{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

assignment solutions

assignment solutions - MATH 135 Assignment 1 This...

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

View Full Document Right Arrow Icon
MATH 135 Assignment 1 This assignment is due at 8:30am on Wednesday January 13, in the drop boxes opposite the Tutorial Centre, MC 4067. 1. Let a, b, c, d be integers. Suppose that c | a and d | b . Prove that cd | ab . Solution: Since c | a there exists an integer q such that a = cq . Since d | b there exists an integer r such that b = dr . Therefore ab = cd ( rq ), so cd | ab by definition (because qr is an integer). 2. Let a and b be integers. Prove that gcd ( a, b ) | gcd ( a + b, a - b ) . Solution: Let d = gcd ( a + b, a - b ). Then we know there exist integers x and y such that ( a + b ) x + ( a - b ) y = d. Therefore a ( x + y ) + b ( x - y ) = d. By definition gcd ( a, b ) | a and gcd ( a, b ) | b . Therefore gcd ( a, b ) | d , as required. 3. Let a, b, c be integers. Suppose that gcd ( a, b ) = 1 and c | ( a + b ). Prove that gcd ( a, c ) = 1. Solution: Let d = gcd ( a, c ). Then by definition d | a and d | c , and d 0. Since
Background image of page 1

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

View Full Document Right Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

Page1 / 3

assignment solutions - MATH 135 Assignment 1 This...

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

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