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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=p

2q

Which tells me that p

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.

2q

2q

q

∴ q

√2 = p/q where p & q are in their lowest terms. Where q is non-zero.

2=p

^{2}/q^{2}2q

^{2}= p^{2}Which tells me that p

^{2}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 2q^{2}is even and as 2q=p^{2}we see that p^{2}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.

2q

^{2}=(2n)^{2}2q

^{2}=4n^{2}q

^{2}=2n^{2}where n^{2}is an integer, as n was an integer from the definition of an even number.∴ q

^{2}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.
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