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PROFESSOR: So let's look at
a last example of applying
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the well ordering principle,
this time to something that we
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actually care about-- a theorem
that really does require some--
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The theorem is the
following famous formula
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for the sum of a geometric
sum-- for a geometric series
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or a geometric sum.
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So the numbers on the left,
the powers of r starting at 1,
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which is r to the 0 followed
by r, which is r to the 1,
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followed by r squared, up
through the nth powerful of r.
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You add all those numbers
up and it turns out
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that there's a nice,
simple, fixed formula that
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doesn't have those
three dots in it that
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tells you exactly what the value
of that sum is in the formula.
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As you can read is r to the n
plus 1 minus 1 is the numerator
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and r minus 1 is
the denominator.
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And the claim is
that this identity
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holds for all
non-negative integers n
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and for all real numbers r
that aren't 1 because I don't
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want the denominator to be 0.
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So how are we going
to prove this?
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Well, I'm going to
prove it by using
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the well ordering
principle, and let's suppose
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that this identity didn't hold
for some non-negative integer
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n.
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So we'll apply the
well ordering principle
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and we'll let m be
the smallest number
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n where this equality fails--
it becomes an inequality.
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Now, what I know
about m immediately
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is that this equality, if
you look at it, when n is 0
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the left-hand side comes down.
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It degenerates to
just r to the 0, or 1.
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The right-hand side,
if you check it,
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is r minus 1 over r
minus 1, which is also 1.
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So equality holds
when n is 0, and that
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means that the least m for
which equality doesn't hold
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has to be positive.
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So, what we know
about the least number
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where this equality fails
is that it's positive.
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And that means in particular
since it's the least one
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[? word ?] fails, if you
go down one to m minus 1,
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the equality holds.
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So we can assume that the sum
of the first m powers of r,
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starting at 0 and ending
at r to the m minus 1,
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is equal to the formula where
you plug in m minus 1 for n
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and you get that formula
on the right, which
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I'm not going to read to you.
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Well, we can simplify
it a little bit.
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If you look at the exponent,
r to the m minus 1 plus 1
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is after all just r to the m.
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So repeating what I've
got is that the sum
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of those first powers of r
up to m minus 1 we can assume
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is equal to the formula r to the
m minus 1 divided by r minus 1
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because m failed and this
was the number one less where
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it had to succeed.
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So now we take the
obvious strategy.
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What I'm interested
in is properties
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of the sum of the
powers up to r to the m.
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Now, the left-hand
side is the powers
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up to r to the m
minus 1, so there's
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an obvious strategy for turning
the left-hand side into what
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I'm interested in.
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Namely, let's add r to
the m to both sides.
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So the left-hand side
becomes just the sum
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that I want and the
right-hand side becomes
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this messy thing, r to the
m minus 1 over r minus 1
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plus r to the m.
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Well, let's just
simplify a little bit.
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Let's put r to the m over
the denominator, r minus 1,
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which I do by multiplying
it by r minus 1.
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And then it comes
out to be r to the m
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plus 1 minus r to
the m over r minus 1.
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And I collect terms
and look what I got.
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I've got the formula r
to the m plus 1 minus 1
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over r minus 1, which means
that the identity that I was
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originally claiming,
in fact, holds
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at m contradicting the assertion
that it didn't hold at m.
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In other words, we've
reached a contradiction
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assuming there was a least
place were equality fails,
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that means there's
no counter example
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and the equality holds for
all non-negative integers n.
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So here's the
general organization
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of a well ordering proof
which we've been using.
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Let's just summarize it
into a kind of template
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for proving things.
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So what you have in mind is that
there's some property, P of n,
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of non-negative integers.
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And what you'd like to
prove is that it holds
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for every non-negative integer.
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So for all n in non-negative
integers, P of n holds.
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And we're going to
try to prove this
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by the well ordering
principle, which
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means that we're going to define
the set of numbers for which P
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doesn't hold.
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That is, the set of
counter examples,
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and call that C. So C is the
set of non-negative integers
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for which not P of n holds.
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Now, by the well
ordering principle
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there's got to be
a minimum element,
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call it m, that's in C.
And at this point the job--
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by assuming that m is the
smallest counter example,
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we have to reach a
contradiction somehow.
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Now, up to this second
bullet it's the template,
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but the third bullet is where
the real math starts and there
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isn't any template anymore.
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How you reach a
contradiction is by reasoning
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about properties of P of n,
and there's no simple recipe.
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But the usual organization
of the contradiction
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is one of two kinds.
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You find a counter example
that's smaller than m--
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you find a C that's in the
set of counter examples
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and C is less than m that would
be a contradiction because m
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is the smallest thing at C.
Or you reach a contradiction
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by proving that P does
hold for m, which means
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it's not a counter example.
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And those are the
two standard ways
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to organize a well
ordering proof.