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PROFESSOR: The elements
that have inverses modulo

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and will be particularly
important to us.

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And so the first
question is how many of

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them are there, which is what
Euler's function tells us.

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So the definition of
Euler's function, phi of n,

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is it's the number of
integers in the remainder

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interval from 0
to n minus 1 such

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that k is relatively prime to n.

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So remember, there's the
notation for the remainder

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interval that includes
0 and excludes n.

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And another way to say
relatively prime to n

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is to say the gcd
of k and n is 1.

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So let's define
that set of numbers

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that we're interested
in-- gcd1 of n [? be ?]

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those numbers that
have a gcd of 1 with n.

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That is, the numbers that have
inverses and the numbers that

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are cancellable modulo n.

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So what it means
is that phi of n

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is precisely equal to
the size of gcd1 of n.

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Now, some authors
call gcd1 n star.

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I didn't find that a
very informative notation

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and so I'm not using it.

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phi of n is also, for
your information, called

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Euler's totient
function, but we'll just

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stick to calling it
phi or Euler's phi.

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So let's look at an
example-- gcd1 of 7.

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The numbers that are
relatively prime to 7

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are all the positive numbers
less than 7 because 7 is prime.

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So it's the set
1, 2, 3, 4, 5, 6.

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gcd1 of 12 is the
numbers that have

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no factor in common with 12.

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They are the numbers
in green below.

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And the other red
numbers do have

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a number in common with 12-- do
have a prime in common with 12.

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The pattern here
is not so apparent.

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Anyway, phi of 7 is the
size of gcd1 of 7-- namely

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the size of the set 1
through 6, which is 6.

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gcd 12 determines
phi of 12. phi of 12

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is the number of green
elements, which is 4.

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OK.

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A simple rule for
calculating phi.

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When phi is prime we've
already indicated,

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namely, everything--
every positive number

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less than p is
relatively prime to p.

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And so phi of p is
simply p minus 1.

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Let's look at a more
important example,

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or illustrative example--
namely, phi of 9.

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Well, OK.

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So there are the
candidate numbers from 0

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through 8, and which ones
are relatively prime to 9?

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Well, it's relatively
prime to 9 if and only

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if it's relatively prime to 3.

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Now, which numbers
in this interval

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are relatively prime to--
are relatively prime to 3,

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or, rather, are not
relatively prime to 3?

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Well, it's every third
number that's divisible by 3.

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So, those are the bad ones.

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If we subtract the bad ones,
we're left with the good ones--

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the ones that are
relatively prime.

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So a phi of 9 is simply
the set of all the numbers

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minus 1/3 of 9, which is the
bad one-- bad one's namely 6.

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This generalizes to
a power of a prime.

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If k is a positive
integer then phi of p

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to the k-- the reasoning is
that a number is relatively

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prime to the p to
the k if and only

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if it's relatively prime to p.

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p divides every pth
number, so one pth

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of the numbers in
the interval are bad,

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which means that phi of p is
the good ones minus 1 pth of p

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to the k.

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Namely, phi of p to the k is
p to the k minus p to the k

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over p, which can
also be expressed

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in a more standard
form-- p to the k

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minus p to the power k minus 1.

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And that knocks off
the story of phi

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to the p for powers of primes.

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Well, suppose you're
dealing with a number that's

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not a power of a prime.

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And there's one very elegant
little fact about phi

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that explains how to deal
with non powers of primes.

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Namely, if a and b
are relatively prime,

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then phi of a b is simply
gotten by computing phi of a

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and multiplying it by phi of b.

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This property of phi is called
multiplicativity, by the way.

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It comes up a lot
in number theory.

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A function is
multiplicative when

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its value at a product of
relatively prime numbers

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is the product of the
values at those two

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relatively prime numbers.

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So phi is multiplicative.

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Now, the proof of that-- one
proof is on problem set 5,

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and there's another proof that
we'll see in a couple of weeks

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when we get into counting the
inclusion-exclusion principle.

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Let's just use this
fact about phi--

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the multiplicity of phi--
multiplicativity of phi

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to see how it lets us calculate
phi of an arbitrary number.

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So, in particular, phi of
12-- which looked complicated

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earlier-- well, 12 is 3 times 4.

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So that means that phi of 12
is phi of 3 times phi of 4.

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But now I'm in great
shape because 3

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is a power of a prime,
namely 3 to the 1.

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And 4 is a power of a
prime, namely 2 squared.

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So applying the power of prime
formulas, I get that phi of 3

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is 3 times 1 times 2 squared
minus 2 to the 2 minus 1,

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which simplifies to 4, which is
the answer that we saw before.

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And the punchline for
why we're examining phi

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is Euler's theorem,
which tells us

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how powers of numbers
in gcd1 of n behave.

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Namely, that if k is
relatively prime to n, then

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if you raise k to
the power phi of n,

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it's congruent to 1 mod n.

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And that will lead us.

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In the next section we
will look at the proof

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of Euler's theorem.