# Irrational numbers

I understand that irrational numbers are those that have an unending decimal tag, numbers such as pi and sqrt 2. Can I determine, using say an 8 figure calculator whether a number is irrational. In other words, how often, if ever, do numbers begin to repeat after 4 or 5 digits. Are there numbers out there that start repeating after 100 or 1000 digits?

The length of the repeating block can be arbitrarily long, and can start after any number of digits. The answer to your question is yes there are such numbers.

To be irrational doesn’t mean it’s decimal expansion never ends, it means the expansion never repeats.

Checking on your calculator alone will never demonstrate whether a number is rational or irrational. It’s easy to construct a rational number whose decimal expansion is some arbitrarily long finite, repeating sequence.

For example, say we want a rational number whose digits only repeat every million digits. OK:

.101001000100001…

(In other words, 1 (one zero) 1 (2 zeros) 1 (3 zeros) 1 (4 zeros)… and so on).

This never repeats. However, I can define a number by taking the first million digits (or however many you like) and then repeating those million digits over and over again. That’s a rational number.

If a number is a repeating decimal it is a rational number. The rational number 1/(10^1000 - 1) has a pattern that repeats every 1000 decimal places.

What Cabbage said. I would venture to guess that it would be very easy to identify rational numbers of any given length of repetition cycle. n/7 and n/14, for example (with the exception of 7/14, which is 0.5), all have a pattern of repetition of the sequence 142857, starting at different points, and, in the case of n/14 where n is odd, with a preceding digit. Like this: 1/7 = .142857142857…; 2/7 = .285714285714…; 3/7 = .428571428571…; and so on. Likewise, 1/14 = .0714285714285, and so on. As I recall n/13 has a 12-digit cycling unit. Probably n/p where n is any natural number and p is any prime other than 2 or 5 (the divisors of ten) will how this pattern in some way.

Remeber that the decimals are not the real numbers, but a represntation of the real numbers, so mathematically the fact that a numbers has no decimal represnetaion of finite length is not necessarily a particualrly interesting property of that number, just a property of representing numbers as decinmals.

How is the number defined? How are you entering it into your calculator?

If you’re dividing one integer (whole number) by another, the result is, by definition, rational. The decimal expansion will eventually either terminate or repeat.

If you’re taking the square root of something that isn’t a perfect square, it’ll be irrational.

There are IIRC numbers whose rationailty/irrationality has not yet been conclusively determined.

If I understand correctly how most calculators store numbers, any number that is actually stored in your calculator must be, technically, a rational number. Calculators don’t have infinite precision, so when the calculator shows you the square root of 2, for example, it’s actually storing a rounded-off, approximate version of this number (though it may have more precision internally than it actually shows you in the display window.)

Check out the series: Period of decimal representation of 1/n, or 0 if 1/n terminates. You can see that, as Polycarp says, the repeating period for 1/7 is 6 digits. For 1/61, the repeating period is 60 digits. (For 1/397, it’s 99 digits.)

Thanks for all the very rapid replies. Sorry, I meant to say non repeating I knew that. The discussion came up with a homework paper my daughter had where she had to label a variety of numbers as rational, irrational, real, integer, whole, or counting (???). While I knew that sqrt 2 and pi were irrational, they also had sqrt 5 and sqrt 7. Looking them up on the internet gives the answer, but I said there was no way to tell with a calculator. My wife thought that if they didn’t repeat in one or two places they would be irrational. You have answered that question. A new question is whether the sqrt x, with x being a whole number, if not an interger, is always irrational?

Yes. Thudlow Boink got it right:

Off the top of my head, I would guess yes. You would need to build on the proof that the square root of 2 must be irrational, and on the unique decomposition of x into prime factors. An integer is a perfect square if and only if that decomposition into prime factors has each prime represented an even number of times.

Well technically not quite right because the sqrt of a number that is the ratio of 2 perefct squares is also a rational number.

Complex Conjugate, it seems that would be just a simple matter of definition. Perfect square need not only apply to integers, we could define a rational perfect square to be any rational q such that q = x[sup]2[/sup] for some rational x. (That’s how I read Thudlow Boink’s comment, anyway).

I thought perfect squares were specifically the squares of integers, is it usual to define them from the rationals? But then it simply says that only the square roots of a number belonging the set of all the squares of rational numbers are rational, which is of course obvious!

I guess “perfect square” is typically used to describe integers; I was only mentioning that it certainly makes sense to speak of “perfect square rationals”, as well (i.e., it’s a well-defined concept). And yes, from that definition of “perfect square rationals” it’s quite obvious (directly from the definition) that the square root of a non perfect square is irrational. I certainly don’t mean to imply that anything I’ve said about it was even the slightest bit deep!

Ayup. In fact, in base e, both sqrt(2) and 1/2 should have non-repeating radix representations.

In base e, all positive integers greater than 2 have a non-repeating representation.

Okay, how’s this: There is no non-integer rational number whose square is an integer.
(By considering prime factorizations, you can show that p[sup]2[/sup]/q[sup]2[/sup] can’t be an integer (i.e. q[sup]2[/sup] doesn’t divide p[sup]2[/sup]) unless q divides p.)