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  • PROFESSOR: Last week we talked about the key components

  • of a proof, propositions, axioms, and logical deductions,

  • and as you probably talked about during recitation,

  • we're not going to worry too much

  • about what axioms or logical deductions that you use.

  • Anything that is reasonable, is fine by us.

  • We're not going to ask you to know what modus ponens is,

  • or to label some law when you make a logical deduction.

  • Just, you know, be reasonable.

  • Any facts you knew coming into this course about mathematics,

  • probably close enough to use as an axiom.

  • Want to make sure your axioms are consistent, but that's OK.

  • Now the exception to this would be, is say we're on an exam,

  • and we ask you to prove some proposition, p.

  • Well you can't say, I already knew p, it's an axiom, check.

  • That's not so good.

  • We're asking you to prove it from some more

  • elementary facts.

  • OK it's also don't want me making wild leaps of faith,

  • or saying it's obvious that, unless it really is obvious.

  • That kind of stuff can get you in trouble.

  • Much better to sort of explain the reasoning in the proof.

  • Now the proofs that we covered last week in recitation,

  • the problems set, were all examples

  • of what are called direct proofs.

  • You start with some axioms, you have some theorems

  • you knew before or you proved along the way,

  • and you make logical deductions until you

  • get to where you want to go, the theorem.

  • We're going to start today with an indirect proof.

  • For example, a proof by contradiction.

  • And this is a little bit different.

  • In a proof by contradiction, you assume the opposite

  • of what you're trying to prove.

  • Then you just take steps for logical deductions

  • forward until you arrive at a contradiction,

  • something where you prove false equals true.

  • Now if you can ever get to the point

  • we approve something is false and true,

  • that means what you assumed at the start had to be wrong.

  • Namely, what you're trying to prove has to be true.

  • So let's write that down, because it

  • can be a little confusing the first time.

  • So to prove a proposition p is true,

  • you will assume that it's false.

  • In other words, that not p is true.

  • And then you use that hypothesis, namely the p

  • is false, to derive a falsehood.

  • In other words, you prove a falsehood is true.

  • And this is called deriving a contradiction.

  • And so it must be that, in fact, p is not false,

  • namely that it's true.

  • Now this works because if you can prove,

  • if not p implies false is true, well from last time,

  • the only way this is a true statement is if this is false,

  • namely p is true.

  • All right?

  • So we can conclude that p is true,

  • if we can show that not P implies a falsehood.

  • Any questions about that?

  • It's sort of a lot of sort of notation,

  • and until you've seen, it can be confusing.

  • So maybe we should do an example.

  • Let's prove that square root of 2 is irrational.

  • Is irrational.

  • OK, everybody knows what an irrational number is?

  • That's something that can't be expressed

  • as the ratio of integers.

  • OK, and probably most people already

  • know that-- how many people have not

  • seen a proof that square root of 2 is a rational before?

  • So most of you have seen a proof of that, good.

  • You know, if you try to do a direct proof for this,

  • it's pretty hard.

  • How do you show there's no way to represent

  • the square root of 2 is as integers a over b?

  • But it's very easy, if we do a proof by contradiction.

  • Now when you're doing a proof by contradiction,

  • always start off by saying, by contradiction.

  • Write that down.

  • And then what you do next is you say,

  • I'm going to assume, for the purpose of contradiction,

  • that p is false and when not p is true.

  • In this case, that would be square root of 2 is rational.

  • So in this case, here's what we're trying to prove.

  • That's p.

  • I'm going to assume not p, namely that square root of 2

  • is irrational number.

  • Then I'm going to get a contradiction or falsehood,

  • and then I'm going to know that p was true after all.

  • All right, so let's see where this leads us.

  • Well, if square root of 2 is rational

  • then we can express it as a over b, where a over b

  • is a fraction in lowest terms.

  • That means a and b have no common divisors.

  • And then I can square both sides,

  • and I get two is a squared over b squared.

  • Then I multiply by b squared, and I get

  • 2b squared equals a squared.

  • And what does that imply about a?

  • What can you tell me about a if it equals--

  • if a squared is 2b squared?

  • Anything special about a?

  • Could a be anything?

  • A squared is even, all right?

  • Because 2b squared is an even number, so a squared is even,

  • and what does that mean about a?

  • A is even.

  • B squared is even, then a is even.

  • So a is even.

  • So we could write that as two divides a.

  • That's the divide symbol.

  • You'll see a lot of that next week.

  • All right if a is even, what do I know about a squared?

  • I know more than just, it's even.

  • What is it?

  • It's multiple of 4, yeah.

  • So that means that four divides a squared.

  • A squared is 2b squared, so that means

  • that four divides 2b squared.

  • Divide each side by 2 means 2 divides b squared.

  • What does that imply about b?

  • B is even.

  • All right, b is even, good.

  • Well, I've got a is even and b is even.

  • I got a contradiction here, somewhere?

  • Yeah a over b was not a fraction in lowest terms,

  • because both and b are even.

  • All right, so that implies a over b is not in lowest terms.

  • And that is a contradiction.

  • Now you'll see that written lots of ways.

  • One is, you can say a contradiction,

  • sometimes you'll see just this sharp symbol written,

  • and that means you've got to a contradiction.

  • Because here, we had it being in lowest terms,

  • and here we have it not in lowest terms.

  • You can't have both at the same time,

  • so you got a contradiction.

  • And that means we've now proved that this assumption was wrong.

  • Square root of 2 is not rational,

  • so it must be irrational.

  • and then we put a little box here at the end,

  • or sometimes you'll see a check, sometimes you'll

  • see QED, that says the proof's done now.

  • It's over.

  • Any questions about that proof?

  • We're going to do a lot of proofs by contradiction.

  • Actually, there's an interesting story behind this proof.

  • As far as we know, it was first discovered by the Pythagoreans,

  • way back when in ancient Greece.

  • And the Pythagoreans were a religious society

  • started by Pythagoras, of Pythagoras theorem frame.

  • Now it sounds weird today, but back then, in ancient Greece,

  • math was a religion, all right?

  • Every once in while you'll see somebody around MIT,

  • and you'll think he must think math is a religion,

  • but back then it really was, and it was ruled by God,

  • because this is ancient Greece.

  • And there were two key gods in this religion, Apeiron

  • and Peros.

  • Now Apeiron was the bad god, and he was the god of infinity,

  • because infinity was considered all that was bad.

  • And I don't think we'll do a lot with infinity this term,

  • but if we do you'll appreciate why that's the case.

  • And Peros was a good god.

  • He was the god of the finite world,

  • and represented everything that was good to the ancient Greeks.

  • Now one of their main axioms, or beliefs,

  • was that there were no irrational numbers.

  • They just didn't exist.

  • Now the reason is, they didn't like irrational numbers.

  • They were bad.

  • Because, well they're infinite.

