I am trying to prove that √2 is irrational using proof by contradiction. Here is my work so far: √2 = p/q where p & q are in their lowest terms. Where q is non-zero. 2=p2/q2 2q2 = p2 Which tells me that p2 is an even number, using the definition of an even number. We can use this definition because we know that q is an integer and an integer squared is also an integer. Hence 2q2 is even and as 2q=p2 we see that p2 is also even. Can we use the knowledge that p is an integer to say that p is also an even number? Or is there another rule that I'm not aware of? If we can use this knowledge, then we can say that p=2n, where n is some integer, using the definition of an even number. 2q2=(2n)2 2q2=4n2 q2=2n2 where n2 is an integer, as n was an integer from the definition of an even number. ∴ q2 is also an even number. If both p and q are even numbers then the fraction p/q can be simplified further and p/q is not in its lowest terms. This shows that √2 is in fact not a rational number and proves that it is irrational.