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According to this question, the binary expansion of any rational number is always either terminate or repeated. And there are some irrational numbers in decimal with an irrational binary representation. For instance let us consider $\sqrt {2}$.

$$\sqrt {2}=(1.41421356237309504880\dots)_{10}\\=(1.01101010000010011110…)_2$$

My question: Is there any irrational number in decimal with a ratinal binary expansion?

Paolo Leonetti
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ASB
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    That's not the definition of a rational, only a consequence. The definition is, a number is called "rational" if it equals an integer divided by another integer. This has nothing to do with base. – Akiva Weinberger Jun 26 '17 at 09:05

2 Answers2

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No.

If a number is rational, then the expansion in any numeral system will repeat, and if it is irrational, then it will not repeat in any system.

The proof given in the question you linked is independent of the numeral system, i.e. it works in any base $b$ where $b\in\mathbb N$.

5xum
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The concept of irrational number is independent of the way we choose of representing them. So, if a number is irrational, its expression in any base will be non-terminating and non-repeating.

José Carlos Santos
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