  • You can't write them as a decimal

  • without repeating forever, and you know, just an infinite sort

  • of decimal representation.

  • They liked rational numbers, you know, like one seventh?

  • That's a good number, because you can write it

  • as 0.142857 repeating.

  • So rational numbers are finite, in that

  • you can always find a repeating pattern of finite length.

  • Irrational numbers are not.

  • They're infinite in that sense of the ancient Greeks.

  • So they said there were no irrational numbers.

  • That was an early axiom.

  • Now they also had an axiom that said

  • that every length of a line was finite, therefore rational.

  • All right, you know the ancient Greeks

  • were good with a compass, and drawing lines

  • with straight edges and stuff.

  • So they said any line that you can construct

  • has a finite length to it, so therefore it

  • has a rational length.

  • That was axiom number two.

  • Now they were good geometers, so they knew, of course,

  • they developed the Pythagorean theorem.

  • But if you took a triangle, and you have side lengths one,

  • the length of the hypotenuse was square root 2.

  • Therefore they had a theorem that said

  • square root 2 was rational.

  • Because you've got the 2 axioms there, right?

  • Now eventually, they discovered a proof like that,

  • that it wasn't rational.

  • It was irrational.

  • This caused quite a stir.

  • First it meant that their axioms were inconsistent.

  • Every theorem they proved was now suspect,

  • once you have inconsistent axioms.

  • Even worse, the devil is in their midst.

  • Square root 2 is infinite is the bad god,

  • and this is the most basic length they had, besides one.

  • So that's a very bad thing.

  • Sort of like, you know, today we were

  • to wake up and discover that there were only nine

  • commandments, and the 10th was planted there by the devil.

  • And you're not sure which one, maybe.

  • That would be a big mess, sacrilege.

  • Well, so what were they to do?

  • This is a disaster of major proportions.

  • So they covered it up, and they denied the result.

  • So they didn't want to publish that proof.

  • So they kept on saying square root 2 was rational.

  • But then, according to legend there was a Deep Throat.

  • Somebody who let out the word and the proof the square root 2

  • is irrational, because that would be very destabilizing

  • for the society, and so they killed them.

  • This is the legend.

  • Now, hard to imagine getting killed

  • over the irrationality of 2.

  • All right, we're going to do a lot more

  • of these kind of proofs in homework

  • and throughout the term.

  • The next proof I want to show you

  • is one of my favorite proof techniques.

  • One of my favorite proofs.

  • And that is a false proof.

  • And we're going to see a lot of these during the term, too.

  • And if we could bring the screens down.

  • Somebody back there to-- yeah.

  • Great.

  • Bring the screens down, I'm going to--

  • and then turn this on for me.

  • So I'm going to prove to you that 90 is bigger than 92.

  • All right?

  • And that hopefully, is not really true.

  • But, you know, watch this proof and see if there's

  • any problems with it.

  • All right, the proof, by PowerPoint.

  • Right away you should be suspicious.

  • I'm going to take two triangles with total area 90

  • and put them together, and then divide them up

  • into four triangles with area greater than 92.

  • And by the conservation of area axiom, which

  • you want to be maybe thinking about,

  • this will imply an area of 90 is larger than an area of 92.

  • And therefore that 90 is bigger than 92.

  • All right, those are my triangles.

  • They are right triangles, 9 by 10.

  • If I put them together, I get a 9 by 10 rectangle.

  • Of course it has area 90, 9 times 10.

  • Now I'm going to slide these triangles across their diagonal

  • so that I get, instead of having 10 on the side,

  • I'm going to slide it so I get a 2 and an 8.

  • OK I had 10 there before, now it's 2 and 8.

  • And then you see I've got this-- I'm

  • going to cut off along the dotted line,

  • and as you can see that dotted line, yeah

  • but won't compute it exactly, but it's

  • a little bit bigger than 2.

  • All right?

  • It's a little bit longer than that 2 there, you

  • could see that.

  • All right, so I'll call it 2 plus, bigger than 2.

  • Now I slice off along the dotted line,

  • and now I'm going to put those two triangles together,

  • and I create a rectangle.

  • Two by two-- little more than two.

  • Now the area of the little rectangle

  • is bigger than 4, cause I got 2 times something bigger than 2,

  • and you can see here I have the 8 left over,

  • and I have 9 plus 2 plus, means it's a little bit bigger

  • than 11 by 8.

  • I got area bigger than 88, add them up, get area bigger

  • than 92.

  • So I started with 90 and I created more than 92.

  • All right, what's the problem here?

  • This would be good if I could do it.

  • I'd get some gold, you know, bars of gold

  • and cut them up, and do the game, and I make more gold.

  • That would be pretty good if I could do that.

  • Because of the conservation of area axiom?

  • Did I assume something there that was too powerful

  • that if I manipulate the area of rectangles like this,

  • that the area needs to be the same?

  • Yeah?

  • My bigger ones aren't closed.

  • Well, let's see, I don't know.

  • Let's go back and see I made those bigger ones.

  • All right, I got my triangles, right?

  • I'm just chopping along the line there, and I got 2 plus and 9.

  • Those are rectangles.

  • They look like rectangles.

  • Yeah.

  • AUDIENCE: [INAUDIBLE]

  • PROFESSOR: Well it looks bigger than 2, doesn't it?

  • Yeah, 2 pluses.

  • Well bigger than 2 is bigger than 2, but, yeah, maybe that

  • should have been a 2 minus.

  • Would that change anything?

  • Yeah, if that had been a 2 minus,

  • then we'd have area less than 4, area less than 88, and then 90

  • would be less than 92.

  • That would be OK.

  • But it sure as heck looks bigger than-- the 2 plus looks

  • bigger than the 2 doesn't it?

  • I don't know, do they got distortion out there?

  • I guarantee you, if I measured on my screen,

  • that 2 plus is bigger than the length of the tube.

  • I guarantee you.

  • We can to take a ruler on my screen,

  • and the 2 plus will be longer on the ruler than the 2.

  • That I guarantee you, but you're on the right track.

  • Yeah

  • Good, all right, now how did you--

  • but it is true, if I measured on my screen, it's bigger.

  • Yeah.

  • AUDIENCE: Are the triangles, like, drawn to scale?

  • It's like, on the SAT they always say, not to scale,

  • right?

  • PROFESSOR: Yeah.

  • Yeah, they're not drawn to scale.

  • In fact, if I go back to the beginning, it says 9 by 10.

  • But in fact, if I measure the nine, it's bigger than the 10.

  • All right?

  • So this is one of the big problems of proofs by picture.

  • Because look at it, you're not thinking which

  • is bigger, 9 or 10 up there?

  • 10's bigger even though it's not, it's shorter.

  • But when you get down into the proof,

  • well, clearly the 2 plus was bigger than the 2,

  • because it looks that way.

  • It is, on the paper.

  • All right?

