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PROFESSOR: It's doubtful if
you really understand something

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if you can explain
why it's true.

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That's what proofs are
about in mathematics

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and in computer science.

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So we're going to be talking
about proofs of lots of things

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that we're trying to understand.

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And in particular,
we're going to look

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at a proof technique now called
proof by contradiction, which

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is probably so
familiar that you never

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noticed you were using it.

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And now we're going to call
explicit attention to it,

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and think about it.

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So let's do an example first
to see what's going on.

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Suppose that I wanted to prove
that the cube root of 1,332

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was less than or equal to 11.

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Or more precisely,
suppose I didn't know

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and I'm asking this question,
is the cube root of 1,332

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less than or equal to 11?

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Well, one way to do it
would be to simply compute

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the cube root of 1,332, which is
a small bother, but manageable.

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But there's a simpler
way than figuring out

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how to compute a cube root
of a four-digit number.

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Let's just suppose
that this inequality

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was true-- that is, that
the cube root of 1,332

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was less than or equal to 11.

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Well, if that was true,
then what I could do

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is cube both sides.

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And I'll conclude that 1,332 is
less than or equal to 11 cubed.

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Now, 11 cubed is a
lot easier to compute

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than the cube root of 1,332.

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As a matter of fact,
11 cubed is 1,331.

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Wait a minute,
I've just concluded

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that 1,332 is less than 1,331.

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That's obviously not
true, which means

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that my assumption that
this inequality held

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doesn't make sense.

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It leads to this
immediate contradiction,

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which means that in fact,
the inequality doesn't hold.

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And I have now precisely
and unambiguously-- I

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hope clearly-- proved that
the cube root of 1,332

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is greater than 11, even though
we never actually computed

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the cube root of 1,332.

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This is kind of a [? toy ?]
and simple-minded example

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to illustrate proof
by contradiction.

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So let's step back and explain,
and say what it is in general.

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If an assertion implies
something false,

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then the assertion
itself must be false.

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That's what's going on here.

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If you're reasoning step
by step, and at every step

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your reasoning is good-- which
means that if you had something

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true and then you reached a
conclusion from it in one step,

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the conclusion that you reached
was also true-- then if you

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start off with some
assumption, you

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keep proving things step by step
in a way that preserves truth,

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and you arrive at
something false,

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it's inevitable that
what you started with

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must have been false.

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Or else the thing you finished
with would have been true.

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OK, let's look at a
real example of this--

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an amazing fact that was
known thousands of years

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ago to the ancient Greeks, which
is that the square root of 2

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is irrational.

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Now, let's remember that a
rational number is a fraction.

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A rational number is a
quotient of integers.

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And the way we're going to
prove that the square root of 2

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is not a quotient of integers
is by assuming that it was.

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So let's assume that
the square root of 2

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was a rational number,
which means that we've

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got integers n and d without
common prime factors, such

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that the square root of
2 is equal to n over d.

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What I'm doing here is
I'm saying squared of 2

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as a fraction, n over d.

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And we know that you can
always reduce a fraction

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to lowest terms,
which means there

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are no common prime factors.

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So let's get that done.

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We have the square root
of 2 is equal to n over

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d, with no prime that
divides both n and d.

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From this assumption,
I'm going to prove to you

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that both n and d are even.

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And that, of course, is an
immediate contradiction,

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because then both n and d
have the common factor 2.

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So all I've got to do
in order to conclude

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that the square root of 2 is
an irrational number-- it's not

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a fraction-- is prove to
you that n and d are both

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even if the square root
of 2 is equal to n over d.

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Let's do that.

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We'll start off with what I'm
assuming-- square root of 2

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is n over d.

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And let's get rid
of the denominator.

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So let's multiply
through both sides by d,

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and get that the square root
of 2 times d is equal to n.

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Let's get rid of
the square root of 2

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now by squaring both sides.

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And I get 2d squared
is n squared.

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Well, that's great, because
look-- the left-hand side is

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divisible by 2.

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There it is.

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Which means that n
squared is divisible by 2.

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The right-hand side is even.

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But if n squared is even, then n
is even, and I'm halfway there.

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I've shown that the
numerator is even.

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OK, let's keep going.

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Now, since n is even, it's
equal to twice something.

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So n is 2k for some number k.

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I don't care what k is.

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Let's square both sides of that,
and conclude that n squared

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is equal to 4k squared.

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Why did I square it?

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So that I could connect up
here with the other question

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that I had about it
n squared-- that n

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squared it was 2d squared.

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So combining these
two, what I get

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is that 2d squared is
equal to 4k squared.

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And of course, I can
cancel 2, and get that d

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squared is equal to 2k squared.

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And again, I've got the
right-hand side divisible by 2.

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So the left-hand side
is divisible by 2.

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d squared is even, and
therefore, d is even.

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And we've completed
the proof as claimed.

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n and d both have 2
as a common factor,

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contradicting the fact
that their in lowest terms.

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Now, I did assume something that
is kind of obvious-- namely,

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that if n squared is
even, then n is even.

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Why is this true?

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Well, you might think
about it for a moment.

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There's a simple way
to see it, and it's

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a proof by contradiction.

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We're going to use the fact
that you can verify easily

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enough by doing a little
arithmetic-- namely,

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the product of two
odd numbers is odd.

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Let's assume that.

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So if the product of
two numbers is odd,

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if I tell you that
n squared is even,

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and suppose that n was not
even, well, that means it's odd.

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But that would mean that n
squared was odd, contradicting

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the fact that n is even.

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Therefore, it's a contradiction
to assume that n is odd.

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It must be even

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That's an ad hoc
proof that has to do

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with evenness and oddness.

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There's a more general way to
understand this that actually

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will come in handy--
namely, that what I know

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is that numbers factor into
primes in a unique way.

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So if I tell you that
n squared is even,

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what I know about n squared
is that all the primes that

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divide n squared come from n.

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So if there's a 2 among the
primes that divide n squared,

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it has to be a 2 that is one
of the prime divisors of n.

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And that would work
even if I told you

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that n squared was
divisible by 3.

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It would follow
by that reasoning

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that n is divisible by 3.

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Now, that's a powerful fact.

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I'm assuming the prime
factorization of integers.

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And it's not obvious at
all that that's true,

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although it's very familiar.

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It's OK to assume.

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In a few weeks we'll
actually look back

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at how to carefully prove that.

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But for now, it's OK to assume.

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And we also have
the simple argument

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that worked based on
properties of even and odd--

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that if n squared is
even, then n is even.

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That's the last gap in the
proof, and so we're done.