# The Square Root Of Two Is Not The Ratio Of Two Integers

The proof requires these facts about integers (which are the whole numbers, 0, 1, -1, 2, -2, 3, -3, etc.).

1. For integers a, if a2 is even, then a is even. The definition of even is that c is even if and only if there is some integer k such that c=2*k.
2. For integers a, if a 2 is odd, then a is odd. The definition of odd is that c is not even; necessarily, this means that there is some integer k such that c=2*k+1.
3. If a number x is the ratio of two integers, then it is the ratio of two relatively prime integers. That is, if x is a ratio of two integers, then there are some integers a and b such that b is not 0, x=a/b, and the greatest common divisor of a and b is 1 (1 and -1 both divide a and b, but no other integers do that).
Suppose that x is the square root of 2 and that x is also the ratio of two integers. Let a and be b relatively prime integers such that b is not 0 and x=a/b. Then
2=x2=(a2)/(b2).
So, 2 * (b2)= a2.
Therefore, a2 is even (because it is twice another integer). By (1) above, a is even.
Thus a=2*k for some integer k, and hence 2 * (b2)= (2k)2
Therefore, 2 * (b2)= 4*(k2).
Divide by two, to discover that b2= 2*(k2).
Therefore b2 is even (because it is twice another integer). By (1) above, b is even.
Therefore, b=2*j for some integer k.
Therefore, a and b are not relatively prime, because each has 2 as a divisor and 2 is bigger than 1 (their supposed greatest common divisor). This is a contradiction.
The contradiction means that something in our "Suppose that x is the square root of 2 and x is also the ratio of two integers" is false. So, if one part of this "and" sentence is true, the other part must be false. In particular, if x is indeed the square root of 2, then x is NOT the ratio of two integers. In mathematical terminology, x is irrational. "Ir" as a prefix means "not". So x is NOT rational where rational means being the ratio of two integers (with the denominator integer non-zero).

This style of proof can be modified to show that n(p/q) is irrational when p and q are relatively prime integers (1 and 2 in the paragraph above) and n is not the qth power of an integer.

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