Mathematical Iductio
Itroductio Mathematical iductio, or just iductio, is a proof techique. Suppose that for every atural umber, P() is a statemet. We wish to show that all statemets P() are true. I a proof by iductio, we show that P(1) is true, ad that wheever P() is true for some, P + 1 must also be true. I other words, we show that the coditioal P P( + 1) is true for all. This creates a logical chai reactio the truth of P(1) implies the truth of P(), which implies the truth of P(3), which implies the truth of P 4, ad so o. O the ext slide, we will restate this so-called priciple of iductio, prove it, ad itroduce some termiology.
The Priciple of Iductio Theorem (Priciple of Iductio): Suppose P() is a statemet for every atural umber. Further suppose that 1. The first statemet, P(1) is true.. For every, the coditioal P P( + 1) is true. The P() is true for all. Proof: Suppose P() is ot true for all. The there must be at least oe for which it is false. Let 0 be the smallest of all these. Sice we kow that P(1) is true, 0. Sice 0 is the smallest for which P() is false, ad 0 1 is a atural umber, P( 0 1) is true. By assumptio, the coditioal P 0 1 P( 0 ) is true as well, hece P( 0 ) is true. That is a cotradictio because P( 0 ) is false. This proves that P() is true for all. I a iductive proof, verifyig coditio 1 is called the base case or basis step. Verifyig coditio is called the iductio step or iductive step. We verify coditio by assumig that P is true (called the iductive hypothesis) for some arbitrary, ad showig that the P + 1 is true as well.
The Reamig Ritual May textbooks, icludig Rose, highlight the fact that the iductive hypothesis is made about some arbitrary value by givig this a ew ame such as k. I the iductive hypothesis, rather tha assumig P for some arbitrary, they assume P for some arbitrary = k" ad the demostrate that P k + 1 must be true as well. This reamig ritual is ot ecessary, ad has the disadvatage that it may cause serious cofusio whe the statemet about already ivolves some variable amed k. The slide titled Commo Mistakes (4) illustrates this.
Example Ia Let us prove by iductio that for all atural umbers, k = ( + 1) (This equatio is the statemet P. ) 1. Base case: we verify P 1. σ1 k = 1 is true.. Iductive step: let us assume P for some arbitrary, i.e. k = + 1 We ow add the ext term of the summatio, which i this case happes to be + 1, to both sides. O the left side, that upgrades the sum to the ext higher upper limit. O the right, we hope that the right side of P( + 1) emerges. (Note that geerally, whe you write a iductive proof, the statemet P( + 1) will NOT be produced by addig + 1 to both sides of P(). That algebraic maipulatio is SPECIFIC to this example.) This statemet, +1 k = ( + 1) + + 1 = +1 k =. + 1 + 1 = ( + 1)( + ) ( + 1)( + ) is the statemet P + 1. We just completed the iductive step: we showed that if P is true, the P + 1 is true as well. That completes the proof by iductio.,
P( + 1) is ot geerally obtaied by addig + 1 to both sides of P(). It bears repeatig: the algebraic maipulatio we used to obtai P( + 1) from P() i the first example is specific to that example ad oly works because the quatity uder examiatio grows by + 1 as we step from the case to the case + 1. Each iductive proof requires its ow, idividual algebraic steps to coclude that P( + 1) must be true if P() is true. The ext page demostrates a slightly differet algebraic perspective for provig summatio formulas. Istead of addig to the summatio formula we are assumig is already true, we ca split the ext summatio ito what we already kow about, plus the ext term.
Example Ia, doe a little differetly Let us prove by iductio that for all atural umbers, k = ( + 1) 1. Base case: we verify the summatio for =1: 1 = 1. This is true.. Iductive step: let us assume that we already kow σ k = for some arbitrary. The we split the ext sum ito the sum we already kow, plus the ext term: +1 k = k + + 1 We ow use the iductive hypothesis o the sum we already kow: +1 +1 k = ( + 1) + + 1 = + 1 + 1 = ( + 1)( + ). This statemet +1 k = ( + 1)( + ), is the case +1 of our summatio formula. That completes the proof by iductio.
