Here is a question about induction proof/pumping lemma. Please give your answer

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Here is a question about induction proof/pumping lemma. Please give your answer with complete steps. I would give you positive feedback if your answer works. Appreciate your help!!

MCS 5.26: A sequence of numbers is weakly decreasing when each number in the sequence is the numbers after it. (This implies that a sequence of just one number repeated is weakly decreasing Here's a bogus proof of a very important true fact, that every integer greater than 1 is a product of a unique weakly decreasing sequence of primes a pusp, for short Explain what's bogus about the proof. Lemma. Every integer greater than 1 is a pusp For example, 252 7. 3. 3. 2. 2, and no other weakly decreasing sequence of primes will have a product equal to 252 Bogus proof. We will prove the lemma by strong induction, letting the induction hypoth esis, P (n) be "n is a pusp So the lemma will follow if we prove that P(m) holds for all n 2. Base Case (n 2): PO2) is true because 2 is prime, and so it is a length one product of primes, and this is obviously the only sequence of primes whose product can equal 2 Inductive step: Suppose that n 2 and that i is a pusp for every integer i where 2 Si n We must show that P(n 1) holds, namely, that (n 1) is also a pusp. We argue by cases: If (n 1) is itself prime, then it is the product of a length one sequence consisting of itself This sequence is unique, since by definition of prime (n 1) has no other prime factors. So (n 1) is a pusp, that is P(n 1) holds in this case Otherwise (n 1) is not prime, which by definition means n 1 krm for some integers k and m such that 2 K k, m n 1. Now by the strong induction hypothesis, we know that k and m are pusps. It follows that by merging the unique prime sequences for k and m, in sorted order, we get a unique weakly decreasing sequence of primes whose product equals n 1. So n 1 is a pusp, in this case as well So P(n 1) holds in any case, which completes the proof by strong induction that POn) holds for all n 2

Explanation / Answer

Suppose you specify another composite number n+1 = r*s, such that k*m and r*s both have different factorizations of n+1. There is no additional information in the proof that shows that the prime factorizations of k and m are unique compared to the prime factorizations of r and s.

Obviously the fundamental theorem of arithmetic holds (i.e. each positive integer has a unique factorization) so we know that individually, k, m, r, and s have unique factorizations.

The question now is: when you merge k and m and when you merge r and s, how can you be sure that the resulting sequences for each pair is the SAME?

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