Proof that the Square Root of 2 is Irrational.

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.

Last edited:

Dick
Homework Helper
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.

All you need is that the square of an even number is even and the square of an odd number is odd. You can prove that. So if p^2 is even, p must be even since it can't be odd.

All you need is that the square of an even number is even and the square of an odd number is odd. You can prove that. So if p^2 is even, p must be even since it can't be odd.

If I take p2 = 6, p = √6 which is why I thought the definition of p as an integer was important, as √6 isn't even.

Dick