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Unformatted text preview: LECTURE 23: INVERSES AND PERMUTATIONS ( 9.69.7) Let all things be done decently and in order. Paul the apostle 1. More on inverses From last time: Theorem 1. Let f : A B be a function. The inverse relation f 1 is a function from B to A if and only if f is a bijection. Corollary 2. If f 1 is a function then f 1 is a bijection. Further, f 1 f = id A and f f 1 = id B . Sometimes it is easy to figure out f 1 because f has only finitely many pairs! In other cases, it might take more work. Example 3. The function f : R{ 2 } R{ 3 } given by f ( x ) = 3 x/ ( x 2) is a bijection. Prove this, and find the inverse. Proof. First we show f is injective. Let x,y R { 2 } and assume f ( x ) = f ( y ). Then 3 x/ ( x 2) = 3 y/ ( y 2). Clearing denominators, we have 3 xy 6 x = 3 xy 6 y . So 6 x = 6 y . Thus x = y . Next we show that f is surjective. Let b R { 3 } . We want to find a R { 2 } so that f ( a ) = b ....
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This note was uploaded on 03/08/2011 for the course MATH 334 taught by Professor Smith during the Spring '11 term at Vanderbilt.
 Spring '11
 Smith
 Permutations

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