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Any number that has a finite representation in the binary system have a finite representation in the decimal system. Why?

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2 Answers 2

up vote 5 down vote accepted

To elaborate:

Again, in this question, you can use what you established in your earlier post:

A real number has a finite representation in the binary system if and only if it is of the form $$ \pm \frac{m}{2^n} \text{ where}\;n \text{ and} \;m \text{ are positive integers.}$$

Likewise,

A real number has a finite representation in the decimal system if and only if it is of form $$ \pm \frac{k}{10^n}\text{ where}\; k, n\text{ are positive integers.}$$

Noting that $$\pm \frac{m}{2^n} = \pm \frac {5^nm}{5^n2^n} =\pm \frac{5^nm}{10^n} = \pm \frac{k}{10^n},\;\text{with}\; k = 5^n m \;\text{ and}\;\,m, n\in \mathbb{Z},\;m>0, \;n>0,$$ we conclude that any number that has a finite representation in the binary system also has a finite representation in the decimal system.

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Love it Amy. +=) –  Babak S. Sep 25 '13 at 3:59

Just because $x/2^m$ = $5^mx/10^m$.

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