## A Leap Day 2016 Mathematics A To Z: Transcendental Number

I’m down to the last seven letters in the Leap Day 2016 A To Z. It’s also the next-to-the-last of Gaurish’s requests. This was a fun one.

## Transcendental Number.

Take a huge bag and stuff all the real numbers into it. Give the bag a good solid shaking. Stir up all the numbers until they’re thoroughly mixed. Reach in and grab just the one. There you go: you’ve got a transcendental number. Enjoy!

OK, I detect some grumbling out there. The first is that you tried doing this in your head because you somehow don’t have a bag large enough to hold all the real numbers. And you imagined pulling out some number like “2” or “37” or maybe “one-half”. And you may not be exactly sure what a transcendental number is. But you’re confident the strangest number you extracted, “minus 8”, isn’t it. And you’re right. None of those are transcendental numbers.

I regret saying this, but that’s your own fault. You’re lousy at picking random numbers from your head. So am I. We all are. Don’t believe me? Think of a positive whole number. I predict you probably picked something between 1 and 10. Almost surely something between 1 and 100. Surely something less than 10,000. You didn’t even *consider* picking something between 10,012,002,214,473,325,937,775 and 10,012,002,214,473,325,937,785. Challenged to pick a number, people will select nice and familiar ones. The nice familiar numbers happen not to be transcendental.

I detect some secondary grumbling there. Somebody picked π. And someone else picked e. Very good. Those are transcendental numbers. They’re also nice familiar numbers, at least to people who like mathematics a lot. So they attract attention.

Still haven’t said what they are. What they are traces back, of course, to polynomials. Take a polynomial that’s got one variable, which we call ‘x’ because we don’t want to be difficult. Suppose that all the coefficients of the polynomial, the constant numbers we presumably know or could find out, are integers. What are the roots of the polynomial? That is, for what values of x is the polynomial a complicated way of writing ‘zero’?

For example, try the polynomial x^{2} – 6x + 5. If x = 1, then that polynomial is equal to zero. If x = 5, the polynomial’s equal to zero. Or how about the polynomial x^{2} + 4x + 4? That’s equal to zero if x is equal to -2. So a polynomial with integer coefficients can certainly have positive and negative integers as roots.

How about the polynomial 2x – 3? Yes, that is *so* a polynomial. This is almost easy. That’s equal to zero if x = 3/2. How about the polynomial (2x – 3)(4x + 5)(6x – 7)? It’s my polynomial and I want to write it so it’s easy to find the roots. That polynomial will be zero if x = 3/2, or if x = -5/4, or if x = 7/6. So a polynomial with integer coefficients can have positive and negative rational numbers as roots.

How about the polynomial x^{2} – 2? That’s equal to zero if x is the square root of 2, about 1.414. It’s also equal to zero if x is minus the square root of 2, about -1.414. And the square root of 2 is irrational. So we can certainly have irrational numbers as roots.

So if we can have whole numbers, and rational numbers, and irrational numbers as roots, how can there be anything else? Yes, complex numbers, I see you raising your hand there. We’re not talking about complex numbers just now. Only real numbers.

It isn’t hard to work out why we can get any whole number, positive or negative, from a polynomial with integer coefficients. Or why we can get any rational number. The irrationals, though … it turns out we can only get *some* of them this way. We can get square roots and cube roots and fourth roots and all that. We can get combinations of those. But we can’t get *everything*. There are irrational numbers that are there but that even polynomials can’t reach.

It’s all right to be surprised. It’s a surprising result. Maybe even unsettling. Transcendental numbers have something peculiar about them. The 19th Century French mathematician Joseph Liouville first proved the things must exist, in 1844. (He used continued fractions to show there must be such things.) It would be seven years later that he gave an example of one in nice, easy-to-understand decimals. This is the number 0.110 001 000 000 000 000 000 001 000 000 (et cetera). This number is zero almost everywhere. But there’s a 1 in the n-th digit past the decimal if n is the factorial of some number. That is, 1! is 1, so the 1st digit past the decimal is a 1. 2! is 2, so the 2nd digit past the decimal is a 1. 3! is 6, so the 6th digit past the decimal is a 1. 4! is 24, so the 24th digit past the decimal is a 1. The next 1 will appear in spot number 5!, which is 120. After that, 6! is 720 so we wait for the 720th digit to be 1 again.

