[MUSIC PLAYING] Can you produce a rational number by exchanging infinitely many digits of pi and e?

5 00:00:15,890 --> 00:00:19,260 The discipline of math has a wonderful sense of community.

Mathematicians love to share their ideas and questions with each other, to problem solve together, just like you all do in the comment section of this show.

There are several online forums where mathematicians post questions and responses.

Recently on one of them, mathematician Erin Carmody asked, if I exchange infinitely many digits of e and pi, are the resulting numbers transcendental.

She received some curious answers, ones which provide an awesome preview into the way mathematicians approach research-- with intuition, collaboration, and a strong desire to explore the unknown.

Transcendental numbers are a very specific kind of irrational number, and certainly the subject of some future episodes.

But we'll focus on irrational numbers in general, which is what most mathematicians in the post did.

Pi and e, also known as Euler's number, are two famous irrational numbers.

So we'll ask, can you exchange infinitely many digits of pi and e to get a rational number.

All the real numbers, any number that appears on the number line, can be divided into two categories-- rational and irrational.

Rational numbers are those that can be written as n divided by m, where n and m are two integers, like 73 over 5, or 1 over 7, or 133 over 300.

An equivalent definition is that rational numbers are those whose decimal expansions terminate or eventually repeat, like 14.6, or 0.142857 repeating, or 0.44333333333.

Even rational numbers that we think of as having terminating decimal expansions, like 14.6, actually have a repeating 0 if we write it as 14.60000000.

Irrational numbers are all the real numbers that aren't rational, meaning their decimal expansions don't repeat.

Some well-known examples include pi, e, the square root of 2, and the golden ratio.

Actually, the proof that the square root of 2 is irrational is a fantastic classic.

You can try it for yourself or check out the link in the description.

Before we return to the original question, let's start with a warm-up question.

Do there exist two irrational numbers such that we can exchange infinitely many digits of them and one of the resulting numbers is rational?

Pause here if you want to think about it.

The answer is yes.

Maybe you came up with a different example, but here's mine.

The number 0.010110111 is irrational.

It's made up of strings of 1's of increasing length with a 0 between each string of 1's.

Since the string of ones becomes longer and longer, it never repeats.

Therefore, it's irrational.

Similarly, the number 0.101001000 and so on is irrational.

But if we exchange infinitely many of their digits, like this, we get two rational numbers, 0 and 1/9.

Or, we could trade these digits of these two irrational numbers.

In this case, one of the resulting numbers is rational and one is irrational.

OK, back to the original question where Carmody is asking about two very specific numbers.

If I exchange infinitely many digits of pi and e, are the resulting numbers irrational?

Here's pi and here's e. There's a lot of different ways we could exchange infinitely many digits.

We could exchange the odd digits to get this pair, or every fourth digit to get this pair.

Or the second, eighth, and ninth digits, and all the digits after the 16th digit, to get these.

With infinite time and patience, we could just write down a big endless list of all the possible pairs of numbers that arise by switching infinitely many digits of pi and e. What kind of numbers appear on that list?

Are there rational numbers?

Are there irrational numbers?

Mathematician Joel David Hamkins, whose work we featured in the infinite chess episode, began his response to Carmody's question with a question of his own.

How often do the digits of pi and e differ-- as in, the first digit after the decimal place is different.

It's one for pi and seven for e. The second digit is different, it's four for pi and one for e. But the 12th digit is the same, it's nine for both.

If we exchange the 12th digit, it doesn't do anything.

Same thing for the 16th and 17th digits.

What if the digits of pi and e were mostly the same?

If there were only finitely many places they differed, what would happen?

Well, then there would only be finitely many new numbers produced by the digit swaps.

And these new numbers would only differ from pi or e at finitely many places.

This means that all the numbers on the list would be irrational.

Why?

Let's just look at pi.

Assume we change finitely many digits of pi.

Let's look at the last digit that we changed.

Everything after it is unchanged.

Since pi is irrational, these digits don't have a repeating pattern, so the modified version of pi also doesn't have a repeating pattern and is therefore irrational.

Notice that if pi and e are only different at finitely many digits, then pi minus e is rational.

OK, so the answer is pretty easy if pi and e happen to only differ at finitely many digits.

The numbers on the list are all irrational.

But it seems more likely that the digits of e and p differ at infinitely many places.

In this case, we can produce infinitely many new numbers by swapping their digits.

