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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.
