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# 23 - LECTURE 23 INVERSES AND PERMUTATIONS(9.6-9.7 Let all...

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LECTURE 23: INVERSES AND PERMUTATIONS ( § 9.6-9.7) 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 . In other words, we want 3 a/ ( a - 2) = b . Our scratch work suggests that a = 2 b/ ( b - 3) should work.

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23 - LECTURE 23 INVERSES AND PERMUTATIONS(9.6-9.7 Let all...

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