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MATH 109 Midterm: Math 109-Winter 2015-Exam 2 Solutions

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turquoisegnu128
31 Jan 2019
School
UCSD
Department
Mathematics
Course
MATH 109
Professor
All

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Math 109 winter 2015 midterm 2 sample solutions. For each part, decide if the statement is true or false, and prove that your answer is correct. (a) (5 pts). X r, y r, x + y < 0. (b) (5 pts). X r, y r, x + y < 0. Given any x r, choose y = x 1. Then taking y = x + 1, we must have x + y < 0. But x + y = x + ( x + 1) = 1 > 0, a contradiction. Let r = r \ {0} be the set of all nonzero real numbers. Consider the function f : r r de ned by f (x) = 2/x. Find a formula for its inverse function f 1 and justify that your formula is correct. We need to show that f is both injective and surjective. R and that f (x1) = f (x2).

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Related Questions

Let φ : G → G′ be an isomorphism from the group G to the groupG′. In particular, φ is bijective so that its inverse φ−1 : G′ → Gexists (see 1.9.1). Prove that φ−1 is anisomorphism.

1.9.1 Theorem. A function f : X → Y has an inverse if and only ifit is bijective.
Proof. Letf:X→Y beafunction.
(⇒) Assume that f has an inverse f−1. (f injective?) Letx,x′ ∈ X and assume that f(x) = f(x′). Then x =f−1(f(x)) = f−1(f(x′)) = x′.

Therefore, f is injective. (f surjective?) Let y ∈ Y . Put x =f−1(y). Then x ∈ X and
f(x) = f(f−1(y)) = y.

Therefore, f is surjective. Since f is both injective andsurjective, it is
bijective.
(⇐) Assume that f is bijective. Define f−1 : Y → X byletting f−1(y) be the unique x in X for which f(x) = y.(Since f is surjective, there is at least one such x and since f isinjective, there is at most one such x.) For x ∈ X, we havef−1(f(x)) = x (since f−1(f(x)) is defined tobe the element that f sends to f(x)). Similarly, for y ∈ Y ,f(f−1(y)) = y (since f−1(y) is defined to be the elementthat f sends to y). Therefore, f−1 is an inverse off.

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