3613-s02

# 3613-s02 - 1 . | c| < r 0 d n a r + qc- = a yltnelaviuqe...

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Solutions to Homework Set 2 (Homework Problems from Chapter 1) Problems from Section 1.1. 1.1.1 Let n be an integer. Prove that a and c leave the same remainder when divided by n if and only if a - c = nk for some k Z . Proof. Suppose a - c = nk . By the division algorithm, there exist unique integers q 1 ,r 1 ,q 2 2 such that a = nq 1 + r 1 ;0 r 1 <n c = nq 2 + r 2 r 2 <n . But then we have a - c = nk +0 a - c = n ( q 1 - q 2 )+( r 1 - r 2 ) Thus r 1 - r 2 = n ( q 1 - q 2 + k ) and so r 1 - r 2 is divisible by n . However, the conditions 0 r 1 2 imply 0 ≤| r 1 - r 2 | But the only non-negative integer divisible by n and less than n is 0. Hence, r 1 - r 2 =0,or r 1 = r 2 . Assume a = nq 1 + r and c = nq 2 + r . Then a - c = n ( q 1 - q 2 ). So a - c is divisible by n . 1.1.2 Let a and b be integers with c = 0. Then there exist unique integers q and r such that ( i ) a = cq + r ( ii )0 r< | c | . Proof. If c is positive, then c = | c | and this is just the statement of the Division Algorithm (Theorem 1.1) so there is nothing more to prove. If c is negative, then - c = | c | is positive and we can apply the Division Algorithm: there exist unique integers q and r such that a = | c | q + r and 0 r < | c | , or, equivalently a = - cq + r and 0 r < | c | . 1

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2 We have thus shown that there exist integers q and r satisfying (i). We must now show this choice of q and r is unique. Suppose we have q,r,q ,r Z such that a = cq + r ;0 r< | c | a = cq + r r < | c | Thenwehave 0= c ( q - q )+ r - r or r - r = c ( q - q ) . ( ) So r - r is divisible by c . But also | r - r | < | c | . Hence, since 0 is the only non-negative number less than | c | that is divisible by c , we must have r - r = 0. But this with ( ) then implies q - q = 0. Hence, q = q and r = r .So q and r are unique. 1.1.3 Prove that the square of any integer a is either of the form 3 k or of the form 3 k + 1 for some integer k . Proof. By the Division Algorithm, any integer a is representable as a =3 q + r with r an integer such that 0 3. That means r ∈{ 0 , 1 , 2 } a has one of three possible forms a q +0 (1) a q +1 (2) a q +2 (3) In the first case, a 2 =9 q 2 = 3(3 q 2 ) is obviously of the form 3 k ,w ith k q 2 . In the second case, a 2 =( 3 q +1) 2 q 2 +6 q ( 3 q 2 q )+1 and so a 2 is of the form 3 k + 1, with k q 2 q . In the last case,
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## This note was uploaded on 11/03/2010 for the course MATH 262-338-20 taught by Professor Gieseker,d. during the Fall '10 term at UCLA.

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3613-s02 - 1 . | c| < r 0 d n a r + qc- = a yltnelaviuqe...

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