# Proof that the Square Root of 2 is Irrational.

1. Jan 9, 2014

### Calu

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: Jan 9, 2014
2. Jan 9, 2014

### Dick

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.

3. Jan 9, 2014

### Calu

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.

4. Jan 9, 2014

### Dick

Of course, it's important that p is an integer. Now I'm not sure what your question actually is.

5. Jan 9, 2014

### Calu

It was really just whether knowing p was an integer was important. Now I know it is, so thank you.