This example is from Discrete Math and its Applications
I understand the steps the author is taking. First he assumes sqrt(2) is rational meaning that there exists integers a, and b such that sqrt(2) = a/b, b!= 0, and that a and b have no common factors other than 1(lowest term). He takes algebraic steps to show that if sqrt(2) = a/b, a will be even and b will be even, meaning that this is a contradiction of the lowest term idea. By proof by contradiction, therefore the author has shown that sqrt(2) is irrational, or rational.
I am trying to apply the the same proof to show that sqrt(20) is irrational. Here is my work so far
I took the same steps as the author but I arrived at 10b^2 = 2c^2 while the author arrived at b^2 = 2c^2. The author was able to conclude that b must be even because the square of an odd must be an odd and the square of an even. I am not so sure I can conclude the same thing. What I can say is that 10b^2 will be even but b could be odd or even because even(10) * ood and even(10) * even is both even. What step should I take instead to show that sqrt(20) is irrational with a proof by contradiction?