6. In this problem we'll prove by contradiction that 11 is irrational. The argument is very similar to our proof that 2 is irrational. First let's establish some definitions: Let I be a positive integer. We say that I is prime if it's divisible by exactly two integers: itself and 1. The smallest prime is 2; 1 technically doesn't fit the definition because it's divisible by just a single integer (itself). Let x and y be positive integers. We say that I is a factor or divisor of y if there exists some integer k such that y = kr. This is also equivalently stated as "I divides y" and written as 1|y. Here's a useful fact involving prime factors of a number: Let I and y be positive integers, and suppose I is prime. If I is a factor of y, then I is also a factor of y. For example, 7 is a factor of 7529536 (which is 27442). Because 7 is prime, we know that 7 must also be a factor of 2744. It's not very hard to prove this fact, but it requires some concepts that we haven't discussed just yet. For now, you may use this fact as a lemma in the rest of this problem without proving it yourself. (a) (4 points) To prove that 11 is irrational, we start by assuming for contradiction that is rational. Then there must exist some integers a and b, b + 0, such that (1) Furthermore, it must be possible to express a/b in simplest form such that ged(a, b) = 1. Using equation (1), prove that 11 must be a factor of a. (b) (4 points) Using equation and the previous results, prove that 11 must also be a factor of b. (c) (2 points) Briefly explain why the conclusion that 11 is a factor of both a and b is a contradiction.