Here it is, for you to critique. This is a proof by contradiction. This is a good example of how I usually go about doing proofs, so if you give me tips on how to improve this particular proof, I'll be able to improve all my other proofs. I just learnt how to do proof by contradiction yesterday, so forgive any ugliness in my proof. 1. The problem statement, all variables and given/known data 1) Prove: r√2 is irrational, where r is rational and r ≠ 0 variables involved: r, rational m, integer n, integer a, integer b, integer a', integer b' integer x, integer y, integer 2. Relevant equations Knowledge of the proof that √2 is irrational was used. That is all. 3. The attempt at a solution 1) r√2 = (m/n)√2 = a/b (this is assuming that Q in the implication P→Q is negated, which is the basis of the proof by contradiction) 2) (m√2)/n = 2a'/2b' = a'/b' a'/b' = x/y, where at least one of x or y is odd (this is where the contradiction will lie later on) 3) (m√2)/n = x/y m√2 = xn/y √2 = xn/ym 2 = ((x^2)(n^2))/((y^2)(m^2)) 2y^2m^2 = x^2n^2 ..............after solving for x.............. x^2 is even therefore x is even 4) x = 2k 2y^2m^2/n^2 = 4k^2 ..............after solving for y............. y^2 is even so y is even So there is a contradiction; both x and y are even when at least one of x or y should be odd. Therefore, the statement that was required to be proved is true.