text: proofs in advanced calc (possibly discrete math)? ok, i need to prove that sqrt(7) is an irrational number, and plug it into a calculator isnt a good answer, it needs to be somethign like the proof to sqrt(2) is irrational...but i am stuck.  basically it needs to be a contridiction.  if anyone can help, point me in a direction, provide a website, anything at all it would be great.  thanks!! proof by contradiction suppose sqrt(7) is rational. then it can be expressed and a fraction in reduced form, that is the numerator and denominator have no common factors.  sqrt(7) = m/n.  thus  7 = (m/n)^2 = m^2/n^2  7n^2 = m^2  since 7 divides both sides, 7 has to divide m^2, so 7 must be of the form m = 7k for some integer k.  7n^2 = (7k)^2 so we can factor a 7 out of both sides  n^2 = 7 k^2. we repeat the arguement, now n has to be divisible by 7 so n = 7q for some integer q.  so n = 7q, m = 7k, that says m and k both have 7 as a common factor a contradiction since m/n was a reduced fraction.  there for the assumptin that sqrt(7) is rational must be false.