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Math 2534 Solutions to Homework 6 Prove the following theorems using PMI Theorem 1: n N , 2 4 6 ... (2n) n2 n Proof: I will verify that the hypothesis is true for at least one value of n N Consider n = 1, we see that 2 = (1)2+(1) Now consider n = 2 and we see that 2 + 4 = (2)2+2 We will now assume that our hypothesis is true for each natural number between n = 1 up to some arbitrary natural number k. ie. 2 4 6 ... (2k ) k 2 k My intent is to prove true for k+1. I will show that 2 4 6 ... 2(k 1) (k 1)2 (k 1) k 2 3k 2 Now for the body of the proof: Consider the k+1 term below. 2 4 6 ... 2k 2(k 1) [2 4 6 ... 2k ] 2( k 1) [k 2 k ] 2(k 1) By the inductive assumption k 2 3k 2 which is the form that I intended to produce to verify that the k+1 term works. Therefore I have assumed true up to k and proved true for k+1, so the hypothesis is true for all natural numbers. Theorem 2: n N , 5n 1 is divisible by 4 Proof: I will verify that the hypothesis is true for at least one value of n N Consider n = 1, 51 – 1 = 4 and is therefore divisible by 4 Now consider n = 2 and we see that 52 – 1 = 24 which is divisible by 4. We will now assume that our hypothesis is true for each natural number between n = 1 up to some arbitrary natural number k. ie. 4 (5k 1) k 1 that 4 (5 My intent is to prove true for k+1. I will show 1) For the body of the proof we will consider the k+1 term, 5k 1 1 5k 5 1 (5k 1) (4)5k by the inductive assumption we know that 4 divides (5k 1) k and 4 also divides (4)5k since 4 is a factor in this term. Therefore 4 (5 1) . I have assumed true up to k and proved true for k+1. The hypothesis is true for all natural numbers by PMI. Theorem 3: n N 5, n2 2n Proof: I will first verify that the hypothesis is true for at least one value of n. Consider n = 5, we have that (5)2 < 25 which is 25 < 32 Now assume that the hypothesis is true up to some arbitrary value k. ie: k 2 2k I will now prove true for k+1. ie: (k+1)2 2k 1 In the main body of the proof we will consider the k+1 term (k+1)2. So (k 1) 2 k 2 2k 1 < 2k (2k 1) by the inductive assumption < 2k 2k Since for all natural numbers greater than 3, 2n + 1 < 2n = 2(2k ) 2k 1 Since I assumed true up to k and proved true for k+1, the hypothesis is true for all natural numbers greater than 4. Theorem 4: If a jigsaw puzzle has n pieces, then it is completed in n-1 fits. (It takes one “fit” to join an additional piece to those that have already been assembled) See solution in PMI Resource sheet 3 on the Class Information Page.