Exam s Final Solutions

# Exam s Final Solutions - MATH 290 FINAL EXAM SOLUTIONS 1....

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

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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....
View Full Document

## This note was uploaded on 03/05/2011 for the course MATH 290 taught by Professor Alligood,k during the Fall '08 term at George Mason.

### Page1 / 3

Exam s Final Solutions - MATH 290 FINAL EXAM SOLUTIONS 1....

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

View Full Document
Ask a homework question - tutors are online