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Unformatted text preview: MATH 290 FINAL EXAM SOLUTIONS 1. (12 pts.) Prove that for all integers a , and b , and all nonzero integers c , ac | bc if and only if a | b . Which of the implications remains true if c = 0? Solution: Let a and b be integers and let c be a nonzero integer. (= ) Suppose that ac | bc . This means that there is an integer m such that mac = bc . Since c 6 = 0, ma = b and hence a | b . ( =) Suppose that a | b . This means that there is an integer m such that ma = b . Multiplying both sides by c gives mac = bc . Hence ac | bc . If c = 0 then the statement ac | bc is always true no matter what a and b are. Therefore the implication a | b = ac | bc remains true. The other implication is false. 2. (12 pts.) Prove that for all real numbers x and y , if x is rational and y is irrational then x + y is irrational. Solution: We will prove this by contrapositive. Assume that x + y is rational. We will show that it is not true that x is rational and y is irrational. In other words, we will show that x is irrational or y is rational. If x is irrational we are done, so assume that x is rational. We will show that y is rational. Since x + y and x are rational, so is y = ( x + y )- x . Hence y is rational....
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