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Suppose I have an imaginary computer, with an infinite binary table, like the one below:

$$ \begin{array}{|c|c|c|c|c|c|c|c|c|} \cdots & 128s & 64s & 32s & 16s & 8s & 4s & 2s & 1s \end{array} $$

If you don't already know how to make integers with a binary table, this is how it's done. To make an integer place a $1$ under the desired columns of the table, then add the values of each column where there is a $1$ together, and you get your number. Here's an example:

$$ \begin{array}{|c|c|c|c|c|c|c|c|c|c|} 128s & 64s & 32s & 16s & 8s & 4s & 2s & 1s & \text{result}\\ 0 & 0 & 0 & 0 & 0 & 1 & 1 & 0 & 6 \end{array} $$

My question is, how is it possible to make every [positive] [real] [whole] number with our imaginary infinite binary table? Will there be numbers that can't be made?

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    Irrational numbers! – Yiyuan Lee Oct 20 '14 at 13:10
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    Any (positive) integer can be represented like this, in a unique way, but for large integers you might need many places. By the way, if you strip the table and just keep the zeros and ones (like $110$ in your example), you've successfully written down your number in base $2$, or binary. – Arthur Oct 20 '14 at 13:14
  • @YiyuanLee Edited question. – Ethan Bierlein Oct 20 '14 at 13:15
  • @Arthur I have the table for clarity. Also, edited my question. – Ethan Bierlein Oct 20 '14 at 13:16
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    With your table, you'd still have trouble with $\frac{1}{3}$ or $\pi$, just as you would have with normal base ten, but as I said, any whole, positive number works completely fine. – Arthur Oct 20 '14 at 13:18

1 Answers1

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In order to represent non integer numbers, you must expand your table to the right, adding $1/2$, $1/4$, $1/8$ and so on. Just like decimal representation. For the sign, you must add an additional slot. Usually $0$ means positive and $1$ negative.

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