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This question is simple but I don't know the exact answer: My Professors' lecture notes say $(0.01)_2 = 0.1 \times 2^{-1}$ and he insists that all numbers are written in base 2. However, I think it's wrong and I interpret this expression in this way: If we convert $(0.1)_2$ to its corresponding number in base 10 and multiply it by $2^{-1}$ then we get the value of $(0.01)_2$ in base 10 or equivalently if we convert $2^{-1}$ to its corresponding number in base 2 and multiply it by $(0.1)_2$ we get $(0.01)_2$. Am I right? Thanks.

Jean-Claude Arbaut
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Emad
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  • $(0.1)_2$ is $\dfrac 1 2$ and not $\dfrac {1}{10}$. Thus, $(0.1)_2 \times 2^{-1}= \dfrac 1 2 \times \dfrac 1 2= \dfrac 1 4= (0.01)_2$ – Mauro ALLEGRANZA Feb 19 '21 at 10:45
  • Your interpretation is almost impossible to interpret! But your professor got it slightly wrong (according to your transcription): it should be $(0.01)_2=(0.1)_2\times 2^{-1}$. – TonyK Feb 19 '21 at 10:45
  • When you "move" to the left of the decimal dot, every place is a power of 10 in decimal representation, while it is a power of 2 in binary rep. If we "move" to the right of the decimal point, every place is a negative power of ten (one-tenth, etc.) in decimal while it is a negative power of 2 (one-half,...) in binary. – Mauro ALLEGRANZA Feb 19 '21 at 10:55
  • @MauroALLEGRANZA I suppose "he insists that all numbers are written in base 2" means $0.1$ is in base $2$. – Jean-Claude Arbaut Feb 19 '21 at 10:57
  • @MauroALLEGRANZA I am aware of that. From what I understand of the question, the OP is trying to convert in base 10, do the computation in base 10 because it's more familiar to him, then convert back in base 2. Which is correct, but an unnecessary complication. We can compute in any base. Maybe it's what bugs him. – Jean-Claude Arbaut Feb 19 '21 at 11:02
  • @MauroALLEGRANZA Thanks for your answer. I understand your first comment and that was what I was thinking. What I say is that notation $0.01 \times 2^{-1}$ is wrong since the first number is in base 2 and the second number is in base 10 but books show this notation for floating point representation :They multiply a number written in an arbitrary base (like $0.345256$) (with finite digits since the memory is limited) by a power written in base 10(like $2^3, 3^5$,etc). Isn't that actually wrong? – Emad Feb 19 '21 at 11:18
  • Numbers should be in the same base when we multiply. – Emad Feb 19 '21 at 11:26
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    Maybe we can reconcile it with what your prof said, noting that $2^{-1}$ is one-half and it is simply not written in floating point notation. In binary notation, dividing by 2 is like dividing by 10 in decimal notation: move the decimal point one position to the left. – Mauro ALLEGRANZA Feb 19 '21 at 11:28
  • Of course you could also write $(0.1)2·(10)_2{}^{-1}$, but then one still has that number symbol in the index that does not exist in binary numbers. So how would one interpret $(2A){16}$, is it "natural" that the base is given in decimal? Single-digit number symbols are unique enough that they can be used independent of base. It is only when you combine multiple digits that the base enters in an essential way. – Lutz Lehmann Feb 20 '21 at 12:11
  • @LutzLehmann Good point. You answered my question. Thanks. – Emad Feb 20 '21 at 19:48

1 Answers1

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I intepret this as a question about notation, and your professor is right. $(0.01)_2$ or sometimes $0.01_2$ are both common notations for binary expansions, so by definition

$$(0.01)_2 = 2^{-2},$$

Edit: Notice that according to your transcription, your professor’s notation is sloppy. He should indeed write

$$(0.01)_2 = (0.1)_2 \cdot 2^{-1}.$$

More generally for any $\beta > 1$. We have a $\beta$-expansion of a number:

$$(d_m \ldots d_1. d_{-1} \ldots d_{-n})_{\beta} = d_m \beta^m + \cdots d_1 \beta + d_{-1} \beta^{-1} + \cdots d_{-n} \beta^{-n}$$

Thusle Gadelankz
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