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PROFESSOR: Now we come to a more
serious application of the fact

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that the GCD is a
linear combination.

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We're going to use it to
prove the prime factorization

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theorem-- which we've
talked about earlier.

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This is the unique prime
factorization theorem.

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So let's begin by looking at a
technical property of primes,

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which is familiar, but we're
going to need to prove it.

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If you believe in prime
factorization, then

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this Lemma-- which says
that if p divides a product,

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it divides one or the other of
the components of the product--

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that's an immediate consequence
of the prime factorization

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theorem.

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But we mustn't
prove it that way,

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because we're trying to use this
to prove prime factorization.

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So how can I prove, based
on the facts of what

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we know about GCDs,
without appealing

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to prime factorization
that if p is a prime,

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and p divides a
product, then it divides

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one of the components
of the product,

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either the multiplier or
the [? multiplicand? ?] OK,

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well here's how to prove that.

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Suppose that p divides ab,
but it doesn't divide a.

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Of course it does
divide a, I'm done.

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So we may as well assume
that it doesn't divide a.

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Now that means that since
the only divisors p are p

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and 1-- the only positive
divisors of p are p and 1--

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that if p doesn't divide
a, the GCD of a and p is 1.

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All right, now comes the
linear combination trick.

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Given that the GCD
of p and a is 1,

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that means that I have a linear
combination of a and p that's

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equal to 1-- sa plus tp is equal
to 1, for some coefficients, s

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and t.

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Cool-- multiply everything
by b on the right.

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So that means that sab plus
tpb is equal to 1 times b.

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But look at what we have now.

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The first term on the
left is something times

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ab, and p divides ab, so that
first term is divisible by p.

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The second term
explicitly has a p in it,

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so it's certainly
divisible by p.

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So the left hand side
is a linear combination

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of multiples of p,
and therefore, itself

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is a multiple of p--
which means the right hand

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side is a multiple of p, and
the right hand side is b.

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So sure enough, p divides b.

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We're done-- a very
elegant little proof

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that follows immediately from
the fact that you can express

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the GCD of two numbers
as a linear combination

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of those numbers.

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Now this is the
key technical Lemma

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that we need to prove
unique factorization.

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A corollary of this that
I'm actually going to need

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is that if p divides a product
of more than two things--

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if p divides a product
of a lot of things--

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it has to divide at
least one of them.

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And this you could
prove by induction,

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with the base case being
that it works for m equals 2.

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But it's not very
interesting, and we're

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going to take that for granted.

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If p divides a
product of any size,

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it divides one of the
components of the product.

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All right, now
we're ready to prove

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what's called the fundamental
theorem of arithmetic, which

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says that every integer greater
than one factors uniquely

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into a weakly decreasing
sequence of primes.

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Now the statement
of weakly decreasing

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is a little bit
technical and unexpected.

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What we want to say is
that a number factors

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into the same set of primes.

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Well that's not quite right,
because the set of primes

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doesn't take into account how
many times each prime occurs.

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You could try to make a
statement about every number

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uniquely is a multiple
of a certain number

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of each kind of prime.

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But a slick way to
do that is simply

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to say, take all the
prime factors, including

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multiple occurrences of
a prime, and line them up

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in weakly decreasing order.

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And when you do that,
that sequence is unique.

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This fundamental
theorem of arithmetic

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is also called the prime
factorization theorem.

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And here's what it
says when we spell it

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out-- without using the
words weakly decreasing.

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It says that every
integer, n, greater than 1

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has a unique factorization
into primes--

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mainly it can be expressed as
a product of p 1 through p k

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is equal to n.

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With p 1 greater
than or equal to p 2,

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greater than or equal
to each successive prime

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in the sequence, with
the smallest one last.

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Let's do an example.

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So there's a number
that was not chosen

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by accident, because I worked
out what the factorization was.

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And it factors into the
following weakly decreasing

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sequence.

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You start with the
prime 53, you followed

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by three occurrences of 37,
two 11s, a 7 and three 3s.

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And the point is that if you
try to express this ugly number

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as a weakly decreasing
sequence of primes,

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you're always going to get
exactly this sequence-- it's

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the only way to do it.

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All right, how are we
going to prove that?

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Well, let's suppose
that it wasn't true.

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Suppose that there was
some number that could be

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factored in two different ways.

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Well, by the
well-ordering principle,

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there's at least one.

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So we're talking about numbers
that are greater than 1,

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so there's a least
number greater than 1

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that can be factored
in two different ways.

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Supposed that it's n.

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So what I have is that n is
a product p 1 through p k.

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And it's equal to
another product, q 1

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through q m, where the p's
and the q's are all prime.

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And these two weakly decreasing
sequences are not the same.

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They differ somehow.

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So we can assume that the
p's are listed in a weakly

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decreasing order, and the
q's are likewise listed

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in weakly decreasing order.

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Now the first
observation-- suppose

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that q 1 is equal to p 1.

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Well that's not really possible,
because if q 1 is equal to p 1,

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then I could cancel the
p 1 from both sides,

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and I would get
the p 2 through p k

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is equal to q 2 through q m, and
these would still be different.

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Since they were different,
and I took the same thing

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from their beginning, I'm left
with a smaller number that

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does not have unique
factorization, contradicting

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the minimality of n.

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So in short, it's not
possible for q 1 to equal p 1.

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So one of them
has to be greater.

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We may as well assume that
q 1 is bigger than p 1.

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So q 1 is bigger than p 1, and
p 1 is greater than or equal

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to all the other
p's, so in fact,

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q 1 is bigger than
every one of the p's.

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Well that's going reach
a contradiction because

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of the corollary.

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What I know is
that q 1 divides n,

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and n is a product of the p's.

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And since q divides the product
of the p's, by the corollary,

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it's got to divide one of them--
q 1 must divide p i for some i,

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but that contradicts the fact
that q 1 is bigger than p i.

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That's not possible
for the larger number

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to divide the smaller number.

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And we're done.

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And we have proved the
unique factorization theorem.