text: Proof by contradiction\nSuppose 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.\n\nsqrt(7) = m/n.\n\nThus  7 = (m/n)^2 = m^2/n^2\n\n7n^2 = m^2\n\nSince 7 divides both sides, 7 has to divide m^2, so 7 must be of the form m = 7k for some integer k.\n\n7n^2 = (7k)^2 so we can factor a 7 out of both sides\n\nn^2 = 7 k^2. We repeat the arguement, now n has to be divisible by 7 so n = 7q for some integer q.\n\nso 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.\n\nThere for the assumptin that sqrt(7) is rational must be false.