  • In fact if you go on a computer, you're probably right,

  • it's 1.8, not 2 plus.

  • But that's how the error crept in and how it was drawn,

  • and then you're going along with a proof,

  • and everything else is just fine.

  • So the mistake is right up front.

  • I drew it wrong.

  • OK.

  • And this happens, you're doing graphs--

  • we'll talk about graphs here in a couple weeks-- over

  • and over again, people do this.

  • They draw it, it looks like this, you accept that

  • and then you're dead.

  • Everything else, there's no hope.

  • Everybody clear what went wrong in this picture?

  • How we got off track?

  • This is sort of a nasty one.

  • OK.

  • Now one of the nice things about proofs

  • is that when there is a bug, if you really write it

  • out step by step, and there's a bug,

  • you can go back and find it.

  • And so this, I didn't sort of really

  • write it out very well step by step,

  • and it was harder to find.

  • Now, all right, so we can pull the screens up.

  • Thanks.

  • OK, so for the rest of today and the rest of this week,

  • we're going to talk about a different kind of proof

  • technique, which is induction.

  • Now, induction is by far the most powerful and commonly used

  • proof technique in computer science.

  • If there's one thing you should know

  • by the time you're done with this class,

  • it's how to do a proof by induction.

  • In fact, if there's one thing you

  • will know by the time we're done with this class,

  • is how to do a proof by induction.

  • And in some sense, when we get to grading the final, how

  • we measure ourselves as instructors,

  • the first test is, can you do the proof

  • by induction question, and do a good proof there?

  • You will see it on the midterm and the final,

  • probably in multiple instances.

  • Now just to make sure you become intimately familiar

  • with induction, we're going to do over a dozen proofs

  • with induction in class, and many more in homework

  • over the next five to six weeks.

  • You'll probably be dreaming about induction,

  • if we're successful here.

  • Soon.

  • The good news is that induction is very easy.

  • Once you get your mind around it, get some practice,

  • it really is not a hard thing.

  • In fact, induction is really just an axiom.

  • So let me state it.

  • Let P(n) be a predicate.

  • If P(0) is true, and for all natural numbers n, p of n

  • implies p of n plus 1 is true.

  • So here I'm saying that this is true for all n.

  • OK.

  • Then for all n, natural numbers, P of n is true.

  • I had another way of saying this without the

  • for alls there, is that if P(0) is true,

  • if P(0) implies P(1) is true, if P(1) implies P(2) is true,

  • and so on, forever, then p of n is true for all that.

  • So then, P(0), P(1), P(2), forever, are true.

  • OK.

  • Now you can sort of see why this is a reasonable axiom,

  • because if P(0) is true, and P(0) implies P(1), then

  • by that modus ponens thing or one of the logical deductions,

  • we know P(1) is true.

  • And if P(1) is true, and P(1) implies P(2) is true,

  • then we know by the same reasoning,

  • P(2) is true, and so on, forever.

  • Now the reason it becomes an axiom is that and so on forever

  • bit is part of it.

  • That's why we need it as an axiom.

  • You could sort of view this as a series of dominoes.

  • You know, I got a domino for each n,

  • and each domino knocks over the next in terms of truth,

  • knocking over corresponds to the truth here.

  • And if I know P(0) is true, that means I knock over P(0),

  • P(0) implies P(1) means P(1) one goes down.

  • P(1) implies P(2) means P(2) goes down, and so forth.

  • Pretty basic.

  • Raise your hand if you think you don't have a lot of experience

  • with induction.

  • All right, yeah, that's pretty typical.

  • About a third to a half of you.

  • So we're going to change that.

  • Let's do a simple proof using induction.

  • So let's prove that for all n bigger than or equal to 0,

  • again natural numbers, that 1, plus 2, plus 3, plus n

  • equals n times n plus 1 over 2.

  • This is actually a useful thing to remember.

  • We're going to use this all term, this identity.

  • Now the first thing, before we prove it,

  • I want to make sure we understand this dot,

  • dot, dot, notation.

  • Because it is the source of a lot of errors.

  • What it means is that you need to fill in the pattern here,

  • which is vague.

  • What it means in this case, you fill in four, five, six all

  • the way up to n minus 1, and then n.

  • That's what it means.

  • Figure out the pattern and fill it in.

  • Pretty risky thing.

  • Now in this case, because that's so vague,

  • there's other terminology we use for this.

  • For example, we would use a big sigma,

  • capital sigma i equals 0 to n of i.

  • That means the sum of i, where i is the integers from 0

  • to n inclusive.

  • Actually, let me put 1 here.

  • Another way to write this-- these are all equivalent ways

  • to write it-- is you could put 1 less than

  • or equal to i, less than or equal to n of i.

  • So you could put something i over the range, from one to n,

  • or you can write it on the bottom if you want.

  • So these are four different ways of writing

  • the same thing, the sum of the natural numbers from 1 to n.

  • And we'll use them all during the term.

  • All right, now there's some special cases

  • that make this a little more interesting.

  • What if n equals 1?

  • I've got 1 plus 2 plus dot, dot, dot, plus n.

  • What do you suppose it equals if n equals 1?

  • 1, because we're summing the numbers from 1 to 1.

  • That's just 1, the number 1.

  • There is no 2.

  • And there aren't two copies of 1.

  • So this notation is very ambiguous.

  • You'll see a 2 here in the sum.

  • If n is 1, you'll see a 1 here and a 1 here.

  • So you've got to be careful.

  • I guarantee you, you'll make a mistake with this.

  • In fact I'm going to show you another false proof later

  • where this comes into play.

  • What about if n is less than or equal to zero?

  • What is it then?

  • Any thoughts?

  • 0.

  • There are no integers to sum, no 1, no 2, no n, because you

  • never get started because-- sorry, n is less than

  • or equal to 0-- because you're summing from 1 to 0,

  • 0 is below.

  • You never-- it doesn't include anything.

  • So these are the conventions to keep in mind with the edge

  • cases here.

  • All right, it's easy enough to check that the theorem is

  • true for certain values of n.

  • For example, if n equals 4, I've got 1 plus 2 plus 3 plus 4.

  • That's 10.

  • And 10 equals 4 times 5 over 2, plugging in the formula.

  • So for any value of n you could check this formula is true.

  • Proving is true for all n.

  • It takes a little more effort unless you use induction.

  • So let's do that.

  • So we'll prove the theorem.

  • Now, whenever you're using a proof by induction,

  • first thing you do is you write down by induction

  • so we know what you're going to do.

  • And the next thing you need to do

  • is figure out, what's your predicate.

  • What's your inductive hypothesis?

  • What's p?

  • So usually p will be the thing you're

  • trying to prove, namely that 1 plus 2 plus 3 up to n

  • is n times n plus 1 over 2.

  • And you state that.

  • You say let p of n be the proposition, the predicate,

  • that the sum i equals 1 to n of i equals n times n

  • plus 1 over 2.

