Reductio ad absurdum 


A proof of the irrationality of the square root of two.  


Reductio ad absurdum.  
Here's a proof of the irrationality of the square root of two. It
uses a reductio ad absurdum argument, a "reduction to absurdity" 
we assume the truth of a statement, see the logical and mathematical
consequences of that assumption, then use it to show that it
contradicts itself, and thus cannot be true. Start with a simple square, with sides that are one unit long. We can find the length of the diagonal by using the pythagorean theorem: 

x^{2} = 1^{2} + 1^{2}  equation 1  
x^{2} = 2  equation 2  
x = 2^{1/2}.  equation 3  
Here comes our assumption: Assume that x = 2^{1/2} is rational, that
it can be represented by the ratio of two whole numbers. We'll call
these numbers p and q:


x = p/q.  equation 4  
Also assume that the ratio p/q has already been reduced to its lowest
form  that is, p and q have no factors in common. We begin by
squaring this equation:


x^{2} = p^{2}/q^{2}  equation 5  
p^{2} = x^{2} q^{2}  equation 6  
p^{2} = 2 q^{2}.  equation 7  
p^{2} must then be some number multipied by two  that is, p^{2} is even.
But the square of an even number is even,
[footnote 1]
and the square of an odd number is odd.
[footnote 2]
Therefore, p must also be even  that is, must also
be some number multiplied by 2 (at least once). Let's consider what
this "some number" is and call it s:


p = 2 s.  equation 8  
We can then put this into the above equation and find that


(2 s)^{2} = 2 q^{2}  equation 9  
4 s^{2} = 2 q^{2}  equation 10  
2 s^{2} = q^{2}.  equation 11  
According to this, q^{2} is also even  and by the same argument as
above, q therefore must also be even. If p and q are both even, then they both have at least one factor of two in common. If this is true, then they have not been reduced to their common factors, which is in contradiction to our assumption. Since it is certainly possible to reduce a ratio to its lowest form with no common factors, our original assumption is what must be wrong. Therefore, the square root of two cannot be rational; it is irrational, which cannot be expressed as the ratio of two integers, no matter how big those integers are. Quod erat demonstrandum. 

Footnotes.  
1. If n is even, then its prime factorization contains a power of 2 in it. Thus n^{2} must have a factor of 2^{2} in its prime factorization. Since it is thus divisible by 2, n^{2} must then be even. Thus the square of an even number is also even.

footnote 1  
2. If n is odd, then its prime factorization does not contain a power of 2 in it. So n^{2} does not contain a power of 2 in its prime factorization either, and thus is not divisible by 2 and so is odd. Therefore, the square of an odd number is also odd.

footnote 2  
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