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Use strong induction to show that every positive integer n can be written as a sum of distinct powers of two, that is, as a sum of a subset of the integers 20 = 1, 21 = 2, 2² = 4, and so on. Let P(n) be the proposition that the positive integer n can be written as a sum of distinct powers of 2. Click and drag the given steps (in the right) to the corresponding step names (in the left) to show that if P(1) is true for all j≤ k, then P(k+ 1) is also true. First prove the above statement when k + 1 is odd and then prove when k + 1 is even. Step 1 If k + 1 is even, then k is even, so 2° was not part of the sum for k. Increasing each exponent by 1 doubles the value and gives us the desired sum for k + 1. Step 2 If k + 1 is even, then (k + 1)/2 is a positive integer, so by the inductive hypothesis (k + 1)/2 can be written as a sum of distinct powers of 2. Step 3 Increasing each exponent by 1 doubles the value and gives us the desired sum for k + 1. Step 4 If k + 1 is odd, then k is even, so 2° was not part of the sum for k . If k + 1 is even, then (k + 1)/2 is a positive integer, so by the inductive hypothesis (k + 1)/2 can be written as a sum of distinct powers of 2. If k + 1 is even, then k is even, so 2° was not part of the sum for k. If k+1 is odd, then (k + 1)/2 is a positive integer, so by the inductive hypothesis (k + 1)/2 can be written as a sum of distinct powers of 2. Therefore the sum for k + 1 is the same as the sum for k with the extra term 2° added. Therefore the sum for k + 1 is the same as the sum for k with the extra term 2° added.

Use strong induction to show that every positive integer n can be written as a sum of distinct powers of two, that is, as a sum of a subset of the integers 20 = 1, 21 = 2, 2² = 4, and so on. Let P(n) be the proposition that the positive integer n can be written as a sum of distinct powers of 2. Click and drag the given steps (in the right) to the corresponding step names (in the left) to show that if P(1) is true for all j≤ k, then P(k+ 1) is also true. First prove the above statement when k + 1 is odd and then prove when k + 1 is even. Step 1 If k + 1 is even, then k is even, so 2° was not part of the sum for k. Increasing each exponent by 1 doubles the value and gives us the desired sum for k + 1. Step 2 If k + 1 is even, then (k + 1)/2 is a positive integer, so by the inductive hypothesis (k + 1)/2 can be written as a sum of distinct powers of 2. Step 3 Increasing each exponent by 1 doubles the value and gives us the desired sum for k + 1. Step 4 If k + 1 is odd, then k is even, so 2° was not part of the sum for k . If k + 1 is even, then (k + 1)/2 is a positive integer, so by the inductive hypothesis (k + 1)/2 can be written as a sum of distinct powers of 2. If k + 1 is even, then k is even, so 2° was not part of the sum for k. If k+1 is odd, then (k + 1)/2 is a positive integer, so by the inductive hypothesis (k + 1)/2 can be written as a sum of distinct powers of 2. Therefore the sum for k + 1 is the same as the sum for k with the extra term 2° added. Therefore the sum for k + 1 is the same as the sum for k with the extra term 2° added.

Advanced Engineering Mathematics
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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Use strong induction to show that every positive integer n can be written as a sum of distinct powers of two, that is, as a sum of a subset of the integers 20 = 1, 21 = 2, 2² = 4, and so on. Let P(n) be the proposition that the positive integer n can be written as a sum of distinct powers of 2. Click and drag the given steps (in the right) to the corresponding step names (in the left) to show that if P(1) is true for all j≤ k, then P(k+ 1) is also true. First prove the above statement when k + 1 is odd and then prove when k + 1 is even. Step 1 If k + 1 is even, then k is even, so 2° was not part of the sum for k. Increasing each exponent by 1 doubles the value and gives us the desired sum for k + 1. Step 2 If k + 1 is even, then (k + 1)/2 is a positive integer, so by the inductive hypothesis (k + 1)/2 can be written as a sum of distinct powers of 2. Step 3 Increasing each exponent by 1 doubles the value and gives us the desired sum for k + 1. Step 4 If k + 1 is odd, then k is even, so 2° was not part of the sum for k . If k + 1 is even, then (k + 1)/2 is a positive integer, so by the inductive hypothesis (k + 1)/2 can be written as a sum of distinct powers of 2. If k + 1 is even, then k is even, so 2° was not part of the sum for k. If k+1 is odd, then (k + 1)/2 is a positive integer, so by the inductive hypothesis (k + 1)/2 can be written as a sum of distinct powers of 2. Therefore the sum for k + 1 is the same as the sum for k with the extra term 2° added. Therefore the sum for k + 1 is the same as the sum for k with the extra term 2° added.
Transcribed Image Text:Use strong induction to show that every positive integer n can be written as a sum of distinct powers of two, that is, as a sum of a subset of the integers 20 = 1, 21 = 2, 2² = 4, and so on. Let P(n) be the proposition that the positive integer n can be written as a sum of distinct powers of 2. Click and drag the given steps (in the right) to the corresponding step names (in the left) to show that if P(1) is true for all j≤ k, then P(k+ 1) is also true. First prove the above statement when k + 1 is odd and then prove when k + 1 is even. Step 1 If k + 1 is even, then k is even, so 2° was not part of the sum for k. Increasing each exponent by 1 doubles the value and gives us the desired sum for k + 1. Step 2 If k + 1 is even, then (k + 1)/2 is a positive integer, so by the inductive hypothesis (k + 1)/2 can be written as a sum of distinct powers of 2. Step 3 Increasing each exponent by 1 doubles the value and gives us the desired sum for k + 1. Step 4 If k + 1 is odd, then k is even, so 2° was not part of the sum for k . If k + 1 is even, then (k + 1)/2 is a positive integer, so by the inductive hypothesis (k + 1)/2 can be written as a sum of distinct powers of 2. If k + 1 is even, then k is even, so 2° was not part of the sum for k. If k+1 is odd, then (k + 1)/2 is a positive integer, so by the inductive hypothesis (k + 1)/2 can be written as a sum of distinct powers of 2. Therefore the sum for k + 1 is the same as the sum for k with the extra term 2° added. Therefore the sum for k + 1 is the same as the sum for k with the extra term 2° added.
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