  • And once you've got that established,

  • now we're going to go verify that p 0 is true

  • and that p n implies p n plus 1.

  • So we always have to write this down.

  • The next thing to do is to check what's called the base case,

  • p zero.

  • So let's do that.

  • So we write down base case.

  • Some people call it the basis step.

  • And we have to check that p 0 is true.

  • Well, what's the sum of i equals 1 to 0 of i?

  • 0.

  • There are no terms in this sum.

  • And if I look over there, n times n plus 1 over 2.

  • If n is 0, it equals 0.

  • So we're done with the base case.

  • We've now proved that p0 is true.

  • And the second part is called the inductive step.

  • And here we have to show, for n greater than or equal to 0,

  • we need to show that pn implies pn plus 1 is true.

  • Now how do we show an implication is true?

  • How do I show this is true?

  • What am I going to do to show that's true in general?

  • Yeah?

  • AUDIENCE: [INAUDIBLE]

  • PROFESSOR: Right.

  • Because an implication is true in every case, except for true

  • implies false.

  • So if pn is false, we're done.

  • This implication is true.

  • The only case we have to worry about is p of n is true.

  • So we assume p of n is true.

  • And now we confirm that that means pn plus 1 is true.

  • So we write that down.

  • Assume pn is true.

  • And you might also write down, "for purposes of induction"

  • or "purposes of it verifying the inductive hypothesis." just

  • to let us know why you're assuming it.

  • In other words, you're not assuming

  • pn is true for purposes of contradiction.

  • You're assuming it for purposes of induction.

  • All right, let's do that.

  • That means, in this case, i.e.

  • we assume 1 plus 2 plus up to n equals n times n plus 1 over 2.

  • And we need to show that the n plus 1 case is true,

  • and the 1 plus 2 plus n plus 1 equals n plus 1 times n

  • plus 2 over 2.

  • It's sort of weird, and this does

  • confuse people that aren't familiar yet with induction.

  • It looks like we just assumed what we're trying to prove.

  • We're trying to prove this is true for all n.

  • And we just assumed it.

  • But we're assuming it in the context of establishing

  • this implication is true.

  • And then we apply the induction axiom to conclude pn

  • is true for all n.

  • All right, well, let's rewrite this as 1 plus 2

  • plus n plus n plus 1.

  • Because I've assumed pn, I can rewrite this

  • as an n plus 1 over 2.

  • And now plus n plus 1 out here.

  • That equals, well, I got n squared plus n, here,

  • over 2, plus 2 n plus 2.

  • And that equals n plus one times n plus 2 over 2,

  • which is what we're trying to show.

  • So we've completed, now, the inductive step.

  • We have shown that for all n, pn implies pn plus one for all n

  • greater than or equal to 0.

  • Any questions about that?

  • So, the proof is done.

  • We've done everything we need to imply induction.

  • We've got p0 is true.

  • And pn implies pn plus 1 for n bigger than or equal to zero.

  • Now induction helped us prove the theorem.

  • Did it help us understand why the theorem is true?

  • Do you have any feel for why the theorem is

  • true after seeing the proof?

  • Not really.

  • I don't think-- sometimes induction

  • will give you an understanding.

  • Sometimes it won't.

  • Here you've got no understanding of why the theorem's true,

  • which is sort of unfortunate.

  • Did induction help you figure out the answer to the sum?

  • Namely, say you were trying to derive

  • this answer from this sum.

  • Did induction give you the answer?

  • No.

  • You had to know the answer, namely this, in order

  • to prove it was true.

  • Now later we'll see examples where induction actually

  • can give you the answer, but often it does not.

  • Often, induction gives you no hints, no answer,

  • just prove that it's right once you had the clever idea that,

  • oh, maybe that's the answer.

  • You'll see that with things like the beaver flu problem.

  • Figuring out the inductive hypothesis or the answer,

  • you know, the details of it is hard.

  • But once you do it, then applying the induction,

  • not so hard.

  • It gives you a concrete proof.

  • OK, let's do another one.

  • In fact, we're just going to spend the rest of today

  • doing induction proofs.

  • So for all natural numbers n, 3 divides n cubed minus n.

  • Means than n cubed minus n is a multiple of three.

  • For example, n is 5.

  • 3 divides 125 minus 5, because that's 120.

  • Let's prove that.

  • And we're going to use induction.

  • What do you suppose pn's going to be?

  • What's my predicate or my inductive hypothesis here?

  • Any thoughts?

  • Yeah?

  • PROFESSOR: [INAUDIBLE]

  • PROFESSOR: Yeah, that's you.

  • Go ahead.

  • AUDIENCE: [INAUDIBLE]

  • PROFESSOR: Yeah.

  • First thing you always want to try

  • is you assume that is your induction hypothesis.

  • So we say, let pn be 3 divides n cubed minus n.

  • What's the next thing we do in our proof?

  • Base case.

  • Always easy to forget, but not a good idea.

  • Base case is n equals 0.

  • And sure enough, 3 divides 0 minus 0.

  • So that's done.

  • What's the next step we do?

  • What is it?

  • Next step?

  • Inductive step.

  • And for that we're going to need to show for n bigger

  • than or equal to 0, we want to show pn implies pn plus 1

  • is true.

  • So to do that we assume pn is true.

  • In other words, we assume that 3 divides n cubed minus n.

  • And we're trying to show that 3 divides n plus 1 cubed minus

  • n plus 1.

  • So we take a look at, we examine n plus 1 cubed minus n plus 1.

  • And we want to show it's a multiple of three.

  • Well, let's expand that out.

  • This is n cubed plus 3n squared, plus 3n plus 1.

  • And I subtract off n plus 1.

  • So I get n cubed plus 3n squared plus 2n.

  • Is this a multiple of 3?

  • I need to show that's a multiple of 3 and then I'd be all done.

  • Is it a multiple of 3?

  • Beats me.

  • It doesn't look like a multiple of 3, necessarily.

  • So maybe we need to massage it a little bit.

  • I do know that this is a multiple of 3.

  • I can use that fact.

  • And in proofs by induction you always

  • want to-- if you're not making use of that fact,

  • then you're not really making use of induction.

  • Sort of a warning sign.

  • So I want to use this fact to prove this.

  • Well, let's get a minus n in here.

  • This equals n cubed minus n plus 3n squared plus--

  • if I subtracted an n I got to add an n-- plus 3n.

  • So I've rewritten it.

  • Now is it clear?

  • Now it's clear.

  • Very simple, because 3 divide this by pn.

  • 3 divides that.

  • And 3 divides that.

  • So this is a multiple of 3, because I've

  • got 3 divides 3n squared, 3 divides 3n, and 3 divides n

  • cubed minus n by pn.

  • Or you could say, by the inductive hypothesis.

  • pn is the inductive hypothesis, another name for it.

