Math F215: Induction April 7, 2013

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1 Math F25: Iductio April 7, 203 Iductio is used to prove that a collectio of statemets P(k) depedig o k N are all true. A statemet is simply a mathematical phrase that must be either true or false. Here are some examples of statemets P(k) that deped o a atural umber; some of these use defiitios we have t formally itroduced yet, but you should uderstad them ayways. Examples of P(k): k. k 2 > k + 0. k is eve. k is either eve or odd. k j= j = ( + )/2. k is a product of prime umbers. 5k <. k j= j = 0. I all of these cases P(k) ca be either true or false. For example, i the last case, P() is false because the sum is ad 0, but P(4) is true because = 0. Exercise : For each statemet i the list above, determie (without proof) the atural umbers for which the statemet is true. The versio of iductio we have used is the followig. Theorem 2.7 (Weak Iductio): Suppose P(k) is a statemet depedig o k N. To show that P(k) is true for all k N, it is eough to show that. P() is true, ad 2. wheever N ad P() is true, the P( + ) is also true. Exercise 2: Which of the statemets P(k) listed above satisfy item? Which of them satisfy item 2? No proof is eeded here. You are just buildig ituitio. I a proof by iductio, showig that P() is true is called the base case, whereas showig item 2 i Theorem 2.7 holds for P(k) is called the iductio step. Schematically, here are the atural umbers:

2 Math F25: Iductio April 7, 203 We will circle iteger if P() is true. So showig P() is true is showig that at least the atural umber has a circle aroud it: There might be other circles, ad we do t care about them (yet). We just eed to kow that the first atural umber,, has a circle. To show that P(k) satisfies item 2 of Theorem 2.7 you eed to show that every atural umbers that has a circle also has the property that the atural umber to its right is also circled. How do you prove that all the elemets of a mathematical collectio have a give property? You cosider a arbitrary member of the collectio ad prove that it has the desired property. If a arbitrary elemet of the collectio does, the they all do. The laguage for doig this is the same as we ve used all alog: you suppose that is a arbitrary circled atural umber. The you prove, from this assumptio ad other kow facts, that + must also be circled. More formally, you suppose that for some N that P() is true. You are assumig that the world looks like this: There may be other circles, but you do t kow about them, ad (for the most part) you do t really care about them either. The assumptio that there is a circle about iteger is kow as the iductio hypothesis. It is importat to realize you do t kow if ay atural umbers are circled. Maybe there are oe. You are simply showig that if oe of them were, the the ext oe would be as well. Here are the iductio hypotheses you would use for the first few examples of statemets P(k) listed above. Suppose for some N that. Suppose for some N that 2 > + 0. Suppose for some N that is eve. Write dow the iductio hypotheses for the remaiig examples of state- Exercise 3: mets P(k). If you proved the base case first, the you do kow there is at least oe circled atural umber, amely. But most proofs by iductio ca prove the base case ad the iductio step i either order; its just traditioal to prove the base case first. Withi the proof of the iductio step, you do ot suppose ad do ot use the fact that has a circle. 2

3 Math F25: Iductio April 7, 203 To complete the iductio step, your job is to show that, that P( + ) must also be true. That is, there must be at least oe more circle: Here s what eeds to be proved to complete the iductio step for the first few example statemets. +. ( + ) 2 > ( + ) is eve. Exercise 4: Fiish this list for the remaiig example statemets P(k). I provig these facts, you are oly allowed to assume that P() is true, alog with ay kow facts established before you started your proof by iductio. Sometimes it ca be awkward or tricky to show that P( + ) is true oly kowig that P() is true. There is aother versio of iductio that allows you to assume more. Theorem 2.7 SI (Strog Iductio): Suppose P(k) is a statemet depedig o k N. To show that P(k) is true for all k N, it is eough to show that wheever N ad P(k) is true for all k N such that k <, the P() is also true. We ll say that P(k) satisfies the strog iductio coditio if it satisfies the bulleted item above. We ll prove this theorem i a momet. For ow, let s see how this versio of iductio is differet. I order to show that P(k) satisfies the strog iductio coditio, you cosider a arbitrary N such that P(k) is true for all k N with k <, ad your job is to show that P() is also true. So the iductive hypothesis i a proof via strog iductio is that for some N the world looks as follows: That is, we get to assume that all the atural umbers to the left of have a circle. Assumig all these facts, to complete the iductio step we show that P() must also be true, i.e. 3

