1.2.7 - Solutions to Week 1 Homework problems from Abbott...

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Solutions to Week 1 Homework problems from Abbott Problems (section 1.2) 1.2.1, 1.2.4, 1.2.7, 1.2.8, 1.2.11 1.2.1 (a) Prove that 3 is irrational. Does a similar argument work to show 6 is irrational? (b) Where does the proof of Theorem 1.1.1 break down if we try to use it to prove 4 is irrational? (a) Proof: Suppose 3 is rational. Then 3 = p q where p and q are positive integers with no common factor. It follows that 3 = p 2 q 2 and hence 3 q 2 = p 2 . Since 3 divides p 2 , 3 must divide p itself, and hence in fact 9 divides p 2 . But in that case, 9 must divided the LHS (left hand side) of the equation 3 q 2 = p 2 also, which means that 3 divides q 2 . This tells us that 3 divides q , and we have arrived at a contradiciton: 3 divides both q and p , even though we assumed that q and p have no common factors. Thus, 3 must be irrational. Essentially the same argument will work for 6. (b) If we start the same procedure with 4, we find that the argument breaks down at the point where we concluded (above) that if 3 divides p 2
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