# Here we use the principle of mathematical induction

• Notes
• 129
• 100% (1) 1 out of 1 people found this document helpful

This preview shows page 11 - 14 out of 129 pages.

##### We have textbook solutions for you!
The document you are viewing contains questions related to this textbook. The document you are viewing contains questions related to this textbook.
Chapter 6 / Exercise 43
Mathematical Excursions
Aufmann Expert Verified
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
##### We have textbook solutions for you!
The document you are viewing contains questions related to this textbook. The document you are viewing contains questions related to this textbook.
Chapter 6 / Exercise 43
Mathematical Excursions
Aufmann Expert Verified
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.
• • • 