Irrational Numbers – How to Know That Irrational Numbers Never Repeat?

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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.”

Best Answer

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.

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