisions of natural numbers are effectively natural
numbers. Any fraction whose denominator is a factor of the numerator
is a natural number. In particular, if the denominator is $1$, or if
the numerator is $0$, the fraction is indeed a natural number. Let
us see some examples. \begin{exa}Since the divisors of $6$ are $1$,
$2$, and $3$, all the fractions
$$\dfrac{6}{1}=6, \quad \dfrac{6}{2}=3, \quad \dfrac{6}{6}=1, $$are natural numbers.\end{exa}
\begin{exa} $\dfrac{6}{3}=2$ is a natural number.\end{exa} \begin{exa}$7=\dfrac{7}{1}$ is a
natural number.\end{exa} \begin{exa}$\dfrac{0}{3}=0$ is a natural
number.\end{exa}
\begin{exa}$\dfrac{4}{0}$ is undefined.\end{exa}
\par
We now try to understand the meaning of $\dfrac{m}{n}$ when it is
not a natural number. We first consider the following two cases when
$m$ and $n$ are non-zero natural numbers with: (i) $m n$.
\bigskip
Let us consider the first case.
\begin{exa}A picture representation of the concept of $\dfrac{3}{4}$ can
be obtained as follows. Call a square a {\em unit}, that is a $1$,
and divide it into four equal pieces. If we shade three of these
four equal squares, we have shaded $\dfrac{3}{4}$ of the area. There
is, of course, nothing special about squares. We may have used pies,
line segments, etc. It is clear that $0<\dfrac{3}{4}<1$. See figures
\ref{fig:3/4-square}, \ref{fig:3/4-pie}, and
\ref{fig:3/4-segment}.\end{exa} \vspace{1cm}
\begin{figure}[!h]
\begin{minipage}{5cm}
\centering \psset{unit=.8pc} \pspolygon(-4,-4)(-4,4)(4,4)(4,-4)
\pscustom[fillstyle=solid,fillcolor=blue]{\pspolygon(0,0)(4,0)(4,4)(0,4)}
\pscustom[fillstyle=solid,fillcolor=blue]{\pspolygon(0,0)(-4,0)(-4,-4)(0,-4)}
\pscustom[fillstyle=solid,fillcolor=blue]{\pspolygon(0,0)(-4,0)(-4,4)(0,4)}
\pscustom[fillstyle=solid,fillcolor=gray]{\pspolygon(0,0)(4,0)(4,-4)(0,-4)}
\meinecaption{2}{$\dfrac{3}{4}$ of a square.}\label{fig:3/4-square}
\end{minipage}
\hfill
\begin{minipage}{5cm}
\centering \psset{unit=1pc} \SpecialCoor \degrees[100]
%%the circle is divided into "100" degrees
%%one start the wedges in intervals: here [0;10][]
\pswedge[fillstyle=solid,fillcolor=blue]{2}{ 0 }{75}
\pswedge[fillstyle=solid,fillcolor=gray]{2}{75}{100}
\meinecaption{2}{$\dfrac{3}{4}$ of a pie.}\label{fig:3/4-pie}
\end{minipage}
\hfill
\begin{minipage}{5cm}
\centering \psset{unit=.8pc}
\psline[linecolor=blue,linewidth=2pt]{|-|}(-1,-1)(2,-1)\psline[linecolor=gray,linewidth=2pt]{|-|}(2,-1)(3,-1)
\meinecaption{2}{$\dfrac{3}{4}$ of a
segment.}\label{fig:3/4-segment}
\end{minipage}
\end{figure}
\clearpage
Let us consider now the second case.