  • So therefore, 3 divides that, which means 3 divides this.

  • So that means 3 divides n plus 1 cubed minus n plus 1.

  • And we are done with the proof by induction.

  • Any questions about that one?

  • So the key steps in induction are always the same.

  • You write down "proof by induction."

  • You identify your predicate.

  • You do the base case, usually n equals 0,

  • but it could be something else.

  • And then you do your inductive step.

  • Now in general, you could start your induction--

  • you don't have to start it at 0, you could start it

  • at some value b, some integer b.

  • Let's take a look at that.

  • So you could have for the base case,

  • you could have p of b is true, not p of 0.

  • And then for your induction step you

  • would have for all n bigger than or equal to b, pn implies

  • pn plus 1.

  • And then your conclusion is that for all n bigger than

  • or equal to b, pn is true.

  • So inductions don't always have to start at 0.

  • You can pick where you start.

  • Just make sure that you verify the starting point

  • and you verify the implication for all n

  • at that starting point and beyond.

  • All right, let's now do a false proof using induction.

  • We're going to prove that all horses are the same color.

  • So let's go through the process.

  • So the proof, we write down "by induction."

  • Now we need our induction hypothesis, the predicate.

  • So we're going to let that be-- we can't let it be this.

  • Why can't this be our predicate?

  • All horses are the same color.

  • What's wrong with making that be the induction hypothesis?

  • You can't plug anything into it.

  • There's no number here.

  • I've got to have a number, n, to induct on.

  • So what I'm going to say is that in any set of n horses--

  • and let's make n bigger than or equal to 1--

  • the horses are all the same color.

  • All right, now if I prove this is true for all n,

  • well then all horses are the same color.

  • Because in any set of n horses, they're all the same color.

  • So all horses must be the same color.

  • What's the next thing I do?

  • What's the next step?

  • Base case.

  • Now, what am I going to use as my base case here?

  • One horse, OK, or n equals 1.

  • So p of 1.

  • That would say that any set of one horses the horses

  • are all the same color.

  • That's true.

  • I've got one horse.

  • It's the same color as itself.

  • So that's easy.

  • It's true since just one horse.

  • All right, what's the next step of the proof?

  • What's the next thing I do?

  • Inductive step.

  • So I'm going to assume that pn is true to prove pn-- and show

  • pn plus 1 is true.

  • All right, so I'm going to assume

  • that in any set of n horses, the horses

  • are all the same color that I start with.

  • And now I look at a set of n plus 1 horses.

  • So we consider a set of n plus 1 horses.

  • And let's call those horses h1, h2, hn plus 1.

  • What do I know about horses h1 to hn?

  • They're the same color, because pn is true.

  • There are a set of n horses.

  • By p1 they're all the same color.

  • Also, what do I know but h2, h3 and hn plus 1?

  • All the same color, because they're a set of n horses.

  • All right.

  • Well, since the color of h1 equals

  • the color of those guys, h2 to hn,

  • I know h1 is the same color as these guys.

  • I also know that hn plus 1 is the same color as those guys.

  • That means that h1 is the same color as hn plus 1.

  • And all n plus 1 are the same color.

  • And that's pn plus 1.

  • That implies pn plus 1.

  • And I'm done.

  • Now a few years ago we assigned this problem as homework.

  • And we asked students to figure out what

  • went wrong with the problem.

  • Why doesn't this work?

  • And the responses were a little discouraging.

  • Half the class responded, effectively as follows,

  • this example just goes to show that induction doesn't always

  • work.

  • A third of the class said, I always

  • knew that you can't trust mathematics.

  • This example just proves it.

  • That really hurt.

  • And most of the rest were something similar to that.

  • Not exactly what we were looking for in the homework.

  • So now we don't leave it to homework.

  • We do it in class.

  • What's the flaw here?

  • What was it?

  • AUDIENCE: P of n.

  • PROFESSOR: P of n?

  • What's wrong with p of n?

  • AUDIENCE: [INAUDIBLE]

  • PROFESSOR: No, this is a real assumption

  • in any set of n horses-- depends on n-- the horses are

  • all the same color.

  • That is a proposition.

  • Because it depends on n, it's a predicate.

  • It's true or it's false.

  • Now we know it's false, because you could get a set of a couple

  • horses with different color.

  • But it is a proposition.

  • Yeah, way back there?

  • AUDIENCE: [INAUDIBLE] for like a certain set of horses.

  • So even though it's the same number of horses,

  • so it was the same number as a different set of horses

  • it's not something you can assume anymore.

  • PROFESSOR: That's a great point.

  • I did gloss over something here, because in the predicate

  • I've got "in any set."

  • So there's really a "for all sets" sitting out here.

  • So I've going to be careful that I

  • establish that when I'm trying to prove pn implies pn plus 1.

  • So to establish pn plus 1 here, I

  • assume pn which means in any set of n horses

  • they're all the same color.

  • That I'm given.

  • That's pn.

  • I've got to look at any set of n plus 1 horses.

  • So consider any set, not a set, any set of n plus 1 horses.

  • So you pick any set you want.

  • Call them this, h1 through hn plus 1.

  • Well then, the first n are a set of n horses, so I can apply pn,

  • therefore they have the same color.

  • The last n horses in the set are a set of n horses,

  • so I can imply pn.

  • So these guys all have the same color, and I'm on my merry way

  • here.

  • So you're right, I had to do a little bit more work,

  • but I can do that work and I still get a proof.

  • Yeah?

  • AUDIENCE: [INAUDIBLE]

  • PROFESSOR: Good question.

  • Is there a problem with the base case?

  • So my base case was here, n equals 1.

  • In any set of one horse, the horses-- let's say in the set

  • just to be really careful-- are all the same color.

  • No, in any set of one horses there's only one horse,

  • so it's the same color as itself.

  • Yeah?

  • AUDIENCE: [INAUDIBLE]

  • PROFESSOR: n plus 1 is not the same as n, yeah.

  • AUDIENCE: [INAUDIBLE].

  • So you can't have the same assumption

  • because it's not a set of n horses.

  • It's a set of n plus 1?

  • PROFESSOR: Well, let's see.

  • So you're arguing I made a mistake here, right?

  • Well, I've got horses-- horse 2, 3, up to n plus 1.

  • How many horses are in this set here?

  • AUDIENCE: Oh, in that set?

  • PROFESSOR: In this set.

  • How many horses are there?

  • AUDIENCE: N.

  • PROFESSOR: N. So I can apply p of n to this set just the same

  • as that set, because p of n is any set of n horses.

  • Well, I'm picking this one now and applying pn to it.

  • And so therefore they're all the same color.

  • Yeah?

  • AUDIENCE: H1, h2, dot, dot, dot.

  • PROFESSOR: Yes.

  • AUDIENCE: Because it does work it there's two or more,

  • but you didn't prove it with just one-- or from one to two.

