Any number that has a finite representation in the binary system have a finite representation in the decimal system. Why?
Finite representation in the binary $\implies$ finite representation in the decimal system
2 Answers
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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0Love it Amy. +=) – 2013-09-25
Just because $x/2^m$ = $5^mx/10^m$.