Commo Mistakes (1) Studets ofte cofuse the statemet P() with the algebraic expressio about which a statemet is beig made. They abuse the otatio P(). I the previous example, we proved the statemet P : = k = ( + 1) for all. Notice the paretheses - P is ot the sigma sum, ad it s ot the expressio (+1). It is the statemet that these two quatities are equal. Therefore, you must ot write equatios like P = (+1) i your iductive proofs, if P() has bee agreed upo to be the statemet you are provig. If you must have a otatio for a expressio that is frequetly referred to, you are free to defie your ow otatio for it, for example, ( + 1) a = but you must ot use P. You must also ot write equatios like this: ( + 1) P = k = Without the critical paretheses, P is o loger the statemet that the middle ad right quatity are equal. It is the commo value of that quatity. So o top of usig wrog otatio, you are assumig the coclusio of the proof with this.
It s best to avoid the abstractio of P() i actual iductive proofs To describe the method of iductio i theory, it was useful to have a otatio P() to refer to the sequece of statemets we are provig. I practice though, usig this otatio is ofte worse tha useless. It s just as easy, ad better writig style, to refer verbally to the statemet we are provig. This way, you avoid ay risk of makig the mistake described o the previous page, of abusig P() to refer to both a quatity ivolved i the statemet we are provig ad the statemet itself. Observe how the followig example proofs all do without the otatio P() ad are better for it.
Let us prove by iductio that for all atural umbers, Example Ib k = ( + 1)( + 1). 6 1. Base case: we verify the summatio formula for =1: 1 = 1 3. This is true. 6. Iductive step: let us assume that we have already proved the summatio for some arbitrary, i.e. k = ( + 1)( + 1). 6 We ow add the ext term of the summatio, which i this case is + 1, to both sides. O the left side, that upgrades the sum to the ext higher upper limit. O the right, we expect that the expressio (+1)(+)(+3) 6 emerge. (Observe that the ext term of the summatio is NOT + 1 here. ) will +1 k = ( + 1)( + 1) 6 + + 1 = + 1 6 + + 1 ( + 1) = + 1 + 6 + 1 6 ( + 1) This statemet, = + 7 + 6 6 + 1 = +1 k =, ( + )( + 3) 6 ( + 1)( + )(( + 1) + 1) + 1 = 6 ( + 1)( + )(( + 1) + 1) 6 Observe that prudet, early factorig, rather tha bruteforce distributig, is the algebraic key to a efficiet, readable solutio here. is the summatio formula for the case +1. We just completed the iductive step: we showed that if the formula holds for oe, the it also holds for the ext. That completes the proof by iductio.
Commo Mistakes () A secod, somewhat commo mistake is to cofuse statig P + 1 does ot prove it. with provig it. Merely statig it The followig variatio of the iductive step of Example 1 illustrates that mistake: Let us assume that P is true for some, i.e. The P + 1 is the statemet This completes the proof by iductio. +1 k = k = ( + 1) ( + 1)( + ) The psychology of this mistake seems to be that merely writig P + 1 required some algebra work. You had to replace each by +1 ad simplify, so it may feel like you did somethig. You did, you just did t prove P + 1. Agai the otatio P() is best avoided completely whe you write iductive proofs. It is used here oly because it is coveiet to describe the mistake i questio.
Commo Mistakes (3) Whe P is a summatio formula, people may cofuse the with the ruig variable i the summatio. We earlier proved that k = ( + 1) is true for all. Notice that both sides are fuctios of aloe: is the ONLY free variable i this formula. You may thik that you are seeig aother free variable, k, o the left, but this halluciatio is easily dispelled by rememberig the meaig of the sigma otatio: it meas simply 1 + + 3 + +. Our formula is 1 + + 3 + + = + 1 There is o k there. If you replace k by k + 1, you simply break the sigma sum ad tur it ito somethig urelated. People who are uclear about the distictio betwee k ad may thik that they eed to icremet k whe they form σ +1 k, either istead of icremetig, or i additio to it:. Neither is equal to σ +1 k. +1 (k + 1 ) or (k + 1). From a programmig perspective, you ca thik of the k iside the sigma otatio as a local variable iside a fuctio that is ivisible from the outside. fuctio sum(it ) it sum = 0; for (it ;k<=;k++) sum +=k;
Commo Mistakes (4) If you practice the reamig ritual of assumig i the iductive step that P is true for some = k rather tha just for some, ad if your P ivolves a sigma sum that uses k as a ruig variable, you re i dager of writig wholly osesical expressios like this: k k If you do t see what s wrog with that, cosider the followig equivalet code: fuctio sum(it k) it sum = 0; for (it ;k<=k;k++) sum +=k; The best way to avoid this is to ot reame the iductio variable ad to assume that P is true for some i the iductive step.
