MATH15Aquiz3sol

# MATH15Aquiz3sol - Name(Last First SOLUTIONS Please circle...

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Unformatted text preview: Name (Last, First) SOLUTIONS Please circle your section: A01 A02 A03 A04 5 PM 6 PM 7 PM 8 PM Math 15A Quiz 3 Briggs Fall Quarter 2004 No books, notes, calculators, or any unauthorized assistance is permitted on this quiz. Read each question carefully, answer each question completely, and show all of your work. Write your solutions clearly and legibly; no credit will be given for illegible solutions. If a question is not clear, ask for clarification. GOOD LUCK! 1. (3.6 #24; 12 points ) Prove the following statement by CONTRADICTION: “For all integers a , b , and c , if a 6 | bc then a 6 | b .” (Recall that the symbol 6 | means ”does not divide”.) Proof: (by contradiction) Suppose that a , b , and c are integers such that a 6 | bc and a | b . By definition of “divides”, there exists an integer k for which b = ak. Multiplying both sides of this equation by c gives bc = akc = a ( kc ) ....
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## This note was uploaded on 04/14/2008 for the course MATH 15A taught by Professor Briggs during the Fall '04 term at UCSD.

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MATH15Aquiz3sol - Name(Last First SOLUTIONS Please circle...

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