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PROFESSOR: Understanding
proofs includes the ability

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to spot mistakes in them.

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And as a test of
that skill and also

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your understanding
of induction, let

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me see if I can put
one over on you.

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I'm going to show you a
bogus proof by induction.

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And I'm going to prove something
that's patently absurd.

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Namely, that all horses
have the same color.

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Say they're all black.

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So, there's a bunch of
black horses with maybe

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some highlighted brown manes.

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I'm going to prove
this by induction.

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And for a start, there's no
n mentioned in the theorem,

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so that's common for
various kinds of things

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that you need to
prove by induction.

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So I have to rephrase
it in terms of n.

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It's going to be
by induction on n.

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The induction
hypothesis is going

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to be that any set consisting
of exactly n horses

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will all have the same color.

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Let's proceed to prove this.

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Now, I'm going to use the base
case n equals 1, just because I

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don't want to distract you with
suspicions that the base case

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n equals 0, that is no
horses, is cheating somehow.

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It would be, in fact, be
perfectly legitimate to start

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with n equals 0 and argue that
all the horses in the empty set

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have the same color,
because there's

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nothing in the empty set.

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However, let's
not get into that.

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We'll start with n equals one.

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And indeed, if you look
at any set of one horse,

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it is the same color
of it as itself.

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And in fact, I've proved
the base case n equals 1.

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Let's keep going.

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Now, in the inductive
step, I'm allowed

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to assume that n horses
have the same color, where

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n is any number greater
than or equal to 0.

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Now right here, students
complain that that's not fair,

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because you're already
assuming something false

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and that's absurd.

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Well, yeah, it is absurd.

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But that can't be the problem.

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I'm just allowed to assume
an induction hypothesis.

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All I have to do is prove
that n implies n plus 1.

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Since it's absurd,
there ought to be

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some problem with the proof.

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Well, let's watch
and see if there's

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a problem with the proof.

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So again, I can assume
that any set of n horses

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have the same color.

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I have to prove that any
set of n plus 1 horses

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have the same color.

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How will I do that?

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Well, there's a set
of n plus 1 horses,

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and let's consider the
first n of those horses.

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Now by induction hypothesis,
the first n of them

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have the same color.

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Black, maybe.

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Also by induction hypothesis,
the second set of n horses--

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that is, all but the first
horse-- have the same color.

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And what that tells me is that
the first and the last horse

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have the same color as all
of the horses in the middle.

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And therefore, in fact, they
all have the same color.

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End of proof, QED.

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So, there had better
be something wrong.

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And what's wrong?

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Well, what's wrong is
that the proof that P of n

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implies P of n plus 1 is wrong.

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It looked good, but
the proof that P of n

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implies P of n plus 1 has to
work for all n greater than

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or equal to the base case.

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And if you look
at this proof, it

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doesn't work in the base case.

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When n is 1 and you're trying
to go from the base case

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to two and so on by implication,
the proof breaks down.

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Because what happens with
our argument in the case

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that we're trying to prove
that P of n implies P of n

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plus 1 in the case that n
equals 1, well in that case,

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there aren't any middle horses.

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N plus 1 is 2, so
there's a first horse.

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And that's the first n horses.

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And then the second half of set
of n horses is the other horse,

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and there are no middle
horses that they both have

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a color in common with.

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So, the proof just broke there.

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But, you might not have
noticed because that was

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the only place it was broken.

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This is a case where we were
misled by ellipsis, by the way.

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Where because I was
drawing n plus 1 horses

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with-- showing a lot of horses
with dots in the middle,

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it looked like there were some.

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But they weren't.

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And again, as I said,
the point though

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is that the only
fallacy in this proof

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was that it didn't
work when n was one.

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But it certainly
worked for implying

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that if all sets of two
horses are the same, that does

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imply that all sets of three
horses are the same color.

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And again, it's a false,
we'll imply anything,

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kind of example.

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But even here, the
proof was logically OK.

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But if it breaks in
one place, if there's

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one domino that's missing from
the line when the one before it

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falls, the rest of
them stop falling

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and the proof breaks down.