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PROFESSOR: So we come now to
the topic of induction, which

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is a standard part of a
high school curriculum

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and you've probably seen
before, but is nevertheless

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worth looking at at
the level that we

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look at things in this class.

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So the idea of induction
can be-- one way

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to explain it is this way.

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Suppose that I plan
to be assigning colors

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to the non-negative integers
like, say, in this example,

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I color zero blue and one red
and two blue and three red

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and four and five green,
and it goes on somehow.

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OK.

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Now I'm going to describe to you
a coloring that I have in mind,

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and we'll see whether you
can figure out what it is.

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Here are the properties
that my coloring has.

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First of all, I've
colored zero red,

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and I've also
continued the coloring

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satisfying the following rule.

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If I have an integer that's
next to a red integer,

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then it's red also.

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Any integer next to a
red integer is red also.

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So what's my coloring?

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Well, you obviously realize
that they're all red.

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They have to be.

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And there they are.

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OK.

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This is actually a statement.

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It can be read as a statement
of the rule of induction.

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It's kind of a self-evident
axiom about numbers,

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but let's state it abstractly.

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So first of all, what
induction is assuming

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is that you have some
property of numbers.

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Call it red, R. And R
of zero you're told,

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and you're also told that
R of zero implies R of 1,

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and that R of 1 implies R of
2, and R of 2 implies R of 3,

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and in general R of n implies
R of n plus 1, and so on.

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So we've written it out
this way as an infinite set

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of implications to
emphasize that that's

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what the rule that I stated,
that if an integer is

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next to a red integer then
it's red, is shorthand for.

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It's really a shorthand
for this infinite number

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of different implications,
each of which

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has to hold in order for
you to be able to apply

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the rule of induction.

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Well, what can you conclude
if all of these things hold?

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Well, then you can conclude
that zero is red and one is red

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and two is red and
n is red, and so on.

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OK.

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Now of course, there's
a much more concise way

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to express both these
antecedents above the line

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and the conclusion below
the line using quantifiers,

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namely the antecedents
could simply

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be said by two predicate
formulas, R of 0,

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comma, for all n, R of
n implies R of n plus 1.

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That's really a summary of what
we said on the first slide,

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that if an integer is less than
a red integer, then it's red.

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That is, n plus 1 is next to n.

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If n is red, then
n plus 1 is red.

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Similarly, the stuff below the
line, that R of zero, R of one

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and so on hold is simply
expressed as for all m, R of m

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holds.

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And this is the form
of the induction rule.

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It's read that if
you've proved R of 0

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and you've proved that,
for every n, R of n

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implies R of n plus 1, then you
can conclude that for every m,

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R of m holds, where the
variables are all ranging

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over the non-negative integers.

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By the way, notice that I
used n for the variable name

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above the line in
the antecedent,

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and m for the variable line
below in the consequent.

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I can use any names that I
like for bound variables,

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just as in when you define
a procedure you can name

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the parameters of the
procedure anything you like,

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because they're local variables.

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And I've used an m in the
bottom and an n in the top just

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to emphasize that those
variables have nothing

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to do with each other, which is
a point that sometimes confuses

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students.

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OK.

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Sometimes the rule
of induction is

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explained in terms of dominoes.

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You have all these dominoes
lined up next to each other.

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You knock one over, it knocks
over the next one, and so on.

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If that helps you think
about and remember dominoes,

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that's fine.

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OK.

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Let's apply
induction-- maybe one

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of the most basic and
standard applications

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would be to prove a
numerical identity.

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So let's prove one that
we've actually seen before.

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This is the formula
that we've previously

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proved using the
well-ordering principle

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for the sum of a geometric--
for a geometric sum.

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The sum of R to the 0, R
to the 1, up to R to the n.

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And the claim is that
that's equal to R

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to the n plus 1 minus
1 divided by R minus 1.

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So this sum of n
plus 1 terms actually

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can be expressed concisely
with a fairly single--

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a single, simple term.

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Of course, this only
works if R is not 1,

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because I can't have
the denominator be zero.

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All right.

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How do we prove it?

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Well, I'm going to do the proof.

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And at the same time
that I do the proof,

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I'm going to show you kind
of a standard template

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that you can pull out and
use for induction proofs.

