# Help understanding this proof - sqrt(2) is irrational

Theorm: $\sqrt{2}$ is irrational

Proof (By contradiction): Assume, on the contrary that $\sqrt{2}$ is rational Then there are natural numbers $p$ and $q$, with no common factors, such that $\sqrt{2} = \frac{p}{q}$

Squaring: $2 = \frac{p^2}{q^2}$
Rearranging: $2q^2 = p^2$

So $p^2$ is even ( it is equal to twice something $\to 2 \times q^2$ ), Hence $p$ is even.

(Why?): Because the square of an even number is even, and the square of an odd number is odd. So the only way to get the square of a number to be even is if the number is even

So $p$ is $2r$ for some $r$
Substituting for $p$: $2q^2 = 2r^2 = 4r^2$
Cancelling $2$: $q^2 = 2r^2$

So $q^2$ is even ( it is equal to twice something $\to 2 \times r^2$ ), Hence $q$ is even (Reason: See Why?)

Deduction: $p$ is even, $q$ is even...

Conclusion: if both $p$ and $q$ are even, then they have $2$ as a common factor; but by the original assumption: ...Then there are natural numbers p and q, with no common factors, this shows that $p$ and $q$ cannot have a common factor. Since this conclusion does not agree with the assumption, this means that $\sqrt{2}$ is irrational. QED

I have a couple questions about the proof:

• If I have an even number $N$, then isn't it possible to factor this number until it is odd? Then wouldn't this mean that although $p$ and $q$ are even, they can be factored until no more factors are possible?

• Is it correct to assume that any rational number can be expressed as the quotient of $2$ other numbers who have no common factors?