# hw9solsmc - a b where a and b have no common factors Then...

This preview shows page 1. Sign up to view the full content.

Solutions to Homework 9 - Math 2000 All solutions except 5.16 may be found in the book. We ﬁrst prove the following lemma: Lemma Let a Z . If 3 | a 2 then 3 | a . Proof . (by contrapositive) Instead we shall prove a Z , if 3 - a, then 3 - a 2 . Suppose 3 - a . Then a = 3 k + 1 for some k Z or a = 3 k + 2 for some k Z . Note that if a = 3 k + 1 then a 2 = (3 k + 1) 2 = 9 k 2 + 6 k + 1 = 3(3 k 2 + 2 k ) + 1 . Thus 3 - a 2 . If a = 3 k + 2 then a 2 = (3 k + 2) 2 = 9 k 2 + 12 k + 4 = 3(3 k 2 + 4 k + 1) + 1 and it follows that 3 - a 2 . (# 4.5.16) Prove that 3 is irrational. Solution . We give a proof by contradiction. Assume, to the contrary, that 3 =
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: a b where a and b have no common factors. Then we have 3 = a 2 b 2 or 3 b 2 = a 2 . Thus 3 | a 2 and by the above lemma 3 | a . Let a = 3 k for some k ∈ Z . Thus 3 b 2 = (3 k ) 2 = 9 k 2 and dividing by 3 we get b 2 = 3 k 2 . Note that 3 | b 2 and by the above lemma 3 | b . Thus b = 3 l for some l ∈ Z . In conclusion a = 3 k and b = 3 l and thus a and b have the common factor 3. This contradicts what we assumed at the beginning and completes the proof. 1...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online