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hw1sol

# hw1sol - HOMEWORK 1 SOLUTIONS 1 By induction on n Base...

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HOMEWORK 1 - SOLUTIONS 1. By induction on n : Base Case ( n = 1): f n = f 1 = f which is surjective by assumption. Inductive Step : Suppose f n is surjective for all 1 n k . We want to show that f k +1 is surjective. Let x X be arbitrary. We want to find y X such that f k +1 ( y ) = x . Well f k +1 ( y ) = f ( f k ( y )) and we know f is surjective, so first pick some z X such that f ( z ) = x . Next, by our inductive hypothesis, we know that f k is surjective so pick some y such that f k ( y ) = z . This choice of y does it, since: f k +1 ( y ) = f ( f k ( y )) = f ( z ) = x 2. (GRADED - EASY). By induction on | X | : Base Case ( n = 0): If | X | = 0, then it’s the empty set and has only one subset, namely itself, the empty set. And indeed: |P ( X ) | = |{∅}| = 1 = 2 0 = 2 | X | Inductive Step : Assume that for all n k and every set X such that | X | = n , |P ( X ) | = 2 | X | . Now suppose X is a set with | X | = k + 1. Let’s write X = { x 1 , . . . , x k +1 } . Now every subset of X either contains x k +1 or it doesn’t, and it’s easy to see that there are an equal number of subsets of X which contain x k +1 as there are which don’t. Thus the number of subsets of X is double the number of subsets of { x 1 , . . . , x k } . This latter set has cardinality k , so we can apply the inductive hypothesis to obtain: |P ( X ) | = 2 · |P ( { x 1 , . . . , x k } ) | = 2 · 2 k = 2 k +1 3. The formula is z - 1 = z * | z | 2 . To check that this works, we need to check that

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