  • PROFESSOR: Exactly.

  • Remember I told you that dot, dot, dot

  • is so reasonable, so easy to use.

  • Everybody uses it.

  • It was going to catch us up.

  • The bug is in the dot, dot, dot.

  • And in particular, it has to do with the case n equal 2.

  • So there's two ways to look at the bug,

  • dot, dot, dot, or we didn't completely

  • do all the inductive steps.

  • So let's look at the case-- actually it's the case n

  • equals 1, here, in this inductive step.

  • Did I prove p1 implies p2?

  • Let's just double check that worked for n equals 1.

  • n equals 1.

  • I've got a set of two horses.

  • This becomes 2.

  • So I've got h1, h2.

  • Nothing in the dot, dot, dot.

  • It's just h1 and h2.

  • Then this becomes h1.

  • So sure enough, h1 is the same color as itself.

  • This becomes-- what does this become, this set of horses?

  • What is it really?

  • h2.

  • All right, so I've got the color of h1

  • equals the color of-- oh man, this

  • is so hard to see this bug.

  • I wrote down h2 dot, dot, dot, to hn because-- take out

  • h1, what's left?

  • What's left is h2 through hn.

  • But this set is only h1.

  • What's really left here?

  • What is this set?

  • What is this set?

  • h2, h1?

  • No.

  • You see, what is the whole set? h1, dot, dot, dot, hn.

  • What is the whole thing?

  • It's just h1.

  • pull out h1 and look at the rest of it,

  • how many horses are here?

  • 0 horses.

  • This is the empty set.

  • There are no horses here to compare to.

  • Even though I got an h2, I got an hn, I got a dot, dot, dot.

  • Because generally, if n is big, this

  • has n minus 1 horses are here.

  • There's n total.

  • I got n minus 1 right here.

  • But n minus 1 is 0.

  • There are no horses here.

  • And so this bridge in the equality totally breaks.

  • I got color of h1 equals-- there's no information here--

  • equals the color of h2.

  • There's no equality here, because there's

  • no horses in this set, all because of that dot, dot, dot.

  • Do you see where the problem is by using the dot, dot, dot?

  • For the case n equals-- it was true for every other case of n.

  • n equals 2, n equals 3, it's all true.

  • In fact, what we proved, we proved the base case of p1

  • is true.

  • This is an argument that p2 implies p3.

  • That is true.

  • p3 implies p4, that is true.

  • And so forth.

  • We proved for all n bigger than or equal to 2,

  • pn implies pn plus 1.

  • That we proved.

  • What is the one missing implication we did not prove?

  • The missing link, p1 implies p2.

  • We did not prove that.

  • Now is that true?

  • No.

  • You can find a set of two horses are not the same color.

  • And just because every horse is the same color as itself,

  • does not give you that.

  • That was missing in this proof.

  • And so we have to be really careful

  • when you're doing these proofs that you establish

  • the inductive step for all n bigger

  • than or equal to the base case.

  • And then make sure, if you're going

  • to use this really convenient dot dot dot notation, that you

  • don't wind up here saying, oh this is horses h2 through hn

  • when there's no horses there, because n is 1.

  • Any questions about this?

  • Yeah?

  • AUDIENCE: [INAUDIBLE]

  • PROFESSOR: Great question.

  • All right, let's fix the proof.

  • Start with the base case of p of 2,

  • and now I've got all this done.

  • So therefore, that's another proof.

  • Yeah?

  • Say it again.

  • AUDIENCE: [INAUDIBLE]

  • PROFESSOR: This is still the same.

  • In any set of n bigger than or equal to two horses,

  • all the horses in the set are the same color.

  • That's what he saying.

  • And he's saying, hey look, the proof worked here.

  • The inductive step is just fine.

  • p2 does imply p3.

  • Yeah?

  • AUDIENCE: The base case for 2 isn't true.

  • PROFESSOR: That's right.

  • The base case fails.

  • That's why you've always got to check the base case.

  • Yeah?

  • AUDIENCE: What it does prove is that if you find any two horse

  • and they're always going to be the same color,

  • then all horses have to be the same color.

  • PROFESSOR: That's correct.

  • That's a great point.

  • We have given a proof that if you look at any pair of horses

  • and they're the same color, then all horses are the same color.

  • That's true.

  • That is true.

  • Of course, there are pairs of horses

  • that aren't the same color.

  • So the base case would fail.

  • So always check the base case.

  • You could prove some great stuff if you

  • don't check the base case.

  • All right, any other questions about that?

  • Yeah?

  • AUDIENCE: Negative number [INAUDIBLE]?

  • PROFESSOR: Yeah.

  • As long as it's an integer.

  • And as long as you prove pn implies

  • pn plus 1 starting there all the way out.

  • Yeah, you can start at negative numbers if you want.

  • Usually there's not many cases where that comes up

  • you want to, but you can.

  • Nothing wrong with that.

  • Any other questions?

  • Yeah, you can see why it was a messy homework problem.

  • So far we've seen examples of how induction

  • is useful in proving the hypothesis is true, but not

  • in solving the problem, per se.

  • Or even figuring out what the hypothesis should be.

  • Now in the last example, we're going

  • to show you how induction can be used

  • to prove there is a solution to a problem,

  • and also how to find the solution.

  • So it's actually going to be a very useful, constructive thing

  • in this case.

  • Now this problem arose in the construction

  • of the status center, the building we're in now.

  • This whole building was originally supposed to cost,

  • completely furnished, under $100 million.

  • That was the goal.

  • But the first mistake they made was the first step,

  • was hiring the architect.

  • They hired Frank Gehry.

  • I think MIT is now in a lawsuit with Frank Gehry.

  • So he was the architect.

  • And costs just went nuts.

  • As you could imagine, all these slanted walls

  • and crazy things happening actually are expensive.

  • And the cost quickly got over $300 million, literally.

  • That that's parts true.

  • Now I'm going to fabricate a little bit.

  • Now actually, fund raising became a huge priority

  • once they're more than $200 million

  • over budget they haven't even bought the furniture yet.

  • So some pretty radical ideas were proposed.

  • And one of them was to build a large 2 to the n by 2

  • to the n courtyard and put a statue

  • of a wealthy, potential donor in the center of the courtyard.

  • So let's draw this.

  • So the courtyard-- and of course,

  • you know, it's computer science, so it has to be a power of 2.

  • So it's 2 to the n by 2 to the n.

  • And I've drawn here the case n equals 2.

  • And we've got to get the statue of the wealthy guy

  • in the middle.

  • And I'm not supposed to reveal his name.

  • So we're just going to call him Bill.

  • So Bill's got to go in the center.

  • Now this would be fine, except to that nothing

  • was easy with Frank Gehry, everything

  • was some weird, weird thing going on.

  • And he insisted on using l-shaped tiles

  • for the courtyard.

  • So the tiles that we're going to use looked like this.

