109_sp2011_ho_funcns_sol

109_sp2011_ho_funcns_sol - 109 Spring 2011 - Quantifiers...

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Unformatted text preview: 109 Spring 2011 - Quantifiers and Functions - Solutions Exercise (II.11). Give a proof or a counterexample for each of the following statements. (i) ∀x ∈ R∃y ∈ R(x + y > 0) True: given x ∈ R define y = 2 − x. (ii) ∀x ∈ R∃y ∈ R(x − y > 0) True: given x ∈ R define y = 2 + x. (iii) ∃x ∈ R∀y ∈ R(x + y > 0) False. We prove that the negation, ∀x ∈ R∃y ∈ R(x + y ≤ 0), is true. Given x, let y = −x. (iv) ∀x ∈ R∃y ∈ R(xy > 0) False. Counterexample: x = 0. (v) ∃x ∈ R∀y ∈ R(xy > 0) False. We prove the negation, ∀x ∈ R∃y ∈ R(xy ≤ 0). Given x, let y = 0. (vi) ∀x ∈ R∃y ∈ R(xy ≥ 0) True: given x ∈ R let y = 0. (vii) ∃x ∈ R∀y ∈ R(xy ≥ 0) True: choose x = 0. (viii) ∀x ∈ R∃y ∈ R(x + y > 0 or x + y = 0) True: given x ∈ R define y = 2 − x. (ix) ∀x ∈ R∃y ∈ R(x + y > 0 and x + y = 0) False: the inner predicate can never be made true. (x) ∀x ∈ R∃y ∈ R(x + y > 0) and ∀x ∈ R∃y ∈ R(x + y = 0) True: for first conjunct, given x define y = 2 − x; for the second conjunct, given x define y = −x. Exercise (II.14). Define functions f and g : R → R by f (x) = x2 and g (x) = x2 − 1. Find the functions f ◦ f , f ◦ g , g ◦ f , and g ◦ g . List the elements of the set {x ∈ R : f g (x) = gf (x)}. f ◦ f (x) = (x2 )2 = x4 (f ◦ g )(x) = f (g (x)) = f (x2 − 1) = (x2 − 1)2 = x4 − 2x2 + 1 (g ◦ f )(x) = g (f (x)) = g (x2 ) = (x2 )2 − 1 = x4 − 1 g ◦ g (x) = (x2 − 1)2 − 1 = (x4 − 2x2 + 1) − 1 = x4 − 2x2 A number is in {x ∈ R : f g (x) = gf (x)} if and only if x4 − 2x2 + 1 = x4 − 1. That is, if and only if Thus, this set is {−1, 1}. −2x2 = −2 i.e. x2 = 1. ...
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This note was uploaded on 08/02/2011 for the course MATH 109 taught by Professor Knutson during the Spring '06 term at UCSD.

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109_sp2011_ho_funcns_sol - 109 Spring 2011 - Quantifiers...

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