10) Identify the mistake in the following “proof" by mathematical induction.(Fake) Theorem: For any positive integer n, all the numbers in a set of n numbers are equal toeach other.Proof (by mathematical induction): It is obviously true that all the numbers in a set consistingof just one number are equal to each other, so the basis step is true. For the induction step,let A = {a1, a2, a3, . .., an41} be any set of n +1 numbers. Form two subsets each of sizen: B = {a1,a2, A3, .numbers in A except an+1, and C consists of all the numbers in A except a2.) By the inductionhypothesis, all the numbers in B equal a1 and all the numbers in C equal a1, since both setshave only n numbers. But every number in A is in B or C, so all the numbers in A equal a1;thus all are equal to each other..., a,} and C = {a1, az, a4, . .. , an+1}. (Precisely, B consists of all the

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Description: 10) Identify the mistake in the following “proof" by mathematical induction.(Fake) Theorem: For any positive integer n, all the numbers in a set of n numbers are equal toeach other.Proof (by mathematical induction): It is obviously true that all the

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10) Identify the mistake in the following “proof" by mathematical induction. (Fake) Theorem: For any positive integer n, all the numbers in a set of n numbers are equal to each other. Proof (by mathematical induction): It is obviously true that all the numbers in a set consisting of just one number are equal to each other, so the basis step is true. For the induction step, let A = {a1, a2, az, ..., an+1} be any set of n +1 numbers. Form two subsets each of size n: B = {q1,a2, az,...,a,} and C = {q1,a3, as, … . .,an+1}. (Precisely, B consists of all the numbers in A except an+1; and C consists of all the numbers in A except a2.) By the induction hypothesis, all the numbers in B equal a1 and all the numbers in C equal a1, since both sets have only n numbers. But every number in A is in B or C, so all the numbers in A equal a1; thus all are equal to each other.

10) Identify the mistake in the following “proof" by mathematical induction. (Fake) Theorem: For any positive integer n, all the numbers in a set of n numbers are equal to each other. Proof (by mathematical induction): It is obviously true that all the numbers in a set consisting of just one number are equal to each other, so the basis step is true. For the induction step, let A = {a1, a2, az, ..., an+1} be any set of n +1 numbers. Form two subsets each of size n: B = {q1,a2, az,...,a,} and C = {q1,a3, as, … . .,an+1}. (Precisely, B consists of all the numbers in A except an+1; and C consists of all the numbers in A except a2.) By the induction hypothesis, all the numbers in B equal a1 and all the numbers in C equal a1, since both sets have only n numbers. But every number in A is in B or C, so all the numbers in A equal a1; thus all are equal to each other.

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

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10) Identify the mistake in the following “proof" by mathematical induction.
(Fake) Theorem: For any positive integer n, all the numbers in a set of n numbers are equal to
each other.
Proof (by mathematical induction): It is obviously true that all the numbers in a set consisting
of just one number are equal to each other, so the basis step is true. For the induction step,
let A = {a1, a2, a3, . .., an41} be any set of n +1 numbers. Form two subsets each of size
n: B = {a1,a2, A3, .
numbers in A except an+1, and C consists of all the numbers in A except a2.) By the induction
hypothesis, all the numbers in B equal a1 and all the numbers in C equal a1, since both sets
have only n numbers. But every number in A is in B or C, so all the numbers in A equal a1;
thus all are equal to each other.
.., a,} and C = {a1, az, a4, . .. , an+1}. (Precisely, B consists of all the

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