321PSsolutions2000

# 321PSsolutions2000 - Solutions to Problem Session Questions Problem(3.3 Let f A-→ B be a function Prove that f is one-to-one if and only if f A 1

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Unformatted text preview: Solutions to Problem Session Questions Problem: (3.3) Let f : A-→ B be a function. Prove that f is one-to-one if and only if f ( A 1 ) ∩ f ( A 2 ) = f ( A 1 ∩ A 2 ) for any subsets A 1 , A 2 of A . We know from Ex. 2 that f ( A 1 ∩ A 2 ) ⊆ f ( A 1 ) ∩ f ( A 2 ). The claim amounts to showing that f is one-to-one if and only if f ( A 1 ) ∩ f ( A 2 ) ⊆ f ( A 1 ∩ A 2 ) for any subsets A 1 , A 2 of A . Assume that f is one-to-one. Let A 1 , A 2 be arbitrary subsets of A . If b ∈ f ( A 1 ) ∩ f ( A 2 ), then b = f ( a 1 ) for some a 1 ∈ A 1 and b = f ( a 2 ) for some a 2 ∈ A 2 . Since f ( a 1 ) = b = f ( a 2 ) and f is one-to-one, a 1 = a 2 , so a 1 ∈ A 1 ∩ A 2 , so b = f ( a 1 ) ∈ f ( A 1 ∩ A 2 ). This proves that f ( A 1 ) ∩ f ( A 2 ) ⊆ f ( A 1 ∩ A 2 ). Assume that f ( A 1 ) ∩ f ( A 2 ) ⊆ f ( A 1 ∩ A 2 ) for any subsets A 1 , A 2 of A . If a 1 6 = a 2 are two distinct elements of A , then { f ( a 1 ) } ∩ { f ( a 2 ) } = f ( { a 1 } ) ∩ f ( { a 2 } ) ⊆ f ( { a 1 } ∩ { a 2 } ) = f ( ∅ ) = ∅ , so f ( a 1 ) 6 = f ( a 2 ). This proves that f is one-to-one. Problem: (7.2) For which m ∈ N is Z m \ { ¯ } a group under multiplication? If m = 1, then Z m = Z 1 = { ¯ } , so Z m \ { ¯ } = ∅ and Z m \ { ¯ } cannot be a group. If m > 1 and m is composite, there are integers a , b satisfying 1 < a < m , 1 < b < m and ab = m . In Z m , these conditions may be written as ¯ a 6 = ¯ 0, ¯ b 6 = ¯ and ab = ¯ 0. Equivalently, ¯ a ∈ Z m \ { ¯ } , ¯ b ∈ Z m \ { ¯ } and ab / ∈ Z m \ { ¯ } . Thus Z m \ { ¯ } is not closed under multiplication and Z m \ { ¯ } cannot be a group. The only remaining possibility is that m is a prime number. Suppose that m is prime. Let’s check whether Z m \ { ¯ } is a group in this case. (i) For any integers a , b , Euclid’s lemma (Lemma 5.15) says that m | ab implies m | a or m | b (because m is prime). Equivalently, if m- a and m- b , then m- ab . Stated differently, in Z m \{ ¯ } , if ¯ a 6 = ¯ 0 and ¯ b 6 = ¯ 0, then ab 6 = ¯ 0. Hence Z m \{ ¯ } is closed under multiplication. (ii) Multiplication in Z m \ { ¯ } is associative by Lemma 6.4(7). (iii) There is ¯ 1 ∈ Z m \ { ¯ } and ¯ a · ¯ 1 = ¯ a for all ¯ a ∈ Z m \ { ¯ } , so ¯ 1 is a right identity in Z m \ { ¯ } . (iv) Let a ∈ Z with ¯ a 6 = ¯ 0. Then a is relatively prime to m and, by Lemma 6.4(9), there is an ¯ x ∈ Z m with ¯ a ¯ x = ¯ 1. So every ¯ a in Z m \{ ¯ } has a right inverse ¯ x in Z m . But in fact ¯ x ∈ Z m \{ ¯ } , for otherwise we would have ¯ x = ¯ 0 and ¯ 1 = ¯ a ¯ x = ¯ a ¯ 0 = ¯ by Lemma 6.4(12), yielding the contradiction m | 1. Thus every ¯ a in Z m \{ ¯ } has a right inverse in Z m \ { ¯ } ....
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## This note was uploaded on 11/08/2011 for the course MATH 321 taught by Professor Ergenc during the Spring '11 term at Boğaziçi University.

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321PSsolutions2000 - Solutions to Problem Session Questions Problem(3.3 Let f A-→ B be a function Prove that f is one-to-one if and only if f A 1

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