MAT137Y1 Lecture : MAT 137Y 2007-08Winter Session, Self Generated Solutions to Problem Set 3.pdf
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Mat 137y 2007-08 winter session, solutions to problem set 3. 2, then ab : if a = b = which is rational. If a = then ab = 2, which is rational. 3 = a/b, where a and b are integers and a and b have no common. Squaring both sides gives us 3 = a2/b2 or a2 = 3b2. 3, then a2 is also divisible by 3. It also follows that a must be divisible by 3: suppose not, then a = 3k + 1 or a = 3k + 2 for some integer k, but. Thus a = 3m for some integer m, so (3m)2 = 3b2, which implies b2 = 3m2. This implies that b2 is divisible by 3, which thereby implies that b is divisible by 3. But since a and b are both divisible by 3, this contradicts the assumption that a and b have no common factors, hence.
