4 Mathematical Induction

Transcription

1 4 Mathematical Iductio Examie the propositios for all, ( + 1) + = for all, + 1 ( + 1) for all How do we prove them? They are statemets ivolvig a variable ruig through the ifiite set Strictly speakig, each oe of these propositios above is a collectio of ifiitely may propositios We ca verify them for a fiite umber of cases where assumes some specific values Thus we might verify for = 1,,3,, ad covice ourselves of the truth of this statemet, but this is far from a proof O the other had, we caot check the truth of ifiitely may statemets withi fiite time So we must resort to some other meas I order to prove propositios about all atural umbers, a axiom is itroduced It is the fifth Peao axiom about (Giuseppe Peao ( ), a Italia mathematicia ad logicia) It is called the axiom of mathematical iductio 41 Axiom (of mathematical iductio): If S is a subset of such that I 1 S, II for all k, if k S, the k + 1, the S is the whole of, ie, S = We ca use this axiom to prove statemets of the form 'p for all ' as follows We let S be the set of all atural umbers for which p First we verify 1 S, that is, we verify that p 1 Secod, we assume that k S ad uder this hypothesis, which is called the iductio hypothesis, we prove that p k+1 So we show that k S implies k+1 S By the axiom of mathematical iductio, S =, so the 30

2 statemet p for all operatioal procedure We formulate the axiom as a 4 Priciple of mathematical iductio: Let p be a statemet ivolvig a atural umber We ca prove the propositio by establishig that for all, p I p 1 II for all k, if p k the p k+1 Proofs by the priciple of mathematical iductio cosist of two steps I the first step, we show that p 1 I practice, this is ofte quite easy, but we should ot eglect it I the secod step, we assume that p k This assumptio is the iductive hypothesis Usig this hypothesis, we prove that p k+1 A proof by iductio will ot be complete (ad valid) if we carry out the first step but ot the secod, or if we carry out the secod step but ot the first 43 Examples: (a) Prove that ( + 1) + = We use the priciple of mathematical iductio 1(1 + 1) I 1 =, so the formula for = 1 II Make the iductive hypothesis that k = We wat to establish k + (k + 1) = hyp) k + (k + 1) = k(k + 1) (k+1)((k+1) + 1) for all k(k + 1) We have + (k + 1) (by iductive = ( k (k + 1)(k + ) =, so the formula for = k + 1 if it for = k Hece + 1)(k + 1) = ( + 1) for all 31

3 (b) Prove that = +1 for all I We have = 1+1, which proves the assertio for = 1 II Assume k = k+1 Now we must prove k + k+1 = (k+1)+1 We have hyp) k + k+1 = ( k+1 ) + k+1 (by iductive = ( k+1 ) = k+, so the assertio for = k + 1 if it for = k Thus = +1 for all (c) Let h 1 be a fixed real umber Prove that (1 + h) 1 + h for all 1 I We have (1 + h) h, so the iequality for = II Let us assume (1 + h) k (1 + h) k (k + 1)h We have (1 + h) k+1 = (1 + h) k (1 + h) 1 + kh We wat to prove that (1 + kh)(1 + h) (by iductive hyp ad 1 + h 0) = 1 + h + kh + kh 1 + h + kh + 0 = 1 + (k + 1)h, so the iequality for = k + 1 if it for = k By the priciple of mathematical iductio, (1 + h) 1 + h for all Sometimes it is coveiet to use the priciple of mathematical iductio i a slightly differet form We assume (ot oly q k, but rather) each oe of, q, q 3,, q k ad the coclude that q k+1 This establishes the truth of q for all, as the followig lemma shows 44 Lemma: Let q be a statemet ivolvig a atural umber Assume that 3

4 i ii for all k The q for all, if, q, q 3,, q k are true, the q k+1 Proof: We prove the lemma by the priciple of mathematical iductio We put p 1 = p k = ad q ad ad q k (for all k, k ) Now iductio I p 1 (by the hypothesis i) II Make the iductive hypothesis that p k The ad q ad ad q k (defiitio of p k ), q,, q k are all true q k+1, q,, q k, q k+1 are all true ad q ad ad q k ad q k+1 p k+1 (truth value of cojuctio) (by the hypothesis ii) Hece, for all k, if p k the p k+1 By the priciple of mathematical iductio, p for all I particular, q for all So ad q ad ad q for all This completes the proof We ca ow formulate a ew form of the priciple of mathematical iductio This form will be used may times i the sequel 45 Priciple of mathematical iductio: Let q be a statemet ivolvig a atural umber We ca prove the propositio for all, q by establishig that I II for all k, if, q,, q k are true, the q k+1 The statemet ' ' is ot true for all atural umbers, but true for all atural umbers 5 The priciple of mathematical iductio 33

5 ca be used to prove this ad similar propositios Let a be a fixed iteger (positive, egative or zero) ad let p be a statemet ivolvig a iteger a We prove the truth of p for all a by showig that 1 p a for all k a, if p k the p k+1 This is easily see whe we put q = p +a-1 for ad use Priciple 4 with q i place of p There is a similar modificatio of Priciple 45 Exercises Prove the assertios i Ex 1-6 for all by the priciple of mathematical iductio ( - 1) = (3 1) + (3 ) = ( + 1)( + 1) = = ( + 1) = ( + 1)( + 1)( ) 30 6 Prove that for all 5, 7 Prove that for all, 8 Prove that, for ay ad for ay positive real umbers,a,, a, a a +a + +a 9 Prove that, for ay ad for ay positive real umbers, a,, a, a1 a a +a + +a (Hit: if m, the choose so that -1 m Put b =( +a + +a m )/m The use Ex 8 with,a,, a m,a m+1,, a, where a m+1 = = a = b) 34