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Let r a real number. Let f(r) be a function of r such that in binary representation of r, the first digit is repeated once, the second digit is repeated twice, the third three times and this is continued ad infinitum.

Couple of examples:

r = 101.011

f(r) = 1 00 111. 0000 11111 111111

r = π (11.0010010 etc)

f(r) = 1 11. 000 0000 11111 000000 0000000 11111111 000000000 etc

My questions are:

  1. Are f(√2) and f(π) irrational ?

  2. Does r is irrational imply that f(r) irrational ?

I may be missing something obvious but have not been able to proceed.

We Pretty
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2 Answers2

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$f(r)$ will be irrational any time $r$ has infinitely many $0$s and $1$s in its expansion, so any time it is not a dyadic rational, one with denominator $2^k$. There will not be any repeating pattern. If $r$ is a dyadic rational, $f(r)$ will be also because it will end in an infinite chain of $0$s or $1$s.

Ross Millikan
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Of course.... just think about it.

If the result is rational than the result has periodic repeating squence. But if the sequence is of length $n$, and as every digit past the $n$th place in the input is repeated $n$ and more times. Which means the sequence of $n$ digits must all be the same digit. And that means that all the digits past the $n$th place in the original are are all the same digit. Which means the original was a rational.

Furthermore it was a terminating rational.

And if it was a terminiating rational then it has a last digit an so does the output.

So $f(\frac j{2^k}) \mapsto \frac m{2^n}$ for some $m,n$..

$f(\frac j{(2m+1)*2^k})\mapsto$ irrational (but one of the weird increasing spaces irrationals)

$f(irrational)\mapsto$ irrational.

fleablood
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