# Simple continued fraction for irrational numbers.

I read it here that: "What you must have read is that a number with an infinite simple continued fraction expansion is irrational. A continued fraction is "simple" if all the partial numerators are ones."

Wolfram has this simple continued fraction for $$\zeta(5):$$ $$[1;27,12,1,1,15,...]$$. But we don't know if $$\zeta(5)$$ is rational or not, so I understood something wrong.

• I think WA is just giving the first few terms of the simple continued fraction. – saulspatz Jan 6 '19 at 4:16

## 1 Answer

The problem is we don't know what the "$$\ldots$$" is. If it terminates after finitely many terms, $$\zeta(5)$$ is rational. If it doesn't, $$\zeta(5)$$ is irrational. We (and WA) can compute as many terms as we want by numerical computation, but there's no sign of a pattern.