klarfeld-sec-4-1-4-2

# klarfeld-sec-4-1-4-2 - Jennifer Klarfeld Week 10 Edition 8...

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Jennifer Klarfeld – Week 10 – Edition 8 Section 4.1 6. (b) Let A be the set {1,2,3}, and let R be the relation on A given by { (x,y) : 3x+y is prime}. Prove that R is a function with domain A. If x = 1, then 3x+y = 3+y. For y = 1, 2, or 3, 3+y is prime if and only if y = 2. If x = 2, then 3x+y = 6+y. For y = 1, 2, or 3, 6+y is prime if and only if y = 1. If x = 3, then 3x+y = 9+y. For y = 1, 2, or 3, 9+y is prime if and only if y=2. For every x A, there is a unique value y A such that 3x+y is prime. So, R is a function with domain A. 1

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Jennifer Klarfeld – Week 10 – Edition 8 6. (d) Let R = { (x,y) N × N : 2 x 2 y = 1 }. Prove that R is a function with domain N . Let x N . Then 2 x 2 1 is a natural number, so (x, 2 x 2 1 = y ) R. Therefore, Dom(R) = N . Now suppose that (x,y) R and (x,z) R. Then 2 x 2 1 =y and 2 x 2 1 =z. So y=z. Therefore, R is a function with domain N . 2
Jennifer Klarfeld – Week 10 – Edition 8 7. Complete the proof of Theorem 4.1.1. That is, prove that if (i) Dom( f ) = Dom( g ) and (ii) for all x Dom( f ), f (x) = g (x), then f = g . Suppose that x Dom( g ). Then (x,y) g for some y Rng( g ). Thus, g (x)=y. Since Dom( f ) = Dom( g ), x Dom( f ). For all x Dom( f ), f (x) = g (x), f (x) = g (x)=y. So (x,y) f and (x,y) g . We conclude that f = g . 3

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Jennifer Klarfeld – Week 10 – Edition 8 9. Let the universe be R , and A = [1,3). Find (a) χ A ( 1 ) χ A ( 1 ) = 1, since 1 A. (b) χ A ( 3 ) χ A ( 3 ) = 0, since 3 A. (c) χ A ( π ) χ A ( π ) = 0. Since π > 3, π A. (d) χ A ( 2 ) - χ A ( 0.2 ) χ A ( 2 ) = 1, since 2 A. χ A ( 0.2 ) = 0, since 0.2 A. Therefore, χ A ( 2 ) - χ A ( 0.2 ) = 1 – 0 = 1. 4
Jennifer Klarfeld – Week 10 – Edition 8 10. Let U be the universe. Suppose A U with A , and A U. Let χ A be the characteristic function of A. Find (a) { x U : χ A (x) = 1}.

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• Spring '16
• Math, DOM, codomain, Jazz in the Domain, Jennifer Klarfeld

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