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So the template, it's just
an organizational method

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to do the proof.

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I'm doing the
template in magenta.

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So that's the part that
really is form, not substance.

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There's no math in it,
it's just the structure

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that we're going to
organize the proofs in,

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at least in the beginning.

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So here we go.

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The first thing you
do is tell your reader

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that you're going to be
using proof by induction.

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That helps them
understand what's coming.

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So you begin with the line
proof by induction on n.

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Now n is not in magenta
because sometimes you

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use different variables,
and sometimes there'll

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be many variables
in the assertion.

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So you need to tell
the reader just which

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one is the one that you're going
to be applying induction to.

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All right.

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That said, the most
important part of the proof,

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the part where there's
usually a mistake-- if there's

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a mistake anywhere, it's usually
in the absence of the statement

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of an induction hypothesis, or a
misguided induction hypothesis.

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So the next part of
the template says

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the induction hypothesis P
of n is-- and in this case,

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our induction hypothesis is
that this equality holds.

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That's what we're
trying to prove.

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So the induction
hypothesis is P of n.

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The objective, then, implicitly,
when we're doing induction

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with this induction hypothesis,
is to prove that for all n,

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P of n calls.

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This identity works for all
non-negative integers n.

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OK.

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Having stated the
induction hypothesis,

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the first thing we have to
do is work on the base case.

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That is, prove it
for n equals 0.

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Now we're telling the
reader that it's n equals

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0 because sometimes it's
convenient to start at n

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equals 1 or n equals
2, and then you're

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just concluding that
the property holds

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for all n greater
than or equal to 1,

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or however-- all n
greater than or equal

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to wherever you started.

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So we're going to start at 0,
which is the standard place.

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And what do we have to check?

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We have to check that
the sum on the left,

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when n goes to-- when n is
0, is equal to the sum--

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to the formula on the
right when n is 0.

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Well the sum on the
left, when n is 0,

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it's really just 1, because
it's going from R to the 0

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to R to the 0.

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The R and the R squared,
they're a little misleading

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because they're not
really there when n is 0.

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So the left-hand side is
one, and the right-hand side

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is R minus 1 over R minus 1,
which is 1 since R is not 1.

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So sure enough, it
checks, and we're OK.

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The case n equals 0
has now been proved.

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So the next thing we have
to do in the template

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is to go to the inductive step.

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And that's where we
assume that P of n holds.

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And we're allowed to use
the P of n assumption

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in order to prove that
P of n plus 1 holds,

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where the only thing
that we know about n

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besides that P of
n holds is that n

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is greater than or equal to 0.

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And our proof has to work
for all possible n that

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are greater than or equal to 0.

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All right.

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Well, now we can start doing
the non-template method that

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has to do with the content
of what we're actually

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trying to prove.

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This is what I want to prove.

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This is P of n plus 1.

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It's gotten by replacing the n's
in the previous equation by n

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plus 1's.

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I'd like to have that be.

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OK.

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How do I get to that?

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Well, I can assume P of n, which
kind of looks like it already.

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It's a good head start to
getting to P of n plus 1.

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So I'm allowed to assume that
this equality holds for n.

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I don't know what n
is except that it's

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a non-negative integer, but
this equality holds for n.

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And I have to prove that
it holds for n plus 1.

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Well, if you look at this,
what I'm trying to prove

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is something about the sum that
goes up to R to the n plus 1.

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So given this equation, I
can turn the left-hand side

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into the sum that
I'm interested in.

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That is, the sum of powers of R
up to the n plus 1st power of R

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simply by adding
R to the n plus 1

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to both sides, an
obvious strategic move,

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or tactical move.

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OK.

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So doing that, I
get this equality,

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which I've now proved from
the induction hypothesis.

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Namely, the sum up
to R to the n plus 1,

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which is what I'm
interested in, is

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equal to this
algebraic expression

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on the right-hand side.

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And if I'm lucky, and
of course I will be,

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the right-hand side
is going to simplify

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to be the target expression,
with n replaced by n plus 1.

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So what happens is--
let's put R to the n

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plus 1 over this common
denominator, R minus 1.

