Consider the following proposed proof, which claims to show that every nonnegative integer power of every nonzero real number is 1. Let r be any nonzero real number, and let P(n) be the equation r" = 1. %3D Show that P(0) is true: P(0) is true because r° zeroth power. = 1 by definition of Show that for every integer k> 0, if P(i) is true for each integer i from 0 through k, then P(k + 1) is also true: Let k be any integer withk20 and suppose that r' = 1 for each integer i from 0 through k. This is the inductive hypothesis. We must show that pk +1 = 1. Now rk +1 = rk+k - (k – 1) pk. rk k - 1 because k + k - (k – 1) = k + 1 %3D by the laws of exponents 1.1 by inductive hypothesis %3D 1 = 1 Thus rk+1 = 1 [as was to be shown]. %3D [Since we have proved the basis step and the inductive step of the strong mathematical induction, we conclude that the given statement is true.] Identify the error(s) in the above "proof." (Select all that apply.) O The inductive step assumes what is to be proved. rk.rk O when r = 0, =r.rk = rk+ 1 = 1. %3D k- O When k = 0, pk+1 pk + k - (k - 1). %3D pk. pk O when k = 0, =r+ 1, unless r= 1. rk-1 ロメ+k-(k- 1)rk.rk
Consider the following proposed proof, which claims to show that every nonnegative integer power of every nonzero real number is 1. Let r be any nonzero real number, and let P(n) be the equation r" = 1. %3D Show that P(0) is true: P(0) is true because r° zeroth power. = 1 by definition of Show that for every integer k> 0, if P(i) is true for each integer i from 0 through k, then P(k + 1) is also true: Let k be any integer withk20 and suppose that r' = 1 for each integer i from 0 through k. This is the inductive hypothesis. We must show that pk +1 = 1. Now rk +1 = rk+k - (k – 1) pk. rk k - 1 because k + k - (k – 1) = k + 1 %3D by the laws of exponents 1.1 by inductive hypothesis %3D 1 = 1 Thus rk+1 = 1 [as was to be shown]. %3D [Since we have proved the basis step and the inductive step of the strong mathematical induction, we conclude that the given statement is true.] Identify the error(s) in the above "proof." (Select all that apply.) O The inductive step assumes what is to be proved. rk.rk O when r = 0, =r.rk = rk+ 1 = 1. %3D k- O When k = 0, pk+1 pk + k - (k - 1). %3D pk. pk O when k = 0, =r+ 1, unless r= 1. rk-1 ロメ+k-(k- 1)rk.rk
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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![Consider the following proposed proof, which claims to show that every nonnegative integer power of every nonzero real number is 1.
---
Let \( r \) be any nonzero real number, and let \( P(n) \) be the equation \( r^n = 1 \).
**Show that \( P(0) \) is true:** \( P(0) \) is true because \( r^0 = 1 \) by definition of zeroth power.
**Show that for every integer \( k \geq 0 \), if \( P(i) \) is true for each integer \( i \) from 0 through \( k \), then \( P(k + 1) \) is also true:**
Let \( k \) be any integer with \( k \geq 0 \) and suppose that \( r^i = 1 \) for each integer \( i \) from 0 through \( k \). This is the inductive hypothesis. We must show that \( r^{k+1} = 1 \). Now
\[
r^{k+1} = r^k \cdot r^{k-(k-1)} = r^k \cdot r^{k-k-(k-1)}
\]
because \( k + k - (k-1) = k + 1 \)
\[
= \frac{r^k}{r^{k-1}}
\]
by the laws of exponents
\[
= \frac{1}{1}
\]
by inductive hypothesis
\[
= 1
\]
Thus \( r^{k+1} = 1 \) [as was to be shown].
[Since we have proved the basis step and the inductive step of the strong mathematical induction, we conclude that the given statement is true.]
---
**Identify the error(s) in the above "proof." (Select all that apply.)**
- [ ] The inductive step assumes what is to be proved.
- [ ] When \( r = 0 \), \(\frac{r^k}{r^{k-1}} = r \cdot r^k = r^{k+1} = 1\).
- [ ] When \( k = 0 \), \( r^{k+1} \neq r^k + k - (k-1) \).
- [ ] When \( k = 0](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fdad58c83-f2ea-4146-b813-eac661fb8c00%2F6211073f-1151-4e55-ad29-40739a8b2a66%2Fz576rdu_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Consider the following proposed proof, which claims to show that every nonnegative integer power of every nonzero real number is 1.
---
Let \( r \) be any nonzero real number, and let \( P(n) \) be the equation \( r^n = 1 \).
**Show that \( P(0) \) is true:** \( P(0) \) is true because \( r^0 = 1 \) by definition of zeroth power.
**Show that for every integer \( k \geq 0 \), if \( P(i) \) is true for each integer \( i \) from 0 through \( k \), then \( P(k + 1) \) is also true:**
Let \( k \) be any integer with \( k \geq 0 \) and suppose that \( r^i = 1 \) for each integer \( i \) from 0 through \( k \). This is the inductive hypothesis. We must show that \( r^{k+1} = 1 \). Now
\[
r^{k+1} = r^k \cdot r^{k-(k-1)} = r^k \cdot r^{k-k-(k-1)}
\]
because \( k + k - (k-1) = k + 1 \)
\[
= \frac{r^k}{r^{k-1}}
\]
by the laws of exponents
\[
= \frac{1}{1}
\]
by inductive hypothesis
\[
= 1
\]
Thus \( r^{k+1} = 1 \) [as was to be shown].
[Since we have proved the basis step and the inductive step of the strong mathematical induction, we conclude that the given statement is true.]
---
**Identify the error(s) in the above "proof." (Select all that apply.)**
- [ ] The inductive step assumes what is to be proved.
- [ ] When \( r = 0 \), \(\frac{r^k}{r^{k-1}} = r \cdot r^k = r^{k+1} = 1\).
- [ ] When \( k = 0 \), \( r^{k+1} \neq r^k + k - (k-1) \).
- [ ] When \( k = 0
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