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Irrationality.

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What is an irrational number?
Well, here's a definition.

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A real number x is irrational, if it can't
be expressed as a rational number.

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If it can't be written as p over q, for p
and q integers.

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You might have already met some irrational
numbers.

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For example, here's a claim, the square
root of 2 is irrational.

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Now, how do I go about proving such a
claim?

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Well, the argument begins

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by supposing that I can write the square
root of 2 as a rational number.

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And, if I can write the square root of 2
as a rational number,

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then I can write the square root of 2 as a
fraction in lowest terms.

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So, I'm supposing that a and b don't have
any factors in common.

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Alright, now once I've written the square
root of 2 as a over b, I

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could square both sides, and then I'd find
that 2 is a squared over b squared.

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And then

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I could multiply both sides of this
equation by b

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squared, and I'd find that 2 b squared is
a squared.

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Now what does that mean?

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Well, that means in particular that a is a
multiple of 2, alright?

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So, that means that a must be even.

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So, I can write a as 2 times some integer.
I'll call that integer k.

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So a is even.
A is twice

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some integer.
Now, I can plug this fact back into here.

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Right?

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Instead of saying 2b squared is a squared,
I can say that

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2b squared is now a, which is 2 times an
integer, squared.

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And, if I expand that out I get that
that's.

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2b squared is 4k squared.

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Now, if I divide both sides of this by 2,

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then I get, that b squared is 2k squared.

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[LAUGH]

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And what does that mean about b?

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Well, that means that b, is also a
multiple of 2.

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That means that b is even.

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But if a and b are both even, then the

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fraction a over b is not in lowest terms,
right?

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I've got a common factor of 2 in the
numerator and denominator, and

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that's the contradiction that can show
that the Square of 2 can't be written.

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As a fraction.

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That's a pretty standard

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method for showing that a number is
irrational.

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You start about by assuming that it's a

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rational number, and then drive some kind
of

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contradiction which reveals that your
original assumption that

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the number was rational, must have been in
error.

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And then you know the number's irrational.

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Now try the same kind of game, but not
with square of 2, but with the number e.

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Well, if e were rational number, then it's
reciprocal,

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1 over e would be a rational number as
well.

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If could write e as a fraction, this would
just be the reciprocal of that fraction.

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I can analyze 1 over e, turns out that 1
over e.

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Is equal to the value of this convergent
and alternating series.

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Now, the quality of 1 over e with the
value

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of this series, isn't something that we're
quite able to do.

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We're going to see that these are equal in
week

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six of this course.

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So, there's a little bit more to do here,
alright?

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But here's the big deal.

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I can analyze this alternating series so,
our goal is going to

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be, to show that the value of this
alternating series is irrational.

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And at some future point we're going to
see that the value of this series

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is 1 over e, which will then mean that e
also must be irrational.

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For the meantime

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let's just show the value of that series
as an irrational number.

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So, let's supposed that the value of this
series, this sum n

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goes from 0 to infinity, negative 1 to be
n over n factorial.

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Let's suppose that this is equal to a
rational number.

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That it is a over b for integers a and b.
Now lets look at the b's partial sum.

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Yeah,

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I really mean this b.
Right.

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The b from the denominator of this
fraction

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that I'm imagining 1 over e is equal to.

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So let's take a look, at that bth partial
sum.

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s sub b.

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Right.

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It's just the sum of the terms, n goes
from 0

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to b, of minus 1 to the n over n
factorial.

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Now, the whole deal here is that I've got
an alternating series.

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And what that means is the true value of
this alternating series, which is 1 over

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e, must be between the s of b and s of b
plus 1 partial sums.

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I, hence the, that the trick about the
error bounds for

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an alternating series, that the true value
of an alternating series.

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Is always between neighboring partial
sums.

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Now, the significance of this is that I
know exactly how

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far apart s and b and s and b plus 1 are.

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S

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and b and s and b plus 1 are exactly 1
over b plus 1 factorial apart.

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' because s and b plus 1 and s and

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b differ by the b plus 1th term in the
series.

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So, I could put these two statements
together to make this claim,

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that the distance between 1 over e, the
true value of the series.

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And the s sub b partial sum, the distance

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between these two numbers is no more than
this,

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1 over b plus 1 factorial.

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Now, I'll multiply both sides by b
factorial.

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Right.

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So, we've got this inequality multiplying
both sides

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of this inequality by the positive number
b factorial.

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Now what do I have?

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Well I've got b factorial over b plus 1
factorial.

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This side can be simplified a bit.

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This is 1 over b plus 1.
And one thing I know about 1 over b plus 1

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is that it's less than one.
I can also simplify the left hand side.

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Well, first of all, I know that 1

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over e isn't actually equal to the partial
summary.

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This is the true value of the series, this
is just the sum of the first b terms.

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These are not the same.

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So, that tells me that I've got 0 less
than this quantity.

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And I can also simplify this a little bit
more.

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Right?

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I can say that this is b factorial times,
well, what's 1 over e?

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I was assuming that I could write 1 over e
as this fraction a over b.

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And now I could distribute this to get
this, right?

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B factorial times 1 over e, which is, I'm
assuming, a over b.

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Minus b factorial, times the bth partial
sum.

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But I also know something about the

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quantity, that I'm taking the absolute
value of.

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Specifically, I know that b factorial
times a over b,

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is an integer.

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Right, this is just b minus 1 factorial
times a.

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That's an integer.

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I also know something about b factorial
times the bth partial sum.

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All right?

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What is that?

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Well, by definition, that's just b
factorial, and

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here I've written out the bth partial sum.

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But now I could put the b factorial inside
there.

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And I'd get, this is the sum, n goes from
0 to b, of negative 1 to

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the n plus 1 times b factorial over n
factorial.

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Now what do I know?

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B is at least as big as n.

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So, b factorial divided by n factorial,
that's an integer.

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This is the sum and difference of
integers.

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That means, that b factorial times the bth
partial sum, is an integer.

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And that, is problematic.

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Because if this is an integer, and this is
an integer,

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that means that quantity inside here is
also an integer.

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That means that there's some integer
that's between zero and one.

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But that is not true.

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Well, since there is no integer between
zero and one, that's a contradiction.

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And consequently, it must be the case than
1 over e is irrational.

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And therefore, e is irrational, which is
what I wanted

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to show.

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I still owe you a proof that that
alternating series evaluates to 1 over e.

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And that proof is coming.

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But in the mean time, it's worth

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reflecting on why this argument worked at
all.

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The cool thing about alternating series,
is that they come with error bounds.

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I know that the true value of

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an alternating series, is between
neighboring partial sums.

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And that's great for doing numerics.

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That's great for doing computation.
But the real point

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of computation isn't numbers.
It's insight.

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And I think this argument, this proof that
1 over e and therefore e is irrational.

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Is a really great example of how a
computational tool, the

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fact that we have these explicit error
bounds, can yield real insight.

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