\begin{exa}A picture representation of the concept of $\dfrac{5}{2}$ can
be obtained as follows. Call a square a {\em unit}, that is a $1$,
and divide it into two equal pieces. Since we only have two pieces,
it is clear that we need three squares. It follows that
$2<\dfrac{5}{2}<3$. If we shade five of these six equal pieces, we
have shaded $\dfrac{5}{2}$ of the area. See figures
\ref{fig:5/2-square}, \ref{fig:5/2-pie}, and
\ref{fig:5/2-segment}.\end{exa} \vspace{2cm}
\begin{figure}[!h]
\begin{minipage}{5cm}
\centering \psset{unit=.25pc}
\rput(-10,8){\pspolygon[fillstyle=solid,fillcolor=blue](-4,-4)(0,-4)(0,4)(-4,4)
\pspolygon[fillstyle=solid,fillcolor=blue](0,-4)(4,-4)(4,4)(0,4)}
\rput(0,0){\pspolygon[fillstyle=solid,fillcolor=blue](-4,-4)(0,-4)(0,4)(-4,4)
\pspolygon[fillstyle=solid,fillcolor=blue](0,-4)(4,-4)(4,4)(0,4)}
\rput(10,-8){\pspolygon[fillstyle=solid,fillcolor=blue](-4,-4)(0,-4)(0,4)(-4,4)
\pspolygon[fillstyle=solid,fillcolor=gray](0,-4)(4,-4)(4,4)(0,4)}
\meinecaption{2}{$\dfrac{5}{2}$ squares.}\label{fig:5/2-square}
\end{minipage}
\hfill
\begin{minipage}{5cm}
\centering \psset{unit=.4pc} \SpecialCoor \degrees[100]
%%the circle is divided into "100" degrees
%%one start the wedges in intervals: here [0;10][]
\rput(-10,8){\pswedge[fillstyle=solid,fillcolor=blue]{2}{ 0 }{50}
\pswedge[fillstyle=solid,fillcolor=blue]{2}{50}{100}}
\rput(0,0){\pswedge[fillstyle=solid,fillcolor=blue]{2}{ 0 }{50}
\pswedge[fillstyle=solid,fillcolor=blue]{2}{50}{100}}
\rput(10,-8){\pswedge[fillstyle=solid,fillcolor=blue]{2}{ 0 }{50}
\pswedge[fillstyle=solid,fillcolor=gray]{2}{50}{100}}
\meinecaption{2}{$\dfrac{5}{2}$ pies.}\label{fig:5/2-pie}
\end{minipage}
\hfill
\begin{minipage}{5cm}
\centering \psset{unit=1pc}
\rput(0,3){\psline[linecolor=blue,linewidth=2pt]{|-|}(-1,-1)(1,-1)\psline[linecolor=blue,linewidth=2pt]{|-|}(1,-1)(3,-1)}
\rput(0,0){\psline[linecolor=blue,linewidth=2pt]{|-|}(-1,-1)(1,-1)\psline[linecolor=blue,linewidth=2pt]{|-|}(1,-1)(3,-1)}
\rput(0,-3){\psline[linecolor=blue,linewidth=2pt]{|-|}(-1,-1)(1,-1)\psline[linecolor=gray,linewidth=2pt]{|-|}(1,-1)(3,-1)}
\meinecaption{2}{$\dfrac{5}{2}$ segments.}\label{fig:5/2-segment}
\end{minipage}
\end{figure}
\bigskip
By doing more of the line diagrams above you may have noticed that
there are multiple names for, say, the natural number $2$. For
example
$$ 2 = \dfrac{2}{1} = \dfrac{4}{2} = \dfrac{6}{3} = \dfrac{8}{4} = \dfrac{10}{5} = \dfrac{12}{6}, $$
etc. This observation is a particular case of the following result.
\begin{thm}[Cancellation Law]
Let $m,n, k$ be natural numbers with $n\neq 0$ and $k \neq 0$. Then
$$\dfrac{mk}{nk} = \dfrac{m}{n}. $$
\label{thm:cancellation-law} \index{cancellation law}
\end{thm}
\begin{pf} We will prove this for $m\leq n$. For $m>n$ the argument
is similar. Divide the interval $[0;1]$ into $nk$ pieces. Consider
the $k$-th, $2k$-th, $3k$-th, \ldots , $nk$-th markers. Since
$\dfrac{nk}{nk}=1$, the $nk$-th marker has to be $1$. Thus the $n$
markers $k$-th, $2k$-th, $3k$-th, \ldots , $nk$-th, form a division
of $[0;1]$ into $n$ equal spaces. It follows that the $k$-th marker
is $\dfrac{1}{n}$, that is, $\dfrac{k}{nk}=\dfrac{1}{n}$, the
$2k$-th marker is $\dfrac{2}{n}$, that is,
$\dfrac{2k}{nk}=\dfrac{2}{n}$, etc., and so the $mk$-th marker is
$\dfrac{m}{n}$, that is, $\dfrac{mk}{nk}=\dfrac{m}{n}$, as we wanted
to prove.\end{pf} Thus given a fraction, if the numerator and the
denominator have any common factors greater than $1$, that is, any
non-trivial factors, we may reduce the fraction and get an equal
fraction. \begin{df}Two fractions such that $\dfrac{a}{b} =
\dfrac{x}{y}$ are said to be {\em equivalent}. If $b**