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November 18, 2021\\
Cedars Math\\
 Mr. Rasmusen 
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{\large {\bf  Proof That the  Square Root of Two Is Irrational  }}
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\noindent
{\bf   Definition 1:}  A {\bf prime number} is a whole number greater than 1 that is evenly divisible only by itself and 1. {\it  Examples:}   2, 3, and 5, but not 1,4, and 6. 


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{\bf   Definition 2:}  A number's {\bf factors}  are the whole numbers by which it is evenly divisible. {\it  Example:}   The factors of 12 are 1, 12, 2, 6, 3, and 4. 


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{\bf   Definition 3:}  The {\bf prime factors}  of a number are the factors that are prime numbers. {\it  Example:}  The prime factors of 12 are 2 and 3. 

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{\bf   Definition 4:}  The {\bf prime factorization}  of a number is how it is formed by multiplying its prime factors.  {\it  Example:}  The prime factorization of 12  is $2\cdot 2 \cdot 3$, or equivalently,   $2^2 \cdot 3$.

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{\bf   Definition 5:}  A  whole number is  {\bf even}  if one of its prime factors is 2. If it is not even it is {\bf odd}.   {\it  Example:}   Twelve  is  even because its prime factors are 2 and 3. Fifteen is odd because its prime factors are 3 and 5. 

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{\bf  Definition 6:} A {\bf lemma}  is a claim that is true but is interesting mainly for proving some other claim. 


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{\bf  Definition 7:} A {\bf rational}  number is a number that can be written as a fraction. If a number is not rational, it is {\bf irrational.}  {\it  Example:}  3.5 is a rational number because it can be written as 7/2; $\pi$  is irrational because it cannot be written as a fraction $a/b$ for any whole numbers $a$ and $b$.


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{\bf Definition 8:} Two mathematical statements are {\bf equivalent} if they say the same thing.   {\it  Example:} The ratio 2:1 is equivalent to the ratio 2 to 1. 

 

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{\bf  Lemma 1a:}  If $X$ is even, so is  $x^2$. Only if $x$ is even  is  $x^2$ also even. 


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{\bf   Lemma 1b:}    If and only if   $x$ is even, so is $x^2$.\

Lemmas 1 and 2 are equivalent. 
  

\noindent
 {\it   Proof:} $ x^2 = x \cdot x.$ \\
 If $x$ has the prime factors $m, n, \ldots q$, then  $x \cdot x$ will have the same prime factors, just with double the exponents in the prime factorization. \\
  	(i)	 If  $x$ is even, 2 is one of  its prime factors. \\
	\hspace*{48pt}	  So $x^2$ will still have 2 as a prime factor, so it is even also. \\
	(ii)	If  $x$ isn't even, it's odd. \\
		\hspace*{48pt}	 If  $x$ is odd, it does NOT have 2 as a prime factor. \\
		\hspace*{48pt}	 Thus, $x^2$, with the same prime factors as  $x$, won't have 2 as a prime factor.  \\
	(iii)	 So if  $x$ is even, $x^2$ is also, but if  $x$ isn't even, $x^2$ isn't either. \\
Q.E.D. 

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{\bf  Theorem 1:}  The square root of 2 is irrational. 

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 {\it Proof.}   Suppose not. \\
   Then  there is some ratio $a/b$ such that	the fraction  $a/b = \sqrt{2}$, where  $a/b$ is the fraction reduced as far as possible. \\
   In that case,  $a^2/b^2 = 2 $, since that just squares both sides.\\
   Then, $a^2 = 2b^2,$ since that just multiplies both sides by $b^2$.\\
    So 		$a^2$ is even. \\
By Lemma 1, since $a^2$ is even, so is $a$, since only an even number can be squared to get an even number. \\ 
    That means we can write $a = 2k$, for some number $ k$.\\ 
    Then, $a^2 = (2k)^2 =  2\cdot k \cdot 2 \cdot k= 4k^2. $\\
    Since $a^2 = 2b^2$, that means $4k^2 = 2b^2.$\\
    Then dividing both sides by two gives us $  2k^2 = b^2. $\\
     So $b^2$ is  even.\\
 By Lemma 1, since $b^2$ is even, so is $b$, just like $a$ was. \\
     But then $a/b$  isn't the reduced fraction, since it has a factor of 2 on both top and bottom. \\
  So  it can't be that $\sqrt{2} = a/b$, reduced, for whole numbers $a$ and $b$.\\
 So  $\sqrt{2}$ is irrational.\\
 Q.E.D. 


 
 
  
  
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