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{\large {\bf   How to Do a Proof }}
\end{center}
 
   A proof is a rigorous explanation in math. In a proof, you first write a statement that you are going to prove. Then you go step by step to prove every part of it. For example, suppose we want to prove that $2*[(2+10)/4 + 8 - 4] = 14$.
   
    \noindent
 {\it Proposition 1: 2*[(2+10)/4 + 8 - 4] = 14.}\\
 The left-hand-side of the equation in the proposition can be rewritten as 2*[12/4 + 8 - 4] by adding the 2 and the 10. In turn, this can be rewritten as 2*[3 + 4]by by dividing 12 by 4. Next, we can rewrite it as 2*(7) by adding 3 to 4.  This equals 2*7, which in turn is equal to 14. Quod erat demonstrandum.
 
 ``Quod erat demonstrandum'' means ``Which was to be shown'' in Latin, and is put at the end of proofs, sometimes abbreviated to ``Q.E.D.''
 
 Proposition 1 is just a math problem. Usually proofs are for more general statements. Proposition 2 is an example. 
 
  \noindent
  {\it Proposition 2: $x^2 * x^4 = x^6$ for any  number $x$. }\\
Proof: $x^2 = x*x$.\\
$x^4 = x*x*x*x$\\
Thus, $x^2 * x^4 = (x*x) (x*x*x*x ) =  x*x*x*x*x*x$   by the associative property,  and  equals $ x*x*x*x*x*x=x^6$. Q.E.D. 

We can make Proposition 2  more general: 

 \noindent
  {\it Proposition 3: $x^a * x^b = x^{a+b}$ for any  number $x$ and non-negative whole numbers $a$ and $b$ . }\\
Proof: $x^a = 1*x*x...$ with $x$ repeated $a$ times on the right-hand-side (so $x^0=1$ and $x^1 = x$).  The same is true for $x^b$ except $ x$ is repeated $b$ times on the right-hand-side. \\
 Thus,  $x^a * x^b$ multiplies $1*x*x...$  with $x$ repeated $a$ times by $1*x*x...$  with $x$ repeated $b$ times.  That means, by the associative property, that $x^a * x^b = 1*x*x...$ with $x$ repeated $(a+b)$ times, which equals $x^{a+b}$. Q.E.D.
 
 We could make Proposition 3 even more general  by allowing $a$ and $b$ to be negative, but that would require a more complicated proof. Let's move on to different topic:what numbers can be divided evenly.   
 
 
 \newpage
 \noindent
{\it Proposition 4:  If and only if the last two digits of a number can be divided by four without remainder, so can the entire number.  }\\
Proof: We can write the number as $xyz$, where   $y$ and $z$ are the last two digits and $x$ is all the other digits. Thus, for $53,678, x =  536, y = 7$, and $z = 8.$\\ 
  We can break the number $xyz$ up into a hundreds part and a number from 0 to 99: 
$xyz = x*100 + yz.$\\
100 is evenly divisible by 4 because $100/4 = 25$.\\
Thus,  $x*100$ also evenly divisible--- $x*100/4$ equals $25x$, and $x$ is  a whole number. \\
 Thus,   $xyz$, which we found equalled $x*100 + yz$ is evenly divisible  by 4  {\bf if}   $yz$ is divisible by 4.  Also, this is true {\bf  only   if} $yz$ is divisible by 4, because if it is not divisible by 4 then $xyz/4 = 25x + yz/4$ and $yz/4$ is not a whole number. 
Q.E.D.

The last proof is harder, but it is the most surprising proposition here.

\noindent
{\it Proposition 5: If and only if the sum of the  digits of a four-digit number can be divided by three without remainder, so can the number itself.   }\\
Proof: Call the digits of the number $a,b,c,d$,  so the number is $abcd$ e.g., 3,523 has $a=3, b=5, c=2, d=3$. It will be true that\\
 $abcd = a*(1,000) + b*(100) + c*(10) + d $\\
 Thus, \\
 $abcd = a*(999+1) + b*(99+1) + c*(9+1) + d $\\
and \\
$ abcd = a*999 + a + b*99 + b + c*9 + c + d$\\
and \\
$abcd = (a*999+ b*99+ c*9) + (a+b+c+d)$\\
  We know that 999 and 99 and 9 are divisible even by 3, so everything in the first set of parentheses is divisible by 3. \\
   Thus, whether $abcd$ is divisible by 3 depends on whether $(a+b+c+d)$ is divisible by 3. 
Quod erat demonstrandum.

It is hard to know how many steps to write out in a proof. Proposition 1 had lots of steps, and it was easy enough that some of them could be combined. Proposition 5 is harder, so having lots of steps make it easier to understand. 

 

  

  

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