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And I get the second
term, and then

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you can do a little bit of
algebraic simplification,

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trivial, and you'll realize
that, sure enough, it

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simplifies to R to the n
plus 1 plus 1 minus 1 over R

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minus 1, which was exactly
the equality that I

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was hoping to prove.

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So in fact, at this
point we can say

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that we've proved P
of n plus 1, and we've

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completed the induction proof.

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We're done.

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OK.

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That is the first basic
example of an induction proof.

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And the whole template
is now visible,

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except maybe there should have
been a QED or a Done statement.

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All right.

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By the way, as an aside, and
we already saw a little problem

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with this, the three dots
that appeared in the sum

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are called an ellipsis.

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Plural is ellipses.

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And they're used
where the writer

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is trying to tell the
reader that there's

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an obvious pattern that the
reader is expected to see,

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which I think is fairly
clear in this case.

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You go-- you know, it's R to
the 0, R to the 1, R to the 2,

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R to the 3, up to R to the n.

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The difficulty is that
sometimes the ellipsis

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can cause some confusion.

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For example, we had to
figure out that when n is 0,

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the left-hand side
actually just meant 1.

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It was just R to the
0, so the R and the R

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squared weren't really there.

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One way to really avoid
that kind of fence post

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00:10:52,940 --> 00:10:58,100
problem where you've shown--
in order to make clear

244
00:10:58,100 --> 00:11:00,680
what the pattern is, you've
shown more than it may-- more

245
00:11:00,680 --> 00:11:03,080
of a pattern than
may always be there--

246
00:11:03,080 --> 00:11:06,620
is to use a precise mathematical
notation where I actually

247
00:11:06,620 --> 00:11:09,980
tell you the pattern
of the i-th term,

248
00:11:09,980 --> 00:11:13,810
and tell you that you should
sum from i equals 0 to n.

249
00:11:13,810 --> 00:11:16,630
So the sigma notation
is shorthand for sum,

250
00:11:16,630 --> 00:11:19,240
and I'm telling you that
the i-th term in the sum

251
00:11:19,240 --> 00:11:23,110
is R to the i, and it's going
to run from i equals 0 to n.

252
00:11:23,110 --> 00:11:25,100
So this is a sort of
mathematical notation

253
00:11:25,100 --> 00:11:29,790
for a for loop or a do loop--
do from i equals 0 to n,

254
00:11:29,790 --> 00:11:36,480
add R to the i plus R to
the i to the accumulator.

255
00:11:36,480 --> 00:11:40,690
And the sum notation is
certainly more precise.

256
00:11:40,690 --> 00:11:42,870
But sometimes, it's
actually harder

257
00:11:42,870 --> 00:11:46,600
to read than simply
showing you the pattern,

258
00:11:46,600 --> 00:11:51,360
because the pattern often
is visible visually.

259
00:11:51,360 --> 00:11:53,460
OK, now let me tell
you a little story.

260
00:11:53,460 --> 00:11:57,230
And it's a made-up story,
but it's kind of fun to tell.

261
00:11:57,230 --> 00:12:01,020
This is the familiar
building, the Stata Center.

262
00:12:01,020 --> 00:12:04,400
And this is actually
a design mock-up

263
00:12:04,400 --> 00:12:09,640
that the architects produced
for the MIT team that

264
00:12:09,640 --> 00:12:13,360
was overseeing the construction
and design of the building

265
00:12:13,360 --> 00:12:16,740
to show what the student
lobby would look like,

266
00:12:16,740 --> 00:12:18,620
the student street.

267
00:12:18,620 --> 00:12:24,170
Now the story goes that part of
the plan for the student street

268
00:12:24,170 --> 00:12:27,260
was to have a plaza
that was going

269
00:12:27,260 --> 00:12:30,070
to be built out of
unit size squares,

270
00:12:30,070 --> 00:12:32,140
but an uncertain number of them.

271
00:12:32,140 --> 00:12:34,230
There was going to be a
parameter that determined

272
00:12:34,230 --> 00:12:35,710
the size of the square.

273
00:12:35,710 --> 00:12:37,670
And the size of the
square was actually

274
00:12:37,670 --> 00:12:40,900
going to be a power of two
by a power of two made out

275
00:12:40,900 --> 00:12:43,940
of unit size tiles.