  • So it's almost a two by two, except you're

  • missing that piece.

  • So what you need to figure out how to do

  • is tie all this courtyard perfectly leaving

  • one spot for Bill using tiles like this,

  • these l-shaped tiles.

  • And this is 2 by 2 here.

  • So that's the task.

  • So let's see if we can do that for n equals 2 here.

  • Let's see, we can do a tile here.

  • We can do a tile here.

  • A tile here.

  • A tile here.

  • And a tile there.

  • So we can.

  • In this case we can tile the courtyard perfectly

  • using these L shaped tiles, leaving one square

  • in the Center for the statue.

  • All right, everyone understand what we're trying to do?

  • The goal?

  • Now I want to do it for n.

  • So let's start proving it by induction,

  • even though we don't know how to do it yet.

  • Because we're going to see how induction's going to help

  • us show it's possible and maybe even show us how to do it.

  • So let's state a theorem.

  • For all n there exists away to tile a 2 to the n by 2

  • to the n region, or courtyard, with a center square missing

  • for Bill.

  • And the proof will be by induction.

  • And our induction hypothesis, the predicate,

  • pn is going to be what we're trying to prove.

  • So this is the induction hypothesis.

  • Almost always when you do the induction

  • you want to start out with that as your hypothesis.

  • What's the next step?

  • Base case.

  • Never ever forget the base case or you'll

  • be thinking all horses are the same color.

  • p0.

  • Well, the courtyard for n equals 0 is just 1 square.

  • And that's for Bill.

  • So you're done.

  • There's no tiles at all to worry about.

  • That's easy.

  • So that's true.

  • And then we do the inductive step.

  • So inductive step.

  • For n bigger than or equal to 0, got

  • to remember to keep track of that now,

  • we assume pn to verify the inductive hypothesis,

  • or to prove pn plus 1.

  • A lot of ways to write this down,

  • but you always want to say what you're assuming and why.

  • So we need to show pn plus 1 is true.

  • Well, so let's look at a 2 to the n by-- 2 to the n plus 1

  • by 2 to the n plus 1 courtyard.

  • So let's draw it out here.

  • 2 to the n plus 1 by 2 to the n plus 1.

  • What are we going to do to use our inductive hypothesis?

  • Yeah?

  • AUDIENCE: [INAUDIBLE]

  • PROFESSOR: For the-- yeah.

  • So you're on a good track there, for sure.

  • But I've got to apply-- I'm not going from 1 to 2,

  • I want to get-- I want to use pn here.

  • So I've got a 2 to the n plus 1 by 2 to the n plus 1 courtyard.

  • How do I use pn?

  • 2 to the n by 2 to the n courtyard.

  • Yeah?

  • AUDIENCE: Oh, never mind.

  • I don't think it works.

  • I was about to say that would be as if 2 to the n

  • would divide that into four blocks.

  • PROFESSOR: Good idea, yeah.

  • Let's divide our courtyard into four blocks.

  • That's a great idea.

  • And now each of these is 2 to the n by 2 to the n.

  • Right?

  • And I can apply the inductive hypothesis there.

  • AUDIENCE: [INAUDIBLE]

  • PROFESSOR: Yeah.

  • Hm.

  • Yeah, something-- yeah, it doesn't quite

  • work because Bill wants to be here.

  • Got a little square for Bill there.

  • But I can't use my inductive hypothesis to tile

  • this, because there's no square missing.

  • In fact, even if there wasn't a square missing, I'm in trouble.

  • If I've got-- say this is size 4 by 4 here.

  • Can I tile a 4 by 4 region with L shaped tiles?

  • No, they're size 3.

  • 3 doesn't divide into 16.

  • Yeah?

  • AUDIENCE: [INAUDIBLE] There's one tile missing

  • from each of those blocks at the corner that [INAUDIBLE].

  • PROFESSOR: There's a great idea.

  • All right, we're making progress now.

  • Take a corner out of each of them.

  • Put my l-shaped tile here.

  • Now I can use the inductive hypothesis

  • to tile each one of these.

  • Yeah?

  • Yeah?

  • AUDIENCE: Well, why can't you put a 4 in the center

  • and then you have a bunch of 2s on the side?

  • PROFESSOR: A 4 in the center--

  • AUDIENCE: Like a 2 by 2 in the center.

  • PROFESSOR: I got that.

  • I got Bill and the tile.

  • AUDIENCE: No, but like make it bigger.

  • Not just like 4 single tiles, but like-- so you

  • have something like that over there, right?

  • Put that in the center.

  • PROFESSOR: Put that in the center, OK.

  • AUDIENCE: And the rest of them will have like [INAUDIBLE].

  • PROFESSOR: Well, the rest of them aren't 2 by 2,

  • because this is n.

  • I've got to use pn here.

  • And pn says that I can-- in a region 2 to the n by 2

  • to the n with a square out of the center, I can tile it.

  • Am I good here so far?

  • I claimed I was sort of done.

  • Yeah?

  • AUDIENCE: No, because p of n tells us-- well,

  • assume p of n tells us that you can set up a 2 to the n by 2

  • to the n courtyard with a centerpiece of--

  • PROFESSOR: Yeah.

  • AUDIENCE: [INAUDIBLE]

  • PROFESSOR: Yeah.

  • And what's the problem for this 2 to the n

  • by 2 to the n region?

  • Bill's not in the center.

  • He wants to be in the center.

  • We put Bill in the corner.

  • So you can't use the inductive hypothesis here.

  • So I went a little too fast, here.

  • This is not a proof so far.

  • I've got a problem.

  • Yeah?

  • AUDIENCE: [INAUDIBLE]

  • PROFESSOR: Great.

  • OK, let's then change the inductive hypothesis.

  • Good.

  • Let's say there exists a way to tile a 2 to the n by 2

  • to the n region with a corner square missing for Bill.

  • All right, but now I got to put-- oh,

  • but I can make this work.

  • Yeah.

  • Yeah.

  • We can make something-- we can prove this now.

  • The inductive step is going to work

  • because I'll put Bill-- say he's in one of the four regions.

  • Then use the l down here.

  • Put the l here.

  • And now I've got a corner out of each one.

  • And now I'm done by induction.

  • I have proved there's a way to tile any 2 to the n

  • by 2 to the n region with a corner square missing.

  • I think that proof works, Yeah?

  • AUDIENCE: [INAUDIBLE] with Bill in the middle,

  • because to prove that you can put Bill in the corner in a 2

  • by 2 case, you can just blow that square off.

  • [INAUDIBLE] just becomes just another square, which is 4 by 4

  • and then you can put him in the middle.

  • PROFESSOR: That is true.

  • So you've jumped ahead.

  • We have successfully now proved this is true for p of n.

  • But Bill didn't want to be in the corner.

  • We're trying to prove there's a way

  • to do it with Bill in the center, which

  • is not what we proved.

