{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

230hw4sol

# 230hw4sol - Math 230 E Homework 4 Fall 2011 Drew Armstrong...

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

Math 230 E Fall 2011 Homework 4 Drew Armstrong Problem 1. (a) Consider a, b, d with gcd( a, d ) = 1. Prove: If d | ab then d | b . (b) Consider a, b, m with gcd( a, b ) = 1. Prove: If a | m and b | m then ab | m . (c) Consider a, b, c, n with gcd( c, n ) = 1. Prove: If ca cb mod n then a b mod n . Proof. First we prove (a). Consider a, b, d with a, d coprime, and suppose that d | ab . We will show that d | b . Since d | ab , there exists k such that dk = ab . Then since a, d are coprime, B´ ezout’s Lemma implies that there exist x, y such that ax + dy = 1. Finally, we multiply both sides of this equality by b to get b ( ax + dy ) = b · 1 , bax + bdy = b, dkx + dby = b, d ( kx + by ) = b. We conclude that d | b . To show (b), consider a, b, m with a, b coprime. Suppose further that a | m and b | m . We will show that ab | m . There are two possible solutions. Solution 1. (From Scratch) There exist k, ‘ such that ak = m and b‘ = m . Then since a, b are coprime, B´ ezout’s Lemma gives us x, y such that ax + by = 1. Multiply both sides by m to get m ( ax + by ) = m · 1 , max + mby = m, b‘ax + akby = m, ab ( ‘x + ky ) = m. Hence ab | m . Solution 2. (Quote Part (a)) There exist k, ‘ such that ak = m = b‘ , hence ak = b‘ . Since b | ak with a, b coprime, part (a) implies that b | k . That is, there exists x such that k = bx . Substitute this into ak = m to get abx = m . We conclude that ab | m . Finally, we will prove (c). Consider a, b, c, n with c, n coprime and ca cb mod n . We will show that a b mod n . Indeed, ca cb mod n means that n | ( ca - cb ), or n | c ( a - b ).

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.

{[ snackBarMessage ]}

### Page1 / 3

230hw4sol - Math 230 E Homework 4 Fall 2011 Drew Armstrong...

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

View Full Document
Ask a homework question - tutors are online