276
00:12:43,940 --> 00:12:45,840
So there would be
2 to the n times 2

277
00:12:45,840 --> 00:12:50,470
to the n unit size tiles
filling up this square plaza.

278
00:12:50,470 --> 00:12:53,430
And the plaza was to be
tiled with these unit tiles,

279
00:12:53,430 --> 00:12:58,100
but one tile space
was to be left blank

280
00:12:58,100 --> 00:13:01,520
so that the statue
of a-- what was then

281
00:13:01,520 --> 00:13:04,780
the potential donor,
Bill, could be placed

282
00:13:04,780 --> 00:13:06,650
in the middle as an
incentive for him

283
00:13:06,650 --> 00:13:09,470
to donate funds for the
completion of the building,

284
00:13:09,470 --> 00:13:12,110
which indeed he did.

285
00:13:12,110 --> 00:13:16,130
So the puzzle, then,
was put forward

286
00:13:16,130 --> 00:13:18,820
by the architect
Frank Gehry, who

287
00:13:18,820 --> 00:13:21,670
many regard, after
Frank Lloyd Wright,

288
00:13:21,670 --> 00:13:25,210
as the greatest architect
of the 20th century.

289
00:13:25,210 --> 00:13:27,670
Gehry specified for
aesthetic reasons

290
00:13:27,670 --> 00:13:30,220
that he wanted to
the square to be

291
00:13:30,220 --> 00:13:35,390
tiled with L-shaped
tiles that were made out

292
00:13:35,390 --> 00:13:36,660
of three unit squares.

293
00:13:36,660 --> 00:13:38,660
He thought that that would
give a pretty design,

294
00:13:38,660 --> 00:13:40,690
and it actually does.

295
00:13:40,690 --> 00:13:44,930
So here's an example of
tiling the n equals 3 case,

296
00:13:44,930 --> 00:13:47,760
[? 2 ?] [? to the-- ?] [? 2 ?]
[? cubed ?] equals 8x8 plaza

297
00:13:47,760 --> 00:13:49,410
with Bill in the middle.

298
00:13:49,410 --> 00:13:54,080
There is the 8x8 plaza tiled
with these L-shaped tiles, each

299
00:13:54,080 --> 00:13:56,570
consisting of three unit tiles.

300
00:13:56,570 --> 00:14:00,340
So the question was that
the exact size of the square

301
00:14:00,340 --> 00:14:03,530
was to be determined by other
architectural considerations.

302
00:14:03,530 --> 00:14:06,640
So it was parametrized by n,
which is going to be 2 to the n

303
00:14:06,640 --> 00:14:07,450
by 2 to the n.

304
00:14:07,450 --> 00:14:11,710
The question was, can you always
find such a tiling no matter

305
00:14:11,710 --> 00:14:15,800
how big the square is and
leave Bill in the middle.

306
00:14:15,800 --> 00:14:18,176
Well, let's try to
prove it by induction.

307
00:14:18,176 --> 00:14:19,800
The induction
hypothesis-- we're trying

308
00:14:19,800 --> 00:14:22,960
to prove a theorem that, for any
2 to the n by 2 to the n plaza,

309
00:14:22,960 --> 00:14:25,220
we can make Bill
and Frank happy.

310
00:14:25,220 --> 00:14:27,400
That is, Bill's happy
when he's in the middle,

311
00:14:27,400 --> 00:14:30,820
and Frank is happy when
the rest of the square

312
00:14:30,820 --> 00:14:32,840
is covered with L-shaped tiles.

313
00:14:32,840 --> 00:14:35,220
By the way, middle is
a little bit ambiguous

314
00:14:35,220 --> 00:14:37,270
because there are really
four middle squares.

315
00:14:37,270 --> 00:14:39,632
But of course, it
doesn't matter which one

316
00:14:39,632 --> 00:14:41,590
you fill, because if you
wanted a different one

317
00:14:41,590 --> 00:14:43,240
you could just rotate
the whole square

318
00:14:43,240 --> 00:14:46,590
and get any one of the
four middle squares

319
00:14:46,590 --> 00:14:50,450
empty for the Bill statue.

