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Irrationality.

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

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

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

You might have already met some irrational numbers.

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

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

Well, the argument begins

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

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

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

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

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

And then

I could multiply both sides of this equation by b

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

Now what does that mean?

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

So, that means that a must be even.

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

So a is even. A is twice

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

Right?

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

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

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

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

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

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And what does that mean about b?

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

That means that b is even.

But if a and b are both even, then the

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

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

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

As a fraction.

That's a pretty standard

method for showing that a number is irrational.

You start about by assuming that it's a

rational number, and then drive some kind of

contradiction which reveals that your original assumption that

the number was rational, must have been in error.

And then you know the number's irrational.

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,

1 over e would be a rational number as well.

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

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

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

Now, the quality of 1 over e with the value

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

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

six of this course.

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

But here's the big deal.

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

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

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

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

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For the meantime

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

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

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

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

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

Yeah,

I really mean this b. Right.

The b from the denominator of this fraction

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

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

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s sub b.

Right.

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

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

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

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

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

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

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

Is always between neighboring partial sums.

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

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

And the s sub b partial sum, the distance

between these two numbers is no more than this,

1 over b plus 1 factorial.

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

Right.

So, we've got this inequality multiplying both sides

of this inequality by the positive number b factorial.

Now what do I have?

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

This side can be simplified a bit.

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

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

Well, first of all, I know that 1

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

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

These are not the same.

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

And I can also simplify this a little bit more.

Right?

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

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

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

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

Minus b factorial, times the bth partial sum.

But I also know something about the

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

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

is an integer.

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

That's an integer.

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

All right?

What is that?

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

here I've written out the bth partial sum.

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But now I could put the b factorial inside there.

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

the n plus 1 times b factorial over n factorial.

Now what do I know?

B is at least as big as n.

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

This is the sum and difference of integers.

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

And that, is problematic.

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

that means that quantity inside here is also an integer.

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

But that is not true.

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

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

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

to show.

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

And that proof is coming.

But in the mean time, it's worth

reflecting on why this argument worked at all.

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

I know that the true value of

an alternating series, is between neighboring partial sums.

And that's great for doing numerics.

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

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

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

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

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

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