That is, the list would be infinitely long.

Remember back to the hierarchy of infinities episode where we talked about different sizes of infinity.

There was a smallest infinity, which is called countable.

And then there were bigger infinities.

Anything bigger than countable infinity is called uncountable infinity.

Well, if pi and e differ at infinitely many digits, then the list of numbers produced by all possible exchanges of infinitely many digits of e and pi is uncountably long, which actually means it's not even possible to write it in a list.

So this is now a metaphorical list.

I'll leave it as a challenge problem for you to show that it's uncountable.

Let us know your solutions in the comments.

Here's why that's important for us.

There are only countably many rational numbers.

That means there are more numbers on the list than there are rational numbers.

So only countably many pairs of numbers have a rational number.

That means, for most of the infinitely many pairs of numbers on the list, both numbers are irrational.

So if pi and e differ infinitely often, most of the numbers on our list are irrational.

But are any of them rational?

Mathematician Terry Tao had an answer, sort of, for that part.

There is a conjecture, which is basically an unproven educated guess, that both e and pi are normal numbers.

A normal number is one where any digit or sequence of digits is equally likely to occur.

So as we look through pi, the number 3 occurs just as often as 4.

And the sequence 1, 2, 3 occurs just as often is the sequence 8, 8, 8.

In other words, the digits of a normal number are balanced.

A crazy fact about normal numbers is that we know almost every number is normal, but have very few concrete examples.

In particular, no one has proved pi and e are normal numbers, even though it's believed to be the case.

Tao pointed out that if pi and e are normal numbers, as we suspect, then it's impossible to produce a rational number by exchanging their digits.

If you want the technical argument, I recommend reading the original post linked in the description.

But here's the intuition-- if you did produce a rational number by exchanging the digits of pi and e, that number would have a repeating part of its decimal expansion, like this.

But normal numbers don't like to repeat themselves.

And blending two of them together doesn't change that fact.

So on the list of all possible numbers generated by exchanging infinitely many digits of pi and e, are there rational numbers?

Probably not.

There's three things I want to highlight about why this question is so cool.

1, it's about e and pi.

2, we just saw two very different answers, one coming from set theory and one coming from number theory.

3, and this is my favorite part, this post is a small and accessible preview into the way math research really happens.

It's a curious question with different perspectives and no known correct answer.

We've reduced a complicated problem to other complicated problems, ones about whether pi and e are normal numbers, or whether their digits differ infinitely often.

And we're playing with mathematical intuition.

I recommend checking out the original post in the description for some of the subtle details that I omitted.

And don't forget to answer the challenge in the comments.

Also, if anyone can prove that e and pi differ at infinitely many digits, that'd be a great fact to know for certain.

See you next time on "Infinite Series."

Hello, I just want to respond to some of the comments about our episode on the Wolverine problem and solving it using graph colorings.

So first, let's get to some important stuff-- my T-shirt.

A lot of you had a lot of opinions about my shirt.

It said, let epsilon be less than 0.

So people usually use the variable epsilon to symbolize a really small positive number.

It's most commonly used in calculus, but also shows up in other areas of math.

So let epsilon be less than 0 is just kind of a goofy joke, maybe funny.

All right, on to more important things.

[INAUDIBLE] has asked, for the Hadwiger-Nelson problem, the colored regions don't have to be the same shape, right?

Correct.

They can be any shape.

And in fact, if you place restrictions on the kind of shapes that the colored regions can be, or what color each of those regions can be, it changes the problem.

It changes the bound on the chromatic number.

And if you check out the Wikipedia article for Hadwiger-Nelson problem, it actually tells you a lot of those results, and they're pretty cool.

So go check it out.

Finally, Brian Willett gave a great answer to our challenge question.

Why is the lower bound on the chromatic number of the plane 4?

Here's his answer.

You can bend the graph, we showed, in the plane such that all edges had length one centimeter.

Any coloring of the plane must have a coloring number bigger than the one at the graph.

So to show that the chromatic number of the graph must be bigger than 3, we need to note that it takes at least two colors to color the outer hexagon.

And then this center vertex must be a third color.

The three vertices of the central triangle must be three different colors, but they are all adjacent to the vertices of the hexagon of the same color.

So these can't be the same three colors from before.

And so we need at least four colors.

An example of a four-coloring is shown.

Good job, Brian.

We'll see you next week.

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