4 Math F25: Iductio April 7, 203 You might be woderig what happeed to the base case. To see why it is t eeded ow, suppose that P(k) is a statemet that satisfies the strog iductio coditio. Is P() is true? Sice P(k) satisfies the strog iductio coditio, we kow that if P(k) is true for all atural umbers with k <, the P() will also be true. Look at the atural umbers: Is it true that every atural umber to the left of has a circle? Yes! Every sigle atural umber to the left of has a circle. Not oe of them fails to have a circle. We kow that if P(k) is true for all atural umbers less tha, the P() must be true. So the world must look at least like: I coclusio: if P(k) satisfies the strog iductio coditio, the P() is ecessarily true. So whe you prove that P(k) satisfies the strog iductio coditio, as a part of this demostratio you will ecessarily be showig that P() is true. Here are some iductive hypotheses you would use for strog iductio. Suppose for some N that k for all k N such that k <. Suppose for some N that k 2 > k + 0 for all k N such that k <. Suppose for some N that k is eve for all atural umbers k <. To complete the iductive step, you the prove the followig:. 2 > + 0. is eve. Exercise 5: For each of the remaiig example statemets P(k) give o page, write dow the strog iductive hypothesis, ad what would have to be proved to complete the strog iductive step. 4

5 Math F25: Iductio April 7, 203 I doig a proof by strog iductio, your iductive hypothesis will be that for some N that P(k) is true for every atural umber less tha. Note that if happeed to be, (ad it might be!) you would t be assumig aythig at all! I this way, your proof by strog iductio will effectively prove the base case. Here s a example of strog iductio i actio. Theorem 2.32 (Well Orderig Priciple): smallest elemet. Suppose A N is o-empty. The it has a Proof. Let P(k) be the statemet that all subsets of N cotaiig the iteger k have a least elemet. If we ca show that P(k) is true for all k N, the we will have show that every oempty subset of N has a least elemet. After all, suppose A N is oempty. The there exists k N such that k A. Sice P(k) is true, A has a least elemet. We show that P(k) is true for all k N by strog iductio. Suppose for some N that P(k) is true for all atural umbers k <. So we kow that if k is a atural umber less tha, ad if a subset A N cotais k, the A has a miimal elemet. Our job is to show that P() is true. That is, we eed to show that if A N, ad if A, the A has a miimal elemet. So let A be a arbitrary subset of N that cotais. Now either A cotais a atural umber less tha or it does ot. If it does, our (strog) iductio hypothesis implies that A has least elemet. Suppose A does ot cotai ay atural umbers less tha. We will show that is the least elemet of A, i.e. that. A, ad 2. a wheever a A. We have assumed that A, so we oly eed to establish item 2. Let a A. The, a N as well. Sice A does ot cotais ay elemets of N less tha, Propositio 2.8 implies that a. So satisfies item 2 as well, ad is hece the least elemet of A. It remais to give the proof of Theorem 2.7 SI. Theorem 2.7 SI (Strog Iductio): Suppose P(k) is a statemet depedig o k N. To show that P(k) is true for all k N, it is eough to show that that P(k) satisfies the strog iductio coditio: wheever N ad P(k) is true for all k N such that k <, the P() is also true. Proof. Suppose P(k) is a statemet satisfyig the strog iductio coditio. 5

6 Math F25: Iductio April 7, 203 Let T = { N : P(k) is true for all atural umbers k < }. We will show that T = N. Suppose for a momet that we have doe this. If N, the + N as well. Sice T = N, + T. This implies that P(k) is true for all k N with k < +. I particular, P() is true. So if we ca show that T = N, the we ca coclude that P(k) is true for all k N. We will show that T = N by applyig Propositio 2.6. That is, we will show that N, ad that wheever T, the + T. To see that T, we recall from Propositio 2.20 (ad Propositio 2.8) that there are o atural umbers less tha. Hece, for all k N such that k <, P(k) is true. The strog iductio coditio the implies that P() is true. Suppose that for some N that T. So for all k N with k <, P(k) is true. We wish to show that + T, i.e. that P(k) is true for all atural umbers k less tha +. Observe that P(k) is true for all atural umbers k <, sice T. Moreover, sice P(k) satisfies the strog iductio coditio, we kow that P() is true. Hece P(k) is true for all atural umbers k. Now if k N ad k < +, Corollary 2.22 implies that k. Hece P(k) is true for all k N such that k < +. So + T, as desired. 6