Here's an incorrect theorem with an incorrect proof.Theorem: All people are the same age.Proof: Let P(n) be the statement "any set of n people all have the same age". If we can prove P(n) for all n, then the theorem is true.Weprove P(n) by induction.Base case: Given any one person, that person has the same age as himself/herself. So P(1) is true.Inductive step: Suppose n > 2 and P(n − 1) is true. Take a set of n people:{P₁, P2, ..., Pn}.-Because P(n − 1) is true, people p₁, P2, ..., Pn-1 all have the same age. Also because P(n − 1) is true, people P2, P3, ..., Pn all have thesame age. Therefore people p₁, P2, …, Pn all have the sa ne age. This works for any set of n people, so P(n) is true.So P(n) is true for all n.Explain in a few sentences what's wrong with the proof.

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Answered: Here's an incorrect theorem with an…

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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Mathematical induction to prove that 21 + 22 + 23 + 24 + •… 2" = 2(2" – 1) •.. is true for all neN
Find mistake in the proof by induction of the statement: All people speak the same language. Base case: for n =1 the statement holds since there is only one person in the group. Induction step: Assume that the statement is true for some natural k. That is, any k people speak the same language. We need to prove the statement for k +1. Consider a set containing k+1 people. Remove any person, say, X from this set. By assumption, the remaining k people speak the same language. Next we return X to the set, and remove another one, say Y. Using induction assumption again, we can show that X speaks the same language as the remaining k – 1 people. We can now return Y to the set, and claim that all of k +1 people speak the same language, as needed. The proof is complete. O The induction assumption is stated incorrectly. O The base case is proven incorrectly. O The proof of induction step does not work for k=2.
number theory
subject: discrete math
Prove by induction: How do you prove that ∀x∈Z,n,m∈N(xn)m = xmn ? (where Z is set of integers and N is set of natural numbers)
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1.) Mathematical induction is useful in determining whether the statement REMAINS TRUE with any natural number as k. Is the statement TRUE or FALSE? 2.) We have only one (1) step that we may utilize in Mathematical induction – the Inductive step. Is this statement TRUE or FALSE? a. Underline only the PART that makes the following statement erroneous: "The Basis step helps in determining whether the first iteration of the statement, which is P(k+10), is TRUE." 3.) In a statement in Mathematical induction, we may only use positive integers or natural numbers for k. Is this statement TRUE or FALSE? 4.) We use k and k+1 for the Inductive step and Basis step respectively. Is this statement TRUE or FALSE? 5.) Prove that given any integer for n, n³ + 2n will be divisible by 3. n²(n+1)? 6.) 13 + 23 + 33 + +n³ ... 4 1 7.) 1(2) 1 + + 2(3) 1 + 3(3) 1 + п(п+1) given that any integer for n is positive. ... n+1' 8.) Write the formal proof that 2"+2 + 32n+1 is divisible by 7 for all positive…
Use mathematical induction to prove that for every integer n ≥ 2, if a set S has n elements, then the number of subsets of S with an even number of elements equals the number of subsets of S with an odd number of elements.
Every positive integer is always divisible by 1 and by itself. That is, for every posi- tive integer n we always have 1|n and n|n. We say an integer p ≥ 2 is a prime number if its only divisors are 1 and p. (By convention, 1 is not a prime number.) (a) List the first ten prime numbers. (b) If a positive integer n is not prime, then it is called a composite number. Show that every positive integer n ≥ 2 can be written as a product of prime numbers. (This is called a prime factorization of n. Notice that some primes may need to be repeated. For example, 180 = 2² × 3² × 5 and the primes 2 and 3 are repeated.) Hint: Consider two separate cases based on whether a given number n is prime or composite. For composite numbers, suppose by strong induction that all smaller integers have a prime factorization.
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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