Subsection 8.3.4 \(\sqrt{\mathbf{2}}\) is Irrational
Let’s now look at examples of the use of proof by contradiction (also called indirect proof) in mathematics.
Recall that a number is rational just in case it can be described as the ratio of two integers. In other words, it has the form a⁄b for some integers a and nonzero b.
We can use a proof by contradiction to show that √2 is not rational. Suppose, to the contrary, that it were. Then there would exist integers a and nonzero b, such that:
[1] √2 = a⁄b
If a and b share any common factors, divide those factors out until a and b are relatively prime (i.e., they share no common factors). Another way to say this is that we reduce a⁄b as much as possible. Squaring both sides of [1], we get:
[2] 2 = a^2⁄b^2 Multiplying by b^2, we get:
[3] \(2b^2 = a^2 \)
So a^2 is even and it can only be so if a itself is even. (We’ll prove that a^2 even implies a even in a couple of pages. Take it as a theorem for now.) Thus, there exists a k such that:
[4] a = 2k.
Substituting 2k for a in [3], we get:
[5] 2b^2 = (2k)2 = 4k2 Dividing by 2, we get:
[6] b^2 = 2k2
So b^2 is even and thus b is even. So a and b are both even and thus both have 2 as a factor. But this cannot be so since we’d reduced a⁄b until there were no common factors. Contradiction.
Exercises Exercises
1.
Try to cut and paste the proof we just did of the irrationality of \(\sqrt{2}\ \) to show that \(\sqrt{4}\ \) is also irrational. What happens?
Everything is fine through [2] but [3] is different and we can’t argue that a is even.
Everything is fine through [5] but [6] is different and we can’t argue that b is even.
It works.
Explanation: First, note that the proof cannot work: √4 is 2, which is rational. The problem is that line [5] becomes 4b2 = 4k2. Now we divide by 4 (instead of 2) and we get:
[6] b2 = k2.
But, from this, we can’t conclude anything useful about the factors of b. And so we don’t derive the contradiction that we got in the original proof.