Commo Mistakes (4a) Practicig the reamig ritual ca make you feel like you assumed somethig i the iductive step whe you assumed othig. Some studets write the followig icorrect iductive hypothesis: Suppose = k. This is meaigless, because it is assumig that a udefied variable is equal to a udefied variable k; or that the udefied variable ow has the ew ame k. It s ot eve a propositio. It s omittig the etire poit of the iductive hypothesis, which is the assumptio that what we are tryig to prove has already bee proved for some arbitrary value. If you isist, you may give this arbitrary value the special ame k (uless k is already beig used as a ruig variable i a sigma sum.) However, you must ot thik that Suppose = k meas the same as Suppose we already kow that P() is true for some arbitrary = k.
Commo Mistakes (5) What s wrog with the followig iductive step? Iductive Step. Let us assume P() is true for all.. I the iductive step, we prove that the coditioal P P( + 1) is true for all, i.e. we prove P P + 1. We do this by assumig that P() is true for some (arbitrary), which sets up uiversal geeralizatio. The we show that P + 1 must be true as well. By uiversal geeralizatio, sice our was arbitrary, that shows P P + 1. I the proof fragmet above, it is assumed that P() is true for all. That is assumig the coclusio. If we already kow that P() is true for all, we do t eed a proof.
Commo Mistakes (6) Some studets who practice the reamig ritual cofuse the act of givig the specific for which we assume P() is already proved i the iductive step a ew ame with the act of assumig that P() has already bee proved. Istead of writig Iductive Step. Let us assume P() for some = k, they write Iductive Step. Assume = k. This is meaigless sice i that assumptio, ad k are just two udefied variables. Eve if was defied, givig a ew ame is ot the same as assumig that P() is true. The geeral theme of this misuderstadig is to ot distiguish properly betwee umbers ad the statemets they refer to. Do t write we proved =1. Write we proved the case =1. Do t write =1 is true whe you mea the first statemet is true. Numbers do t have truth values. Likewise, i the iductive step, do t write assume. is a iteger ad therefore ca t be assumed. Assume is a arbitrary iteger is still wrog as the iductive hypothesis, because it leaves out what we re actually assumig about that iteger. Correct: assume P() for some arbitrary iteger.
Commo Mistakes (7) Do ot cofuse the coclusio of the base case with the reaso you reach this coclusio. For example, i example 1a, the summatio formula P(1) reduces to the idetity 1=1. That 1=1 is a kow fact, a premise, ad ot i eed of proof. P(1) is the coclusio that follows from that. The followig, wrog base case reverses this relatioship: Base case: for =1, the summatio formula reduces to 1 = 1. This is true. Therefore, 1=1. We already kow that 1=1, ad we are ot provig that a umber is equal to itself. If you are goig to write a formal coclusio to the base case, it must be that the statemet you are provig is true i that case. Correct versio: Base case: for =1, the summatio formula reduces to 1 = 1. This is true. Therefore, P(1) is true.
Example IIa Theorem: For ay positive iteger, 4 1 is divisible by 3. Proof by iductio: Base Case: for = 1, 4 1 = 4 1 = 3, which is divisible by 3. Iductive Step: suppose we have already proved that 4 1 is divisible by 3 for some positive iteger. Our goal is to show that 4 +1 1 = 4 4 1 must the be divisible by 3 as well. There are differet ways to proceed from here. Variatio 1: We ca split four times 4 ito three times that quatity, plus oe times that quatity: 4 4 1 = 3 4 + (4 1). Accordig to the iductive hypothesis, 4 1 is a multiple of 3. Sice 3 4 is a multiple of 3 as well, so is the sum of the two terms, which is 4 +1 1. Variatio : Accordig to the iductive hypothesis, 4 1 is a multiple of 3, i.e. 4 1 = 3k for some iteger k, or 4 = 3k + 1. Substitutig that ito 4 +1 1, we get Thus 4 +1 1 is a multiple of 3. 4 +1 1 = 4 3k + 1 1 = 1k + 3 = 3(4k + 1).
