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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Question
![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.
Show that P(0) is true: P(0) is true because r° =1 by definition of
zeroth power.
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 with k 2 0 and suppose that r' = 1 for each integer
i from 0 through k. This is the inductive hypothesis. We must show that
k+1= 1. Now
rk +1 = rk + k - (k – 1)
rk.pk
k- 1
because k +k - (k - 1) = k + 1
by the laws of exponents
%3D
1.1
by inductive hypothesis
%3D
= 1
Thus rk+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.)
O The inductive step assumes what is to be proved.
rk.pk
- =r.rk = rk + 1 - 1.
rk-1
O when r = 0,
O When k = 0, rk +1 rk + k – (k – 1)
rk.pk
O When k = 0,
=r+ 1, unless r= 1
ロrk+k-(k-1)rk.rk
rk-1](/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" = 1.
Show that P(0) is true: P(0) is true because r° =1 by definition of
zeroth power.
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 with k 2 0 and suppose that r' = 1 for each integer
i from 0 through k. This is the inductive hypothesis. We must show that
k+1= 1. Now
rk +1 = rk + k - (k – 1)
rk.pk
k- 1
because k +k - (k - 1) = k + 1
by the laws of exponents
%3D
1.1
by inductive hypothesis
%3D
= 1
Thus rk+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.)
O The inductive step assumes what is to be proved.
rk.pk
- =r.rk = rk + 1 - 1.
rk-1
O when r = 0,
O When k = 0, rk +1 rk + k – (k – 1)
rk.pk
O When k = 0,
=r+ 1, unless r= 1
ロrk+k-(k-1)rk.rk
rk-1
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