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So now we begin on four
classes on number theory.

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The purpose of taking it
up now is that we're still

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practicing proofs.

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And number theory is
a nice self-contained

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elementary subject
as we'll treat it,

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which has some
quite elegant proofs

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and illustrates contradiction
and other structures

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that we've learned about.

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A little bit of induction, and
definitely some applications

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of the well-ordering principle.

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The ultimate punchline
of the whole unit

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is to understand the RSA
crypto system and how it works.

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Along the way, we will--
today, actually-- establish

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one of those mother's
milk facts that we all

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take for granted about unique
factorization of integers

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into primes.

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But in fact, that's
a theorem that

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merits some proof as an
example, and the homework

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shows where we exhibited
a system of numbers

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which didn't factor uniquely.

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And finally, we will be able
to knock off the Die Hard

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story once and for all.

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So let's begin by stating
the rules of the game.

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We're going to assume all
of the usual algebraic rules

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for addition and
multiplication and subtraction.

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So you may know
some of these rules

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have names like
the first equality

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is called distributivity of
multiplication over plus--

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of times over plus-- and
then the second rule here

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is called commutativity
of multiplication,

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and here are some
more familiar rules.

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This is called associativity
of multiplication.

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This is called the
additive identity.

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a minus a is 0-- or
actually additive inverse.

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0 is the additive identity and
minus a is the inverse of a.

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a plus 0 equals a is
the definition of 0

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being an additive identity.

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a plus 1 is greater than a.

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So these are all
standard algebraic facts

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that we're going to take for
granted and not worry about.

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And one more fact that
we also know and we're

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going to take as an
axiom, if I divide

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a positive number-- sorry.

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If I divide a number a
by a positive number b,

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then when we're talking about
integers, what I'm going to get

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is a quotient and a remainder.

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What's the definition of the
quotient and a remainder?

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Well, the division
theorem says that if I

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divide a by b, that means if
I take the quotient times b

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plus the remainder I get a.

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And in fact, there's a
unique quotient of a/b

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and there's a unique
remainder of a/b

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where the remainder--
what makes it unique

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is the remainder is
constrained to be

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in the interval greater
than or equal to 0

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and less than the divisor b.

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So we're going to take
this fact for granted too.

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Proving it is not worth
thinking about too hard,

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because it's one of those facts
that's so elementary that it's

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hard to think of other things
that would more legitimately

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prove it.

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I'm sure it could be
proved by induction,

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but I haven't really
thought that through.

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OK.

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A key relation that we're
going to be looking at today

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is the relation of
divisibility between integers.

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So by the way, all of the
variables for the next week

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or so are going to be understood
to range over the integers.

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So when I say number,
I mean integer.

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When I talk about
variables a and c and k,

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I mean that they're
taking integer values.

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So I'm going to define c divides
a using this vertical bar

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notation.

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It's read as divides. c
divides a if and only if a is

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equal to k times c for some k.

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And there are a variety
of synonyms for a divides

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b, like-- a is a-- a
divides c-- sorry--

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c divides a is to say
that a is a multiple of c

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and c is a divisor of a.

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OK.

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Let's just practice this.

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So 5 divides 15?

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Well, because 15 is 3 times 5.

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A number n divides 0.

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Every number n divides 0.

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Even 0 divides 0, because
0 is equal to 0 times n.

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So 0 is a multiple
of every number.

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Another trivial fact that
follows from the definition

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is that if c divides a, then c
divides any constant times a.

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Well, let's just
check that out, how

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it follows from the definition.

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If I'm given that
c divides a, that

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means that a is equal to k
prime c for some k prime.

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That implies that if I multiply
both sides of this equality

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by s, I get that s a is
equal to s k prime c,

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and if I parenthesize the s k
prime, I can call that to be k,

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and I have found, sure enough,
that s a is a multiple of c.

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That's a trivial proof,
but we're just practicing

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with the definitions.

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So we have just
verified this fact

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that if c divides a, then c
divides a constant times a.

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As a matter of fact,
if c divides a and c

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divides b, then c
divides a plus b.

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Let's just check that one.

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What we've got is c
divides a means that a

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is equal to k1 times c.

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And c divides b means that
b is equal to k2 times c.

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So that means that a plus b
is simply k1 plus k2 times

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c, where what I've done here
is used the distributivity law

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to factor c out and used
the fact that multiplication

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is commutative so that I can
factor out on either side.

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OK.

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Let's put those facts together.

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If c divides a and c divides
b, then c divides s a plus t

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b, where s and t are
any integers are all.

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So a combination of two
numbers, a and b, like this

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is called a linear
combination of a and b--

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an integer linear combination,
but since we're only

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talking about
integers, I'm going

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to stop saying integer
linear combination

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and just say linear combination.

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A linear combination
of a and b is

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what you get by of multiplying
them by coefficients s and t

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and adding.

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OK.

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So we've just figured
out that in fact

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if c divides a and
c divides b, then

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c divides an integer
linear combination of b.

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When c divides two numbers,
it's called a common divisor

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of those two numbers.

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So we could rephrase
this observation

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by saying common divisors
of a and b divide integer

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linear combinations of a
and b, which is a good fact

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to just file away in your head.

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Now, what we're going
to be focusing on

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for the rest of today is
the concept of the greatest

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common divisor of a and be,
called the GCD of a and b.

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The greatest common
divisor of a and b

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exists by the
well-ordering principle,

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because it's a set of
non-negative integers

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with an upper bound.

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Namely, a is an upper bound
on the greatest common divisor

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of a and b.

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So as we did in
an exercise, or I

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think in the text, that implies
that there will be the greatest

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one among all the
common divisors,

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assuming there are any.

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But 1 is always
a common divisor,

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so there are
guaranteed to be some.

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Let's look at some examples.

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The greatest common
divisor of 10 and 12.

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You can check.

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It's 2.

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Mainly because 10 factors
into 2 times 5 and 12 factors

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into 2 times 6, and the 6 and
the 5 have no common factors.

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So the only one that
they share is 2.

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The GCD of 13 and 12 is 1.

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They have no common
factors in common.

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You can see that
because 13 is a prime,

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and so it has no factors
other than 1 and 13,

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and 13 doesn't divide
12 because it's too big.

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So it's got to be 1.

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The GCD of 17 and 17 is 17.

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That's a general phenomenon.

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The GCD of n and n is always n.

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The greatest common
divisor of 0 and n

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is equal to n for
any positive n.

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That's because everything
is a divisor of 0

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and it means the GCD
of 0 and n is simply

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the greatest divisor of n,
which is of course n by itself.

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One final fact to set things
up for the next segment

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is to think about the GCD
of a prime and a number,

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and it's either 1 or p.

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The reason is that the
only divisors of a prime

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are plus/minus 1
and plus/minus p.

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So if p divides a, the GCD is
p, and otherwise the GCD is 1.