320
00:14:50,450 --> 00:14:54,530
So an induction proof
would proceed by induction

321
00:14:54,530 --> 00:14:55,690
on something or other.

322
00:14:55,690 --> 00:14:57,550
And the obvious
thing is the n that's

323
00:14:57,550 --> 00:14:59,320
in the statement of the theorem.

324
00:14:59,320 --> 00:15:02,090
And the induction hypothesis
would straightforwardly

325
00:15:02,090 --> 00:15:04,260
be that we can
tile to 2 to the n

326
00:15:04,260 --> 00:15:07,760
by 2 to the n plaza
with Bill in the middle.

327
00:15:07,760 --> 00:15:08,260
OK.

328
00:15:08,260 --> 00:15:10,410
The base case is n equals 0.

329
00:15:10,410 --> 00:15:12,770
That's a 2 to the
1 by 2 to the 2.

330
00:15:12,770 --> 00:15:15,470
It's a 1x1 square.

331
00:15:15,470 --> 00:15:17,640
OK, well, not a problem.

332
00:15:17,640 --> 00:15:19,600
You just put Bill
in the one square,

333
00:15:19,600 --> 00:15:21,950
and you tile the rest
with no L-shaped tiles.

334
00:15:21,950 --> 00:15:23,190
That fits the rules.

335
00:15:23,190 --> 00:15:27,590
The base case n equals
1-- n equals 0 is covered.

336
00:15:27,590 --> 00:15:28,320
All right.

337
00:15:28,320 --> 00:15:32,090
So now we come to the
double size square,

338
00:15:32,090 --> 00:15:34,380
the square that's of
size 2 to the n plus 1

339
00:15:34,380 --> 00:15:36,180
by 2 to the n plus 1.

340
00:15:36,180 --> 00:15:38,650
I have to tile that
with Bill in the middle,

341
00:15:38,650 --> 00:15:40,840
but I have a fairly
powerful induction

342
00:15:40,840 --> 00:15:43,660
hypothesis that I'm
allowed to assume, namely

343
00:15:43,660 --> 00:15:47,560
that I can tile the half
size square, the 2 to the n

344
00:15:47,560 --> 00:15:50,840
by 2 to the n square, and
get Bill in the middle.

345
00:15:50,840 --> 00:15:53,360
So obviously, the
double size square

346
00:15:53,360 --> 00:15:58,370
is made out of four
half-sized squares.

347
00:15:58,370 --> 00:16:03,216
And so I can try to fill
up the whole square that's

348
00:16:03,216 --> 00:16:05,840
2 to the-- the whole full-sized
square, or double-sized square,

349
00:16:05,840 --> 00:16:08,300
2 to the n plus 1 by
2 to the n plus 1,

350
00:16:08,300 --> 00:16:14,070
by working with my ability to
tile them with L-shaped tiles,

351
00:16:14,070 --> 00:16:19,130
leaving Bill in the middle for
each of those four subsquares.

352
00:16:19,130 --> 00:16:24,710
So I can assume that,
and now I'm stuck really.

353
00:16:24,710 --> 00:16:25,500
What do I do?

354
00:16:25,500 --> 00:16:29,330
How do I use this ability to
put Bill in the middle of each

355
00:16:29,330 --> 00:16:32,450
of those four quadrants
in order to color--

356
00:16:32,450 --> 00:16:35,710
to fill in the whole
thing with N-shaped tiles?

357
00:16:35,710 --> 00:16:37,510
I'm stuck.

358
00:16:37,510 --> 00:16:41,970
And the point of this
example is to show you

359
00:16:41,970 --> 00:16:46,580
the way to get unstuck,
which is kind of unexpected.

360
00:16:46,580 --> 00:16:49,170
I'm actually going
to get unstuck

361
00:16:49,170 --> 00:16:51,640
by proving something stronger.

362
00:16:51,640 --> 00:16:55,130
I'm actually going to prove
that we can find a tiling using

363
00:16:55,130 --> 00:16:59,650
L-shaped squares with Bill
placed in any specified square

364
00:16:59,650 --> 00:17:00,600
that you like.

