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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 onetoone 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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 Fall '08
 PARRY
 Math, Algebra

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