  • And you've come up with a way-- yeah, that might be doable.

  • First prove you can put Bill in the corner.

  • And do it in 2 to the n by 2 to the n's.

  • I don't know about this, actually.

  • I got Bill in the corner.

  • There's some way to do it.

  • But it might have involved doing this.

  • And now I can't rotate that.

  • I don't think we have-- I don't think that proof necessarily

  • works.

  • AUDIENCE: [INAUDIBLE] prove that he could be here.

  • But likewise, we could prove that if the block just

  • rotates you can [INAUDIBLE].

  • PROFESSOR: Yes, I agree with you and I liked it

  • when I first heard it.

  • And you might still convince me.

  • But all we proved is this, there's a 2 to the n by 2

  • to the n-- any 2 to the n by 2 to the n region,

  • we could tile it with a corner, Bill

  • in a corner, any corner square missing.

  • Good, so now you want to say, I can get Bill in the middle.

  • And what you want to say is OK, I

  • tile this region with Bill here, in a 2 to the n by-- maybe

  • you're right.

  • So then I would apply the theorem here, with Bill here.

  • Oh, Bill here.

  • No, I think I like it now.

  • And then I would take these out.

  • AUDIENCE: [INAUDIBLE] just takes up up three blocks.

  • And that big square is just like a zoomed out version

  • of a little square.

  • And since you know that little square fits in a 2 by 2--

  • like for example in the 4 by 4 case.

  • We proved the 2 by 2 case.

  • In the 4 by 4 case you have a big size.

  • It's just like a-- you have four squares, right?

  • And each of those four squares has [INAUDIBLE].

  • So you take that top right square and you put Bill

  • within the smaller square exactly where you

  • want him to be.

  • And then you fill those other-- all those empty spaces.

  • PROFESSOR: I agree.

  • You could make a proof.

  • So in this case you'd make a Lemma

  • that uses induction that says you can do it

  • with Bill in the corner.

  • And then as a corollary or a theorem

  • you'd take that and apply it to four sub-squares.

  • Put Bill in here.

  • Take these out.

  • And then now you'd have your result of Bill in the center.

  • So that is is a way to do it.

  • It ends up being more complicated.

  • There is a simpler approach.

  • But that is a way that works.

  • It is a natural thing you would do if you had this on homework.

  • Is you'd think of a different thing to prove by induction,

  • then use that as a Lemma to get you where you wanted to go.

  • There's another way to do it without having that first step.

  • And that's-- yeah?

  • AUDIENCE: That courtyard where n equals 2.

  • Divide each cell into a 2 by 2--

  • PROFESSOR: Yeah.

  • AUDIENCE: Something.

  • PROFESSOR: Well, yeah.

  • You want to do it bottoms up.

  • You want to take it and make that your inductive step, 2

  • by 2's inside.

  • I haven't thought about that approach.

  • AUDIENCE: [INAUDIBLE] So if you could

  • do that then you're all set except for the 2

  • by 2 where Bill was.

  • PROFESSOR: I'm worried about a lot of 2

  • by 2's with a corner missing if I do it that way.

  • I'll think about that.

  • There is-- let me get to another approach here.

  • We couldn't make it work with the center.

  • We could make it work with a corner.

  • But then we had to do more work after.

  • One general technique to use with induction,

  • when you're having struggling, what's the induction

  • hypothesis to use.

  • If what you've got doesn't work, you could pick a different one,

  • but better to pick a stronger one.

  • Yeah?

  • AUDIENCE: [INAUDIBLE]

  • PROFESSOR: Yes.

  • You could.

  • And that is a good thing to do.

  • Make the induction hypothesis be much stronger.

  • I had trouble proving there's a way to do it

  • with a center square missing.

  • Turns out to be easier to prove it

  • where you say any square could be missing.

  • Seems like this should be harder, right?

  • We had a hard enough time just showing

  • this square missing was doable.

  • But by assuming something-- by trying to prove a harder

  • problem, assuming something stronger in pn,

  • it gets easier to prove.

  • All right, now we better check the base case, p0,

  • but that's easy.

  • There's only one square Bill can be.

  • But now let's look at pn implies pn plus 1.

  • So we go back to this.

  • We've got a 2 to the n plus 1 by 2 to the n plus 1 courtyard.

  • And now Bill can be anywhere.

  • Put him here.

  • There's Bill.

  • Now I'm going to place my first l-shaped tile here,

  • in the other three regions.

  • And now I apply pn to each region.

  • A pn says I can do it with any square missing.

  • So I'll pick this one out here, this one out here,

  • that one there, that one there.

  • And now I'm done.

  • That was easy.

  • No extra steps.

  • Now, how is it possible that it was easier to prove something

  • that was harder?

  • Yeah?

  • AUDIENCE: First need to be in the center [INAUDIBLE],

  • you're putting another constraint on yourself.

  • PROFESSOR: Yes.

  • That is true.

  • But by allowing him to be anywhere,

  • I could have started-- I have to-- that's a possibility.

  • See, what I'm trying to do here, this inductive step

  • is all about proving pn implies pn plus 1 is true.

  • That's what I'm trying to show.

  • Now, how is it useful for me if pn is stronger?

  • Has more?

  • AUDIENCE: You grow it.

  • You prove that he can be in a corner.

  • But when you grow it, the corner moves.

  • But since we proved that it can be in any,

  • it's fine if it moves.

  • PROFESSOR: Exactly.

  • That's exactly right.

  • ph got more powerful, which means

  • I get more to assume here.

  • In the recursive problem, Bill can be anywhere now.

  • It gives me more power.

  • I can tile any courtyard with any square missing.

  • This is more powerful.

  • So this got more powerful.

  • So did this.

  • So what it means is that my tool set

  • is bigger with a stronger pn.

  • And what I'm trying to construct or prove got harder.

  • And sometimes, if I've got more tools,

  • it becomes easier to prove, even a harder thing.

  • And so a general rule with induction

  • is if you don't first succeed, try, try again.

  • Well, the rule with induction is if you don't succeed at first,

  • try something harder.

  • All right?

  • And it's amazing, but it actually works a lot of time,

  • as it did here.

  • If I don't assume something strong enough,

  • and pn is some little weak thing like Bill just in the center,

  • it's not enough to go anywhere.

  • But if I could assume a lot more, like Bill

  • could be anywhere he pleases.

  • Then I could prove lots of things here.

  • So it's all the art-- and you're going

  • to learn this over the next several weeks--

  • it's all the art of what's your induction hypothesis.

  • Picking a good one, life is easy.

  • Picking the wrong one, very painful,

  • as you'll see with beaver flu.

  • OK, that's it for today.

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Lec 2 | 麻省理工學院 6.042J 計算機科學數學,2010年秋季。 (Lec 2 | MIT 6.042J Mathematics for Computer Science, Fall 2010)

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