I am self-studying elementary analysis and am learning how to prove things. I have come up with a proof that √3 is irrational, and I believe it is valid, but I am unsure of my logic, as I have not seen it proved in just this way, and I don't have a prof to ask! So if anyone could just take a look at my argument and tell me if it is valid, I would appreciate it very much. Here goes. Theorem. √3 is irrational, i.e., there is no pair of integers p,q (q ≠ 0) satisfying (p/q)2 = 3. Proof (by contradiction). Assume, for contradiction, that there are two integers p,q (q≠ 0) such that (p/q)2 = 3 Then p/q is odd (it is easily proved by contradiction that if n2 is odd, then n must be odd for all integers n), i.e., there is an integer n such that p/q = 2n + 1 and therefore p = q(2n + 1). This can be substituted into the first equation, giving (2n + 1)2 = 3 which expands to 2n(2n + 1) + 2n + 1 = 3 → 4n2 + 4n = 2 → 2(n2 + n) = 1 Since n2 + n is an integer (closure under addition and multiplication), this last statement implies that 1 is an even integer, which it is not. Therefore there is no pair of integers p,q (q ≠ 0) satisfying (p/q)2 = 3, i.e., √3 is irrational, which is what was required to be shown. That's the proof I came up with. My specific question is this: Most proofs of this theorem are similar to the proof of the irrationality of √2, i.e. they start by assuming p and q have no common factor and end by deriving a contradiction of this assumption. My proof does not derive a contradiction of a hypothesis but instead derives a contradiction of a known fact, i.e. that 1 is an odd integer. Is this a 'legal' move, or not? Thanks for taking time to help.