230hw3sol

# 230hw3sol - Math 230 E Homework 3 Solutions Fall 2011 Drew...

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Math 230 E Fall 2011 Homework 3 Solutions Drew Armstrong Problem 1. To prove the following properties of Z you may assume axioms (A1) through (D) from the handout. (a) Prove that 0 a = 0 for all a Z . (b) Recall that - a is the unique integer such that a + ( - a ) = 0. Prove that for all a,b Z we have ( - a )( - b ) = ab . (Hint: Show that ( - a )( - b ) + a ( - b ) = ab + a ( - b ). You will need the result from part (a).) First I’ll isolate a useful lemma. Cancellation Lemma: Given a,b,c Z with a + c = b + c we have a = b . Proof. Suppose that a + c = b + c . By (A4) there exists some d Z such that c + d = 0. Then we have a + c = b + c ( a + c ) + d = ( b + c ) + d a + ( c + d ) = b + ( c + d ) (A2) a + 0 = b + 0 (A4) a = b (A3) ± Now I’ll prove (a) and (b). Proof. First we apply axioms (A3) and (D) to get 0+0 a = 0 a = (0+0) a = 0 a +0 a . Then we apply the Cancellation Lemma to get 0 = 0 a , proving (a). Next we apply axioms (A1), (A3), (A4), (M1), (D) and part (a) to get ( - a )( - b )+ a ( - b ) = (( - a )+ a )( - b ) = 0( - b ) = 0. Similarly, we have ab + a ( -

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## This note was uploaded on 01/08/2012 for the course MATH 461 taught by Professor Armstrong during the Fall '11 term at University of Miami.

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230hw3sol - Math 230 E Homework 3 Solutions Fall 2011 Drew...

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