Homework 11 Solution

Homework 11 Solution - MATH 74 HOMEWORK 11 1. Since gcd(a1...

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

View Full Document Right Arrow Icon
MATH 74 HOMEWORK 11 1. Since gcd( a 1 ,b 1 ) = 1, a theorem from class gives us integers M and N with Ma 1 + Nb 1 = 1 . Multiplying through by b 2 we obtain Ma 1 b 2 + Nb 1 b 2 = b 2 , and using the fact that a 1 b 2 = a 2 b 1 we get Ma 2 b 1 + Nb 1 b 2 = b 2 . so that b 1 ( Ma 2 + Nb 2 ) = b 2 . We conclude that b 1 | b 2 and thus b 1 b 2 by a theorem from class. Since gcd( a 2 ,b 2 ) = 1, a theorem from class gives us integers A and B with Aa 2 + Bb 2 = 1 . Multiplying through by b 1 we get Aa 2 b 1 + Bb 2 b 1 = b 1 , and using the fact that a 2 b 1 = a 1 b 2 we obtain Aa 1 b 2 + Bb 2 b 1 = b 1 , so that b 2 ( Aa 1 + Bb 2 ) = b 1 . We conclude that b 2 | b 1 and thus b 2 b 1 by a theorem from class. We conclude that b 1 = b 2 . Cancelling it from both sides of a 1 b 2 = a 2 b 1 we conclude that a 1 = a 2 . 2(a). The elements of D (216) are 1, 2, 4, 8, 3, 6, 12, 24, 9, 18, 36, 72, 27, 54, 108, 216, and each of these multiplied by - 1. That’s 32 divisors in all. 2(b).
Background image of page 1

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

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

Page1 / 2

Homework 11 Solution - MATH 74 HOMEWORK 11 1. Since gcd(a1...

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