And what is this Liouville number 0.110 001 000 000 000 000 000 001 000 000 (et cetera) used for, besides showing that a transcendental number exists? Not a thing. It’s of no other interest. And this plagued the transcendental numbers until 1873. The only examples anyone had of transcendental numbers were ones built to show that they existed. In 1873 Charles Hermite showed finally that e, the base of the natural logarithm, was transcendental. e is a much more interesting number; we have reasons to care about it. Every exponential growth or decay or oscillating process has e lurking in it somewhere. In 1882 Ferdinand von Lindemann showed that π was transcendental, and that’s an even more interesting number.

That bit about π has interesting implications. One goes back to the ancient Greeks. Is it possible, using straightedge and compass, to create a square that’s exactly the same size as a given circle? This is equivalent to saying, if I give you a line segment, can you create another line segment that’s exactly the square root of π times as long? This geometric problem is equivalent to an algebraic one. That problem: can you create a polynomial, with integer coefficients, that has the square root of π as a root? (**WARNING:** I’m skipping some important points for the sake of clarity. DO NOT attempt to use this to pass your thesis defense without putting those points back in.) We want the square root of π because … well, what’s the area of a square whose sides are the square root of π long? That’s right. So we start with a line segment that’s equal to the radius of the circle and we can do that, surely. Once we have the radius, can’t we make a line that’s the square root of π times the radius, and from that make a square with area exactly π times the radius squared? Since π is transcendental, then, no. We can’t. Sorry. One of the great problems of ancient mathematics, and one that still has the power to attract the casual mathematician, got its final answer in 1882.

Georg Cantor is a name even non-mathematicians might recognize. He showed there have to be some infinite sets bigger than others, and that there must be more real numbers than there are rational numbers. Four years after showing *that,* he proved there are as many transcendental numbers as there are real numbers.

They’re everywhere. They permeate the real numbers so much that we can understand the real numbers *as* the transcendental numbers plus some dust. They’re almost the dark matter of mathematics. We don’t actually know all that many of them. Wolfram MathWorld has a table listing numbers proven to be transcendental, and the fact we can list that on a single web page is remarkable. Some of them are large sets of numbers, yes, like for every positive whole number d. And we can infer many more from them; if π is transcendental then so is 2π, and so is 5π, and so is -20.38π, and so on. But the table of numbers proven to be irrational is still just 25 rows long.

There are even mysteries about obvious numbers. π is transcendental. So is e. We know that at least one of π times e and π plus e is transcendental. Perhaps both are. We don’t know which one is, or if both are. We don’t know whether π^{π} is transcendental. We don’t know whether e^{e} is, either. Don’t even ask if π^{e} is.

How, by the way, does this fit with my claim that everything in mathematics is polynomials? — Well, we found these numbers in the first place by looking at polynomials. The set is defined, even to this day, by how a particular kind of polynomial can’t reach them. Thinking about a particular kind of polynomial makes visible this interesting set.

## howardat58 3:26 pm

onWednesday, 13 April, 2016 Permalink |I like this stuff. I cut my mathematical teeth on Paul Halmos ” Naive Set Theory”.

So I thought about “dense set”, “Canto’s middle third”, “countable”, and then realized that U is next. How about “Uncountable” ?

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## Joseph Nebus 3:43 am

onFriday, 15 April, 2016 Permalink |I like the idea, and, done. Watch this space … well, this space on the front page.

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## A Leap Day 2016 Mathematics A To Z: Uncountable | nebusresearch 3:01 pm

onFriday, 15 April, 2016 Permalink |[…] I’m drawing closer to the end of the alphabet. While I have got choices for ‘V’ and ‘W’ set, I’ll admit that I’m still looking for something that inspires me in the last couple letters. Such inspiration might come from anywhere. HowardAt58, of that WordPress blog, gave me the notion for today’s entry. […]

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