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Unformatted text preview: M114S, Solutions to HW #5 x5.3. Both properties of exponentiation are proved by induction on k . First we show that n ( m + k ) = n m ยท n k . If k = 0, then n ( m + k ) = n m , and n k = n = 1. Now it is a general fact that for all p , p ยท 1 = p . (Proof: By (54) and Lemma 5.13, p ยท S (0) = ( p ยท 0)+ p = 0 ยท p = p .) It follows that n ( m +0) = n m = n m ยท n . This proves the first fact in the case when k = 0. Next, assume that n ( m + k ) = n m ยท n k . Using this, n ( m + S ( k )) = n S ( m + k ) = n ( m + k ) ยท n = ( n m ยท n k ) ยท n . By 5.2 , this last number is n m ยท ( n k ยท n ) = n m ยท n S ( k ) . This completes the proof. The second fact is to check that n m ยท k = ( n m ) k . When k = 0, this is true because n m ยท = n = 1 = ( n m ) . So assume this for k , and note that n m ยท S ( k ) = n ( m ยท k )+ m = n ( m ยท k ) ยท n m = ( n m ) k ยท n m = ( n m ) S ( k ) . Note that we used the first part of the problem in one of these steps, just before we appealed to the induction hypothesis. x5.4. The canonical isomorphism afii9843 : N 1 ึ โ N 2 constructed in the proof of the Uniqueness Theorem 5.4 satisfies the following two properties: afii9843 (0 1 ) = 0 2 , afii9843 ( S 1 ( m )) = S 2 ( afii9843 ( m )) ( m โ N 1 ) . These and the recursive definitions of addition and subtraction suffice to prove that afii9843 preserves these operations....
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 Winter '08
 MOSCHOVAKIS
 Set Theory, #, Natural number, 1 m, one g

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