Here we use the principle of mathematical induction

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Chapter 6 / Exercise 43
Mathematical Excursions
Aufmann
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Here we use The Principle of Mathematical Induction. Note that PMI has two parts which we denote by PMI (a) and PMI (b). We let P ( n ) be the statement 2 n > 5 n . For n 0 we take 5. We could write simply: P ( n ) = 2 n > 5 n and n 0 = 5 . Note that P ( n ) represents a statement , usually an inequality or an equation but sometimes a more complicated assertion. Now if n = 4 then P ( n ) be- comes the statement 2 4 > 5 · 4 which is false! But if n = 5, P ( n ) is the statement 2 5 > 5 · 5 or 32 > 25 which is true and we have established PMI (a). 3
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Mathematical Excursions
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Chapter 6 / Exercise 43
Mathematical Excursions
Aufmann
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4 CHAPTER 2. PROOF BY INDUCTION Now to prove PMI (b) we begin by assuming that P ( n ) is true for 5 n k. That is, we assume 2 n > 5 n for 5 n k. (2.1) The assumption (2.1) is called the induction hypothesis . We want to use it to prove that P ( n ) holds when n = k + 1. So here’s what we do. By (2.1) letting n = k we have 2 k > 5 k. Multiply both sides by two and we get 2 k +1 > 10 k. (2.2) Note that we are trying to prove 2 k +1 > 5( k + 1). Now 5( k + 1) = 5 k + 5 so if we can show 10 k 5 k + 5 we can use (2.2) to complete the proof. Now 10 k = 5 k + 5 k and k 5 by (2.1) so k 1 and hence 5 k 5. Therefore 10 k = 5 k + 5 k 5 k + 5 = 5( k + 1) . Thus 2 k +1 > 10 k 5( k + 1) so 2 k +1 > 5( k + 1) . (2.3) that is, P ( n ) holds when n = k + 1. So assuming the induction hypothesis (2.1) we have proved (2.3). Thus we have established PMI (b). We have established that parts (a) and (b) of PMI hold for this particular P ( n ) and n 0 . So the PMI tells us that P ( n ) holds for n 5. That is, 2 n > 5 n holds for n 5. I now give a more streamlined proof. Proposition 2.2. If n 5 then 2 n > 5 n .
5 Proof #2. We prove the proposition by induction on the variable n . If n = 5 we have 2 5 > 5 · 5 or 32 > 25 which is true. Assume 2 n > 5 n for 5 n k (the induction hypothesis) . Taking n = k we have 2 k > 5 k. Multiplying both sides by 2 gives 2 k +1 > 10 k. Now 10 k = 5 k + 5 k and k 5 so k 1 and therefore 5 k 5. Hence 10 k = 5 k + 5 k 5 k + 5 = 5( k + 1) . It follows that 2 k +1 > 10 k 5( k + 1) and therefore 2 k +1 > 5( k + 1) . Hence by PMI we conclude that 2 n > 5 n for n 5.
The 8 major parts of a proof by induction: 1. First state what proposition you are going to prove. Precede the state- ment by Proposition, Theorem, Lemma, Corollary, Fact, or To Prove: . 2. Write the Proof or Pf. at the very beginning of your proof. 3. Say that you are going to use induction (some proofs do not use induc- tion!) and if it is not obvious from the statement of the proposition identify clearly P ( n ), the statement to be proved, the variable n and the starting value n 0 . Even though this is usually clear, sometimes these things may not be obvious. And, of course, the variable need not be n . It could be represented in many different ways.

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