365
00:17:00,600 --> 00:17:02,190
Wherever you want
to put him, I can

366
00:17:02,190 --> 00:17:03,990
tile the rest with
L-shaped tiles

367
00:17:03,990 --> 00:17:08,349
and leave the specified
single square blank for Bill

368
00:17:08,349 --> 00:17:11,700
to be inserted, for a statue
of Bill to be put there.

369
00:17:11,700 --> 00:17:13,605
So what's unintuitive
about this is that I'm

370
00:17:13,605 --> 00:17:14,730
proving something stronger.

371
00:17:14,730 --> 00:17:16,339
It ought to be harder
to prove, right?

372
00:17:16,339 --> 00:17:20,760
But because I'm trying to prove
a conclusion that's stronger,

373
00:17:20,760 --> 00:17:24,190
I also have a stronger
induction hypothesis

374
00:17:24,190 --> 00:17:27,060
to work with in
conducting the proof.

375
00:17:27,060 --> 00:17:29,950
And the net proof
actually, as you'll see,

376
00:17:29,950 --> 00:17:31,810
is going to be easier.

377
00:17:31,810 --> 00:17:35,000
So let's do it with the
stronger induction hypothesis.

378
00:17:35,000 --> 00:17:38,400
The theorem is, again, for any
2 to the n by 2 to the n plaza,

379
00:17:38,400 --> 00:17:40,580
so we can make Bill
and Frank happy.

380
00:17:40,580 --> 00:17:42,100
Prove by induction on n.

381
00:17:42,100 --> 00:17:44,690
But with a revised
induction hypothesis--

382
00:17:44,690 --> 00:17:46,540
I'm calling it P
of n again-- which

383
00:17:46,540 --> 00:17:51,200
is I can tile the square
with Bill anywhere.

384
00:17:51,200 --> 00:17:54,870
So the base case, n equals
0, is the same as before.

385
00:17:54,870 --> 00:17:55,810
It's just 1x1.

386
00:17:55,810 --> 00:17:58,925
So I put Bill in the
only tile that there

387
00:17:58,925 --> 00:18:02,190
is, which is both the middle and
the corner and everything else.

388
00:18:02,190 --> 00:18:04,470
And the base case
doesn't change.

389
00:18:04,470 --> 00:18:08,830
For the inductive step, now
I have a more powerful thing

390
00:18:08,830 --> 00:18:11,350
that I can assume is the
induction hypothesis.

391
00:18:11,350 --> 00:18:14,810
I can assume that, in any
given square of a 2 to the n

392
00:18:14,810 --> 00:18:18,190
by 2 to the n-- any given tile
location, unit square, of a 2

393
00:18:18,190 --> 00:18:21,380
to the n by 2 to
the n plaza, I can

394
00:18:21,380 --> 00:18:23,560
tile the rest with
L-shaped squares

395
00:18:23,560 --> 00:18:27,040
and get Bill where
I wanted him to be.

396
00:18:27,040 --> 00:18:30,220
And I have to use
that hypothesis

397
00:18:30,220 --> 00:18:35,765
to show that I can get Bill
anywhere that's required in a 2

398
00:18:35,765 --> 00:18:38,630
to the n plus 1 by 2
to the n plus 1 square.

399
00:18:38,630 --> 00:18:43,620
So suppose that we want to
tile Bill in that designated

400
00:18:43,620 --> 00:18:47,680
arbitrary square of the 2 to
the n plus 1 by 2 to the n

401
00:18:47,680 --> 00:18:50,910
plus 1 plaza where we
happen to choose a location

402
00:18:50,910 --> 00:18:53,590
where Bill is in the
upper right quadrant,

403
00:18:53,590 --> 00:18:56,610
in the upper right
half size square.

404
00:18:56,610 --> 00:18:57,270
All right.

405
00:18:57,270 --> 00:19:01,230
So my hypothesis, I can
fill in the purple square,

406
00:19:01,230 --> 00:19:03,120
that quadrant, with
L-shaped tiles,

407
00:19:03,120 --> 00:19:07,730
leaving Bill in the place
that he's supposed to be.

408
00:19:07,730 --> 00:19:08,940
Well, here's the trick.

