psol02 - Math 55 Solutions to Homework 2 Chapter 1 Section...

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Math 55 Solutions to Homework 2 Chapter 1 Section 6 6. A. Yes B. No C. Yes D. No E. No F. No 12.Let A={1} and B={1,{1}} . (Answers will vary). 16.Yes. Suppose A B. Then there exists x A such that x B or vice versa. Then {x} P(A) but {x} P(B) (or vice versa). Hence P(A) P(B). By the contrapositive, if P(A) = P(B) then A=B. Section 7 22.A. No, consider A={1}, B={2}, C={1,2,3} B. No, consider A={1,3}, B={2,3}, C={3} 40.A. 0011100000 B. 1010010001 C. 0111001110 48.n+1 Section 8 10.A. Yes B. No, f(a)=f(b) C. No, f(a)=f(d) 12.A. Yes B. No, f(-1)=f(1) C. Yes D. No, f(1)=f(2) 28.f(g(x)) = (x+2) 2 +1 = x 2 +4x+5 g(f(x)) = (x 2 +1)+2 = x 2 +3 30.f(g(x)) = a (cx+d) + b = acx+ad+b g(f(x)) = c (ax+b)+d = acx+bc+d So f(g(x)) = g(f(x)) when acx+ad+b = acx+bc+d which simplifies to ad+b = bc+d 34.A. {-1,1} B. {x | -1<x<1, x 0}
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C. {x | x<-2 or x>2} 64.We wish to show f is injective f is surjective ( ) f is injective. So | F(A) | = | A |. (Ideally, this fact should be proved by mathematical induction, but that isn't covered until 3.3).
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This note was uploaded on 10/10/2009 for the course MATH 55 taught by Professor Strain during the Spring '08 term at Berkeley.

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psol02 - Math 55 Solutions to Homework 2 Chapter 1 Section...

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