## Ask Professor Puzzler

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"Hi Professor Puzzler, in last week's post you said that if a number is irrational, it doesn't terminate or repeat. I wonder how you know that? Of all the irrational numbers, how do you know there isn't one that terminates or repeats?"

This question comes from Trevor in Georgia.

Hi Trevor, in answer to your question, I'll give you a short proof (with a bit of hand-waving over some details, just to keep the write-up from being too long!). What we'll do is, we'll start with numbers written in decimal notation, which we'll assume are irrational (which means, as you remember from the post about the square root of two, that it can't be written as a ratio of integers) and then we'll arrive at a contradiction, which will prove that they aren't irrational. So let's get started!

## Terminating Decimal

Let n be an irrational number with a terminating decimal. This means that after a certain number of decimal places (let's call that k), the decimal ends.

Since the decimal terminates after k decimal places, the number 10^{k}n is an integer (every time you multiply by 10, you shift the decimal one place to the right, so multiplying n by 10^{k} results in all the digits being to the left of the decimal).

Now we can write n = (10^{k}n)/(10^{k}), which is ratio of two integers, and that contradicts our original assumption that n was irrational. Thus, if a number in decimal form terminates, it is *not *irrational.

## Repeating Decimal

Let n be an irrational number with a repeating decimal. This means that after a certain number of decimal places (let's call that k), the decimal begins repeating every h digits, where h is some integer.

For example, if the number is 3.125727272..., k = 3, h = 2.

To simplify our math, let's multiply our number by 10^{k}, which allows us to get the non-repeating part of the decimal over to the left of the decimal. We understand that when we're done with our manipulations, we have to divide by 10^{k} to get our number back, but if we've proven that our result is rational, then dividing by 10^{k} still gives us a rational number.

Similarly, everything to the left of the decimal can then be subtracted, with the understanding that when we're done we must add it back in. But the same argument holds; if we've proved the result is rational, adding an integer to that rational number will result in another rational number.

In other words, we can, without loss of generality, simplify our starting point to a decimal which looks like this:

n = 0.a_{0}a_{1...}a_{h}a_{0}a_{1...}a_{h}a_{0}a_{1...}a_{h...}

Let m be the integer formed by the following digits: a_{0}a_{1...}a_{h}

Then n = m/(10^{h}) + m/(10^{2h}) + m/(10^{3h}) + ...

This is an infinite series with common ratio 1/(10^{h}), and its sum is:

S = [m/(10^{h})]/[1 - 1/(10^{h})]

S = m/(10^{h} - 1)

Since m is an integer, and 10^{h} - 1 is an integer, we've shown that our number can be expressed as a ratio of integers, which contradicts our assumption that it is irrational. From this we've shown that if a decimal repeats or terminates, it is not irrational.

Thanks for asking, Trevor!