Example IIb Theorem: 1 divides 4 +1 + 5 1 for all atural umbers. Proof by iductio: we first verify the statemet for = 1: 1 divides 4 + 5 1 = 1. Now suppose that 1 divides 4 +1 + 5 1 for some atural umber. This meas 4 +1 + 5 1 = 1k for some iteger k. Let us cosider 4 + + 5 (+1) 1 = 4 + + 5 +1 By usig laws of expoetiatio, we ca rewrite this as 4 4 +1 + 5 5 1. It would be ice here if we could just factor out a commo factor i order to substitute the iductive hypothesis, but that is ot possible. So we split the secod term to make at least a partial factorizatio possible: Therefore, 4 4 +1 + 5 5 1 = 4 4 +1 + 4 5 1 + 1 5 1 4 + + 5 (+1) 1 = 4 4 +1 + 5 1 + 1 5 1 = 1 4k + 5 1. We have demostrated that 4 + + 5 (+1) 1 is divisible by 1. This completes the proof.
Example III Theorem: if S is a fiite set with S =, the P S =, for = 0,1,,3, Proof by iductio: 1. Base Case (=0): if S is a set with S = 0, the P S = 1. This statemet is true, because S = 0 implies S =, hece P S =, so P S = 1.. Iductive Step: suppose we already kow for some arbitrary that if a set has elemets, its power set must have elemets. We must the verify that if S is a set with S = + 1, the P S = +1. To verify this coditioal, we will assume its premise ad show that the coclusio must be true. Let S be a set with S = + 1. Let us sigle out a arbitrary elemet of S ad call it x. Defie R = S {x}. The R =. Now let us cout how may subsets S has. The umber of subsets of S equals A + B, where A = the umber of subsets of S that do t cotai x, B = the umber of subsets that do cotai x. A subset of S that does t cotai x is a subset of R, hece A = P R iductive hypothesis. = by the A subset of S that cotais x is the uio of a subset of R with {x}, hece B = P R =. Therefore, P S = A + B = + = +1. That completes our proof.
Example IV Theorem: < for all itegers 5. Proof by iductio: we first verify the statemet for = 5: 5 < 5 because 5 < 3. This was the base case. Iductive step: Now suppose that < for some atural umber 5. We will show ( + 1) < +1. Sice( + 1) = + + 1, we oly eed to show that + 1 < for 5. The, + + 1 < + = +1 by the iductive hypothesis ad thus ( + 1) < +1. It remais to show that + 1 < for all 5. We do this oce agai by iductio. Base case: + 1 < is true for = 5 because 11 < 3. Iductive step: suppose + 1 < for some 5. Our goal is to show ( + 1) + 1 < +1. Let us add two to the iequality of the iductive hypothesis: + 3 < +. Sice < for, ad certaily for 5, + 3 < + < + = +1. Therefore, ( + 1) + 1 < +1.
A Example of a Icorrect Proof by Iductio False Theorem: for all oegative itegers, =. This statemet is of course absurd. = is true for oly oe umber: = 0. Now let us cosider the followig (icorrect) proof by iductio. Defie P for all oegative itegers to be the statemet =. Base case: for = 0, the statemet = is correct. Iductive Step: suppose = for some arbitrary oegative iteger has already bee show. Multiply both sides of this equatio by +1 to get + 1 = + 1. That completes the proof. It is clear that this proof is icorrect, because ay alleged proof of a false statemet must be icorrect. Let us try to pipoit the mistake. The base case was correctly verified. 0 = 0 is a true statemet. The argumet of the iductive step is correct for all except the first. Whe = 0, the the argumet asks us to multiply by the udefied quatity 1 0. Ideed, ay argumet that claims to prove P P( + 1) for all must at least be ivalid for = 0, because P 0 is a true statemet ad P(1) is a false statemet, so the coditioal P 0 P 1 is false. (The followig coditioals, P 1 P, etc, are actually true, ad our proof above showed that. )