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 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 / 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