# Non-integer bases and irrationality

When it comes to properties like prime, irrational, rational, divisible by 2, etc., nothing changes when you change base.

But I'm not sure about the rational/irrational one.
If you use non-integer bases, even integer numbers can become ugly expressions (See 3 in base 2.5 in Wolfram)
The point is: is that number irrational, or it just has a lot of repeating decimals? Either way, how could you prove it?

• You are aware that "irrational" is not defined as "has an infinite non repeating decimal expansion", right? – Jack M Jun 4 '13 at 18:20
• I was not aware. Now I am. Thank you! – kikones34 Jun 4 '13 at 18:25

The distinction you are missing is that there are numbers, which have certain properties, and then there are numerals, which are sequences of symbols that we use to represent numbers.

The base is a choice about how to represent a number as a numeral. For example, in base 10, the number 100 is represented with the symbol 100. In base 7, it is represented as 202; in base 13 it is represented as 79. In all cases, it is a perfect square; it is an even number; it is a multiple of 5, and so forth, because it is the same number. It is still equal to $$36+64$$, regardless of whether we write that equation as 36 + 64 = 100 or as 51 + 121 = 202 or as 2A + 4C = 79.

The only questions that change with the base have to do with the representation, with the particular symbols that are used to write down the numeral. The base-10 representation of the number 100 ends with a zero, and the base-7 representation of the same number does not. A number is divisible by 10 if and only if its base-10 numeral ends with a zero, so you can conclude that 100 is divisible by 10; it is divisible by 7 if and only if its base-7 numeral ends in a 0, and so you can conclude that 100 is not divisible by 7.

3 is a rational number and an integer, regardless of how it is written or in what base. It is true that a rational number will have a repeating or terminating representation in an integer base, and an irrational number will not, but that is separate from the question of whether it is a rational number.

Writing 3 in "base $$2.5$$" means observing:

$$3 = \color{red}1\cdot 2.5 + \color{red}0\cdot 1 + \frac{\color{red}1}{2.5} + \frac{\color{red}0}{(2.5)^2} + \cdot\frac{\color{red}1}{(2.5)^3} + \cdots \\ = \color{red}{10.101\ldots}_{(2.5)}$$

but the fact that this representation does or doesn't repeat has nothing to do with whether the number 3 is an integer.

• Very clean explanation, now I got the point. Many thanks! P.S.: You edited your answer a lot o_o – kikones34 Jun 4 '13 at 18:28
• Yes, that's the way I write. You should see the drafts of my book. I tinkered with the text for years. – MJD Jun 4 '13 at 18:29
• One thing you didn't touch on that I think @Adrià Rico Blanes was asking about is that in every integer ($>2$) base the rational numbers are the ones with finite or periodic expansions and the irrational numbers are the ones with no finite/periodic expansion. The proof sketch for base 10 on Wikipedia (en.wikipedia.org/wiki/…) works fine for base 2, base 7, etc. – Mark S. Jun 6 '13 at 2:17
• I did say "It is true that a rational number will have a repeating or terminating representation in an integer base". – MJD Jun 6 '13 at 2:21