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Forgive me if I have set this up wrong, I haven't done proofs in a long time. I was thinking at lunch today what it would be like if we could write binary behind a decimal point. Imagine

$$110.11 = 6.75$$

This made me think, are there numbers in base-10 decimal that cannot be represented rationally in base-2 decimal? Maybe, but since 2 evenly divises 10, maybe not. How about 3? $n$?

This is my conjecture:

$$\forall ( x_1,l_1,n \in Z | x_1 < 10^{l_1} ) \exists i_2,l_2 (\frac{x_1}{10^{l_1}} = \frac{x_2}{n^{l_2}} ) $$

How would I go about proving it?

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    A number is rational if and only if it is a quotient of two integers; the representation base is irrelevant. Do you mean, will the numbers that have a non-repeating $n$-ary representations be exactly the irrationals?2012-06-22
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    Yes, this is what I mean.2012-06-22
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    Except that the title that you edited does not agree with what you say in your previous comment. In your title you are asking about *terminating* expansion, but above you agreed that you were talking about *non-repeating* expansions. Which is it?2012-06-22
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    Both mean the same thing. If there's a different set of terminating numbers there will also be a set of non-terminating numbers. Correct?2012-06-22
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    :, No, they don't mean the same thing. A terminating expansion is one that is eventually equal to ....0000000....; a *repeating* expansion is one that eventually repeats (like ...03820382038203820382...) Every terminating expansion is repeating. Not every repeating expansion terminates. And **neither** of them is a **non**-repeating expansion. The set of terminating numbers may be different, but the set of **non repeating** expansions is the same regardless of the base.2012-06-22

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