solutions 3

# Discrete Mathematics and Its Applications with MathZone

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ICS 6D ( Discrete Mathematic): Solution to Selected Problems on Homework #3 Section 3.4 2. 1 | a because there exists integer n , namely n = a , s.t. a = n 1. a | 0 because there exists integer n , namely n = 0, s.t. 0 = n a . 6. By assumptions that a | c and b | d , there exist integers n 1 , n 2 s.t. c = n 1 a and d = n 2 b . Therefore cd = ( n 1 n 2 ) ab , i.e. there exists integer n , namely n = n 1 n 2 , s.t. cd = n ( ab ), and therefore ab | cd . 8. This is not true. For example 8 | (4 2) but it’s not true that either 8 | 4 or 8 | 2. 12. Note that we showed the opposite direction in class! This direction appears a bit more complicated. Let’s show this by arguing the counterpositive, i.e. that if ( a mod n ) negationslash = ( b mod n ) then a negationslash≡ b (mod n ). By the division algorithm theorem there exist unique remainders r 1 and r 2 in Z N s.t. n | ( a r 1 ) and n | ( b r 2 ). (The theorem also says that the quotients are unique, but we don’t need it here.) Let q 1 , q 2 be integers s.t. nq 1 = a r 1 and nq 2 = b r 2 . Therefore n ( q 1 q 2 ) = ( a b ) + ( r 2 r 1 ). Therefore we have
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