Does the decimal part of \( \frac{\sqrt{5}-1}{2} \) have an infinite number of integers that is less than 5? If yes, can you prove it?

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

## Comments

Sort by:

TopNewestIt's equivalent to proving that \(\{\dfrac{1}{2}\cdot 10^x\sqrt{5}\} < \dfrac{1}{2}\) has infinitely many solution in positive integers \(x\).

Kronecker's Approximation Theorem seems too weak... This reminds me of the problem of proving if an irrational is a Normal Number which is not an easy task, although the problem statement seems to be quite a bit looser.

Anybody have any ideas? – Daniel Liu · 1 year, 9 months ago

Log in to reply

– Brian Charlesworth · 1 year, 9 months ago

If \(\sqrt{5}\) is normal then the answer would be yes, but as you point out, such a strong condition may not be necessary here. That said, if there were only a finite number of digits less than \(5\) in the expansion, then there would be some digit such that all subsequent digits are \(5\) or greater. If we then divided (or multiplied) by \(2,\) then some, (and probably an infinite number), of these subsequent digits would "switch" to being less than \(5.\) It's hard to see how such a simple operation could so drastically change the density of the digits of the decimal expansion of an irrational number, (which is most likely normal in the first place), but I'll have to read more about normal numbers to do more than just wave my hands and say "Clearly, ....". :)Log in to reply

– Daniel Liu · 1 year, 9 months ago

Oh wait, right the density of numbers in \(\sqrt{5}\) is definitely not the same as \(\dfrac{\sqrt{5}}{2}\).Log in to reply