409
00:19:08,940 --> 00:19:12,190
With the other three, since
I can tile them with Bill

410
00:19:12,190 --> 00:19:15,230
anywhere, I'm going
to tile them with Bill

411
00:19:15,230 --> 00:19:20,320
in the respective corners of
those three other subsquares,

412
00:19:20,320 --> 00:19:25,990
which meet in the center of the
full-size plaza, as shown here.

413
00:19:25,990 --> 00:19:28,010
And having done that,
now it's obvious

414
00:19:28,010 --> 00:19:30,840
how to fill up the
whole 2 to the n plus 1

415
00:19:30,840 --> 00:19:34,770
by 2 to the n plus 1 plaza,
because I pull those four

416
00:19:34,770 --> 00:19:38,130
separate pieces together to form
the full-size 2 to the n plus 1

417
00:19:38,130 --> 00:19:39,480
by 2 to the n plus 1 plaza.

418
00:19:39,480 --> 00:19:42,770
And look, I just put an
L-shaped tile in the middle

419
00:19:42,770 --> 00:19:46,390
to fill up those three
corner Bills, and I'm done.

420
00:19:46,390 --> 00:19:48,130
And the proof is complete.

421
00:19:48,130 --> 00:19:52,310
We have just proved by
induction that indeed you

422
00:19:52,310 --> 00:19:58,110
can tile any power of 2 by power
of 2 square putting-- leaving

423
00:19:58,110 --> 00:20:01,150
Bill wherever you want
him, and the rest filled

424
00:20:01,150 --> 00:20:03,480
with L-shaped tiles.

425
00:20:03,480 --> 00:20:06,570
Now notice that a
part of this process

426
00:20:06,570 --> 00:20:09,910
actually is implicitly
defining a recursive procedure

427
00:20:09,910 --> 00:20:12,060
to actually do the tiling.

428
00:20:12,060 --> 00:20:14,020
If you watch the
way the proof went,

429
00:20:14,020 --> 00:20:16,290
if I was going to write a
recursive procedure to do

430
00:20:16,290 --> 00:20:18,850
the tiling, what I
would do is say OK,

431
00:20:18,850 --> 00:20:20,710
you give me input
n plus 1, which

432
00:20:20,710 --> 00:20:23,460
are the dimensions-- the
specification of the dimensions

433
00:20:23,460 --> 00:20:26,560
of the plaza, input
n plus 1-- or input n

434
00:20:26,560 --> 00:20:28,850
means it's 2 the
n by 2 to the n.

435
00:20:28,850 --> 00:20:29,950
How do I do that?

436
00:20:29,950 --> 00:20:32,310
Well, you tell me
where you want Bill

437
00:20:32,310 --> 00:20:34,940
to be as another
parameter, and then I

438
00:20:34,940 --> 00:20:43,370
will call myself recursively
on four half size squares.

439
00:20:43,370 --> 00:20:46,340
So that is, call
myself to do squares

440
00:20:46,340 --> 00:20:50,180
with dimension parameter n minus
1 four times for each quadrant,

441
00:20:50,180 --> 00:20:52,780
each time specifying
in those quarters

442
00:20:52,780 --> 00:20:54,130
where I want Bill to be.

443
00:20:54,130 --> 00:20:58,530
The recursive procedure will
return an L-shaped tiling

444
00:20:58,530 --> 00:21:02,960
of those four pieces, and then
I take those, fit them together,

445
00:21:02,960 --> 00:21:07,080
tile that middle, and
I wind up with a tiling

446
00:21:07,080 --> 00:21:08,212
of the whole region.

447
00:21:08,212 --> 00:21:09,670
So what I've just
talked through is

448
00:21:09,670 --> 00:21:13,190
the description of a very easily
written recursive procedure

449
00:21:13,190 --> 00:21:17,060
that would print out a picture
of an L-shaped tiling given,

450
00:21:17,060 --> 00:21:19,750
as input, any number n.

451
00:21:19,750 --> 00:21:25,320
And that's, in fact, that
how we got the 8x8 tiling,

452
00:21:25,320 --> 00:21:28,300
although we did it by hand
rather than writing a program.

453
00:21:28,300 --> 00:21:31,250
And that's enough
of two examples

454
00:21:31,250 --> 00:21:34,790
of basic mathematical induction.