21 postage theorem every amount of postage that is at

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Unformatted text preview: (1) is true. Induction step: Show that (A(1) ⋀... ⋀A(n)) → A(n+1) holds for all n >=1. 21 Postage Theorem: Every amount of postage that is at least 12 cents can be made from 4-cent and 5cent stamps. 22 Postage Proof by induction on the amount of postage. Induction Basis: If the postage is 12 cents = use three 4 cent stamps 13 cents = use two 4-cent and one 5-cent stamp. 14 cents = use one 4-cent and two 5-cent stamps. 15 cents = use three 5-cent stamps. 23 Postage Inductive step: Suppose that we have shown how to construct postage for every value from 12 up through k. We need to show how to construct k + 1 cents of postage. Since we’ve already proved the induction basis, we may assume that k + 1 ≥ 16. Since k+1 ≥ 16, we have (k+1)−4 ≥ 12. By inductive hypothesis, we can construct postage for (k + 1) − 4 cents using m 4-cent stamps and n 5-cent stamps for some non-negative integers m and n. In other words ((k + 1) − 4) = 4m + 5n; hence, k+1 = 4(m+1)+5n. 24 Quiz Why did we need to establish four cases in the induction basis? Isn’t it enough to remark that the postage for 12 cents is given by three 4 cents stamps? 25 Another Example: Sequence Theorem: Let a sequence (an) be defined as follows: a0=1, a1=2, a2=3, ak = ak-1+ak-2+ak-3 for all integers k≥3. Then an ≤ 2n for all integers n≥0. P(n) Proof. Induction basis: The statement is true for n=0, since a0=1 ≤1=20 P(0) for n=1: since a1=2 ≤2=21 P(1) for n=2: since a2=3 ≤4=2 P(2) 26 2 Sequence (cont’d) Inductive step: Assume that P(i) is true for all i with 0≤i<k, that is, ai ≤ 2i for all 0≤i<k, where k>2. Show that P(k) is true: ak ≤ 2 k ak= ak-1+ak-2+ak-3 ≤ 2k-1+2k-2+2k-3 ≤ 20+21+…+2k-3+2k-2+2k-1 = 2k-1 ≤ 2k Thus, P(n) is true by strong induction. 27...
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This note was uploaded on 03/24/2014 for the course CSCE 222 taught by Professor Math during the Fall '11 term at Texas A&M.

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