Now we assume that P(r) is true, that is, r can be written as a sum of 3s and 5s, for any cases, P(9) and P(10), by the equations 24. Prove that any amount of postage greater than or equal to 8 cents can be built using only 3-cent and 5-cent stamps. Here we let P(n) be the statement that only 3-cent and 5-cent stamps are needed to build n cents worth of postage, and prove that P(n) is true for all n2 8. The basis step is to establish P(8), which is done by the equation 8 = 3 + 5 For reasons that will be clear momentarily, we'll also establish two additional ses, P(9) and P(10), by the equations 9 = 3 + 3 + 3 10 = 5 + 5 is Ssrsk, and consider P(k + 1). We may assume that k+1 is at least 11, because we bave already proved P(r) true for r = 8, 9, and 10. If k + 1 2 11, then (k + 1)- 3 = t- 2 2 8. Thus k-2 is a legitimate r value, and by the inductive hypothesis, P(k – 2) %3D %3D is true. Therefore k– 2 can be written as a sum of 3s and 5s, and adding an additional 3 - gives us k+1 as a sum of 3s and 5s. This verifies that P(k+ 1) is true and completes the proof. a. Why are the additional cases P(9) and P(10) proved separately in Example 24? Do This b. Why can't the first principle of induction be used in the proof of Example 24?
Now we assume that P(r) is true, that is, r can be written as a sum of 3s and 5s, for any cases, P(9) and P(10), by the equations 24. Prove that any amount of postage greater than or equal to 8 cents can be built using only 3-cent and 5-cent stamps. Here we let P(n) be the statement that only 3-cent and 5-cent stamps are needed to build n cents worth of postage, and prove that P(n) is true for all n2 8. The basis step is to establish P(8), which is done by the equation 8 = 3 + 5 For reasons that will be clear momentarily, we'll also establish two additional ses, P(9) and P(10), by the equations 9 = 3 + 3 + 3 10 = 5 + 5 is Ssrsk, and consider P(k + 1). We may assume that k+1 is at least 11, because we bave already proved P(r) true for r = 8, 9, and 10. If k + 1 2 11, then (k + 1)- 3 = t- 2 2 8. Thus k-2 is a legitimate r value, and by the inductive hypothesis, P(k – 2) %3D %3D is true. Therefore k– 2 can be written as a sum of 3s and 5s, and adding an additional 3 - gives us k+1 as a sum of 3s and 5s. This verifies that P(k+ 1) is true and completes the proof. a. Why are the additional cases P(9) and P(10) proved separately in Example 24? Do This b. Why can't the first principle of induction be used in the proof of Example 24?
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Transcribed Image Text:Now we assume that P(r) is true, that is, r can be written as a sum of 3s and 5s, for any
24 Prove that any amount of postage greater than or equal to 8 cents can be built using only
cases, P(9) and P(10), by the equations
3-cent and 5-cent stamps.
Here we let P(n) be the statement that only 3-cent and 5-cent stamps are needed to
build n cents worth of postage, and prove that P(n) is true for all n> 8. The basis step is
to establish P(8), which is done by the equation
8 = 3 + 5
For reasons that will be clear momentarily, we'll also establish two additional
9 = 3 + 3 + 3
10 = 5 + 5
%3D
8<rsk, and consider P(k+ 1). We may assume that k+ 1 is at least 11, because we
kave already proved P(r) true for r = 8, 9, and 10. If k + 1 > 11, then (k + 1) – 3 =
t- 22 8. Thus k – 2 is a legitimate r value, and by the inductive hypothesis, P(k – 2)
%3D
is true. Therefore k – 2 can be written as a sum of 3s and 5s, and adding an additional 3
gives us k+ 1 as a sum of 3s and 5s. This verifies that P(k+ 1) is true and completes the
proof.
a. Why are the additional cases P(9) and P(10) proved separately in Example
24?
Do This
b. Why can't the first principle of induction be used in the proof of Example
24?
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