Infinite binary having infinite decimal

Question

Can a finite decimal has an infinite binary representation? I have come to a conclusion that it may not be possible based on what I have read from the following:

What cannot happen is that the decimal is infinite and the binary is finite.

https://www.quora.com/Is-it-true-that-some-binary-numbers-have-infinite-digits-after-the-point-or-will-they-ever-end-for-example-decimal-0-1-to-binary

Comparing infinite binary fractions to infinite decimal fractions

number with finite binary representation and infinite decimal representation

Am I right or wrong. Any reasons or proofs to facilitate the comprehension will be helpful

Answer 1

$\frac13$ – and more generally any unit fraction $\frac1n$ where $n$ contains some prime factor other than $2$ or $5$ – is non-terminating in both binary and decimal.

Answer 2

$$\frac13=0.010101\ldots_{\text{two}}=0.333\ldots_{\text{ten}}$$

Answer 3

We have $$\frac{1}{2} = .5 = (.1)_2$$ $$\frac{1}{3} = .333\ldots = (.010101\ldots)_2$$ $$\frac{1}{5} = .2 = (.001100110011\ldots)_2$$

So it can happen that

• both expansions are finite
• both are infinite
• the decimal is finite and the binary is infinite.

What cannot happen is that the decimal is infinite and the binary is finite. May be that's what you are trying to prove.

In other words we have the implication:

• Decimal infinite $\Rightarrow$ binary infinite.

Or equivalently

• Binary finite $\Rightarrow$ decimal finite.

To prove this observe that if $x = (.b_1b_2\ldots b_n)_2$ then $$x = \frac{m}{2^n} = \frac{m\, 5^n}{10^n}$$ which has a finite decimal expansion.