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Unformatted text preview: 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 then 3 must divide...
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This note was uploaded on 02/01/2011 for the course MATH 208 taught by Professor Brown during the Spring '10 term at Mt. SAC.
- Spring '10