MATH15Aquiz3sol

MATH15Aquiz3sol - Name (Last, First) SOLUTIONS Please...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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 ) ....
View Full Document

Page1 / 2

MATH15Aquiz3sol - Name (Last, First) SOLUTIONS Please...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online