# How to know that irrational numbers never repeat?

How would you respond to a middle school student that says: “How do they know that irrational numbers NEVER repeat? I mean, there are only 10 possible digits, so they must eventually start repeating. And, how do they know that numbers like $\pi$ and $\sqrt2$ are irrational because they can't check an infinite number digits in the decimal form to see whether there is a repetition.”

• Maybe explain in more detail what "repeats" and "never repeats" mean. For example, $0.1221112222111112\dots$ "never repeats." – André Nicolas Sep 3 '13 at 1:02
• Yeah... that's tough, with middle schoolers. I tend to prove this to my calculus students once they've learned sequences and series. – Nick Peterson Sep 3 '13 at 1:03
• In this context "irrational numbers never repeat" doesn't mean "no digit occurs more than once" but "no finite sequence of digits occurs periodically". – walcher Sep 3 '13 at 1:03
• Rather than repeating, mention periodic. It is more precise, unambiguous, and yet easy to grasp. Something very simple to show is that if $x$ has a periodic, or eventually periodic decimal expansion, then $x$ is rational. For example, if $x=0.5632632632...$, then $1000x=563.2632632632...$, so $999x=562.7=5627/10$, etc. Thus, if $x$ has an expansion that is not eventually periodic, it must be irrational. – Andrés E. Caicedo Sep 3 '13 at 1:24
• As for $\sqrt 2$, this is a very nice site: cut-the-knot.org/proofs/sq_root.shtml – Andrés E. Caicedo Sep 3 '13 at 1:25

We know that irrational numbers never repeat by combining the following two facts:

• every rational number has a repeating decimal expansion, and
• every number which has a repeating decimal expansion is rational.

Together these facts show that a number is rational if and only if it has a repeating decimal expansion.

Decimal expansions which don't repeat are easy to construct; other answers already have examples of such things.

I think the most important point to make with regard to your question is that nobody determines the irrationality of a number by examining its decimal expansion. While it is true that an irrational number has a non-repeating decimal expansion, you don't need to show a given number has a non-repeating decimal expansion in order to show it is irrational. In fact, this would be very difficult as we would have to have a way of determining all the decimal places. Instead, we use the fact that an irrational number is not rational; in fact, this is the definition of an irrational number (note, the definition is not that it has a non-repeating decimal expansion, that is a consequence). In particular, to show a number like $\sqrt{2}$ is irrational, we show that it isn't rational. This is a much better approach because, unlike irrational numbers, rational numbers have a very specific form that they must take, namely $\frac{a}{b}$ where $a$ and $b$ are integers, $b$ non-zero. The standard proofs show that you can't find such $a$ and $b$ so that $\sqrt{2} = \frac{a}{b}$ thereby showing that $\sqrt{2}$ is not rational; that is, $\sqrt{2}$ is irrational.

• Schizophrenic numbers would tend to disagree – Klangen Dec 18 '19 at 12:14

What about the number $0.1010010001000010000010000001 \dots$? The runs of zeroes always get longer so clearly it never repeats. I think that might satisfy the objection that the whole concept of a non-repeating decimal doesn't make sense.

The equivalence between having a representation as a fraction and having a representation as an eventually-repeating decimal is somewhat deeper and for a middle school student it might be best to take it on faith. A place to start on that topic is Euler's theorem. One direction of it (that all eventually-repeating decimals can be represented as fractions) can be shown with a convergent series but I'm just not sure about a simple way to explain the other.

• It's not that deep: If $x$ is eventually repeating, then we can assume (by multiplying with some power of $10$) that the iteration starts immediately after the decimal point. Let $n$ be the length of the repeating sequence of digits, then $10^nx$ has the same sequence of numbers occurring after the decimal point, so $10^nx-x=m$ is an integer. Thus, $x=\frac m{10^n-1}$ is a rational. – walcher Sep 3 '13 at 1:11
• I just edited when I realized the same thing but what about the other direction? – Dan Brumleve Sep 3 '13 at 1:12
• For a student who is familiar with long division, it's easy to understand why a rational number has a repeating decimal expansion: once you get past the initial digits of the numerator, the digits repeat whenever the remainder is the same. Unfortunately, long division is out of fashion in many middle schools these days. – Robert Israel Sep 3 '13 at 1:13
• This can be shown using long division: If $x=\frac mn$, then there are only finitely many numbers that can occur as the remainder when divided by $n$, so the sequence behind the decimal point must repeat eventually. – walcher Sep 3 '13 at 1:16

Show them a simple decimal like 0.11000111100000111111... which has 1 zero, 2 ones, 3 zeros, 4 ones, 5 zeros, and so on.

This never repeats.

Also, say that numbers are proved to be irrational by assuming them to be rational and deriving a contradiction.

Perhaps you can then give one of the many proofs that $\sqrt{2}$ is irrational.

You can also use the fact that a fraction must repeat, so the first number above can not be a fraction since it has arbitrarily long sequences of zeros and ones.

Looks like you've missed a few symbols/letters in your question, but the spirit is there. Depends what you mean by 'repeat'. If you mean eventually terminate in an $N$-digit sequence that is iterated infinitely, then you can write out the fraction that corresponds to that 'irrational' number $\implies$ it's not actually irrational. E.g., $$0.8333 = 8/10 + 3/90 = 15/18.$$

There's also the canonical proof that $\sqrt 2$ isn't rational. Assume so. Then $\sqrt 2=p/q$ for $p,q\in\mathbb Z$ and $(p,q)=1$. Then $$2=\frac{p^2}{q^2}\iff 2q^2=p^2\implies 2\mid p\implies 4\mid p^2.$$ So write $p^2=4r^2$. Thus $2q^2=4r^2\implies 2\mid q$, which contradicts $(p,q)=1$. That's the rough proof at least.

• What does (p, q) = 1 mean? – Defacto Feb 25 '19 at 1:54

One way of disproving their suggestion that "there are only possible digits, so they must eventually start repeating" would be to show them the string: 0.10110111011110111110111111011111110.... and convince them that they can always choose a sub-string (for example '0'+'1'-(repeated n times)+'0') that will never be repeated further along the string. Thus, the string can never repeat-in-full (which is what you are really trying to show). If you can convince them of that then you should have no trouble convincing them that with 10 digits itis even easier to find non-repeating infinite strings of digits.