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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 4cent 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 4cent and one 5cent stamp.
14 cents = use one 4cent and two 5cent stamps.
15 cents = use three 5cent 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 4cent stamps and n 5cent
stamps for some nonnegative 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 = ak1+ak2+ak3 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= ak1+ak2+ak3 ≤ 2k1+2k2+2k3
≤ 20+21+…+2k3+2k2+2k1
= 2k1 ≤ 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.
 Fall '11
 math

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