ahw1 - Math 3124 Thursday, September 1 First Homework...

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Unformatted text preview: Math 3124 Thursday, September 1 First Homework Solutions 1. We have f : R≥0 → R≥0 defined by f (x) = (x − 1)2 . √ √ (a) f is onto. Indeed given a ∈ R≥0 , then a is well defined, 1 + a ∈ R≥0 , and √ f (1 + a) = a. (b) f is not one-to-one because, for example, f (0) = f (2) = 1. (c) √ claim that f (R≥2 ) = R≥1 . If x ≥ 2, then f (x) ≥ 1. Conversely if a ≥ 1, then We √ √ a is well defined, 1 + a ≥ 2, and f (1 + a) = a. 2. Exercise 2.23 on page 19. Prove that if α : S → T , β : T → U , γ : T → U , α is onto, and β ◦ α = γ ◦ α , then β = γ . We are given β ◦ α = γ ◦ α and that α is onto. We need to prove that β = γ , equivalently β (t ) = γ (t ) for all t ∈ T . But if t ∈ T , we may write t = α (s) for some s ∈ S because α is onto (of course, s depends on t ). We now have β (t ) = β ◦ α (s) = γ ◦ α (s) = γ (t ) which completes the proof. 3. Exercise 3.4 on page 23. Does the following define an operation on the set of integers? If yes, is it associative, is it commutative, and does it have an identity element? We have ∗ defined on Z according to the formula m ∗ n = mn2 . (a) ∗ is an operation because if m, n ∈ Z, then so is mn2 . (b) ∗ is not associative. We have 1 ∗ (1 ∗ 2) = 1 ∗ 4 = 16, but (1 ∗ 1) ∗ 2 = 1 ∗ 2 = 4. (c) ∗ is not commutative. We have 1 ∗ 2 = 4, but 2 ∗ 1 = 2. (d) ∗ does not have an identity. Suppose x is an identity for ∗. Then x ∗ 2 = 2 (by definition of identity), yet x ∗ 2 = 4x (using the formula for ∗), and we have a contradiction. ...
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