Question 1. Explain what is wrong with the following proofs: . (4 points) Theorem: Every even...

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Question 1. Explain what is wrong with the following proofs: . (4 points) Theorem: Every even integer greater than two is the sum of two primes. Proof. Let's call the integer in question n. We prove this theorem by induction on even integers n. The base case is n = 4 as 4 is the first even integer greater than two. Base case n = 4: 4=2+2, so the theorem holds in this case. Inductive step: Let n be an even integer greater than two. Assume that n is the sum of two primes p and q. Then n +2 is the sum of the two primes p+2 and q, so the next even integer n + 2 is also a sum of two primes. It follows by induction that every even integer greater than two is the sum of two primes. . (4 points) Theorem: Let 3 be an arbitrary positive real number. Then, for all integers n ≥ 1 it holds that 3-1 = 1. Proof: By induction on n. For n = 1, we have 3-1 =3⁰= 1, so the base case holds. Inductive step: Suppose that the theorem is true for n=k. That is, 3k-1 = 1. For n = k + 1, we have 3(k+1)-1 = 3k = (3k-1¹)1 = 1²¹ = 1 which completes the inductive step. Question 1. Explain what is wrong with the following proofs: . (4 points) Theorem: Every even integer greater than two is the sum of two primes. Proof. Let's call the integer in question n. We prove this theorem by induction on even integers n. The base case is n = 4 as 4 is the first even integer greater than two. Base case n = 4: 4=2+2, so the theorem holds in this case. Inductive step: Let n be an even integer greater than two. Assume that n is the sum of two primes p and q. Then n +2 is the sum of the two primes p+2 and q, so the next even integer n + 2 is also a sum of two primes. It follows by induction that every even integer greater than two is the sum of two primes. . (4 points) Theorem: Let 3 be an arbitrary positive real number. Then, for all integers n ≥ 1 it holds that 3-1 = 1. Proof: By induction on n. For n = 1, we have 3-1 =3⁰= 1, so the base case holds. Inductive step: Suppose that the theorem is true for n=k. That is, 3k-1 = 1. For n = k + 1, we have 3(k+1)-1 = 3k = (3k-1¹)1 = 1²¹ = 1 which completes the inductive step.

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