811Solutions1

# 811Solutions1 - Homework 1 Solutions Kyle Chapman 9.1 True...

This preview shows pages 1–3. Sign up to view the full content.

Homework 1 Solutions Kyle Chapman October 14, 2011 9.1 True, True, False, True, False, False, True, False, True 9.2 a D and E b You can only deduce that it is not D. 9.3 a Not valid, would be D C then ¬ D ⇒ ¬ C . b Valid, this is contraposition. C D then ¬ D ⇒ ¬ C c Valid, M / W.S. M / E,M W.S. M I.S.,GBM / I.S., therefore GBM / E 9.4 a d e 9.5 a True 1

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

View Full Document
b False, consider n = - 1 c False, same as part b 11.6 a (1 + 2) 2 = 1 + 2 2 + 2 = 3 + 2 2. Since x is an increasing function and 9 / 4 > 2 we can observe that 3 / 2 > 2. This means that (1+ 2) 2 < 3+2(3 / 2) = 6. Since both sides are positive we can take the square root of both sides and rearrange to get the desired result. b Two methods. First we can use contraposition. It then suﬃces to show that n odd implies n 2 is odd, and we use that n odd means n = 2 k + 1 for some k . Then n 2 = 4 k 2 + 4 k + 1 = 2(2 k 2 + 2 k ) + 1 = 2 m + 1. This means
This is the end of the preview. Sign up to access the rest of the document.

## This note was uploaded on 12/26/2011 for the course MATH 8 taught by Professor Bigelow during the Fall '08 term at UCSB.

### Page1 / 3

811Solutions1 - Homework 1 Solutions Kyle Chapman 9.1 True...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online