Prove that numbers of the form \[\frac{a_{1}}{1!}+\frac{a_{2}}{2!}+\frac{a_{3}}{3!}+\cdots,\] where $0 \le a_{i}\le i-1 \;(i=2, 3, 4, \cdots)$ are rational if and only if starting from some $i$ on all the $a_{i}$'s are either equal to $0$ ( in which case the sum is finite) or all are equal to $i-1$.
Problem
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Tags: factorial, rational numbers
Peter
30.08.2007 00:06
I just wonder how this can be correct... if all are equal to $ i-1$ gives rational number $ q\in\mathbb{Q}$, then putting all equal to $ \frac{i-1}{2}$ instead simply gives you $ \frac{1}2 q\in\mathbb{Q}$...
ideahitme
30.08.2007 06:34
Peter wrote: I just wonder how this can be correct... if all are equal to $ i-1$ gives rational number $ q\in\mathbb{Q}$, then putting all equal to $ \frac{i-1}{2}$ instead simply gives you $ \frac{1}{2} q\in\mathbb{Q}$... I don't have the book right now, I'm gonna check it later!
Peter
15.12.2007 00:20
ideahitme wrote: Peter wrote: I just wonder how this can be correct... if all are equal to $ i - 1$ gives rational number $ q\in\mathbb{Q}$, then putting all equal to $ \frac {i - 1}{2}$ instead simply gives you $ \frac {1}{2} q\in\mathbb{Q}$... I don't have the book right now, I'm gonna check it later! Have you had time to check already?
ZetaX
15.12.2007 00:23
The $ a_i$ have to be integers (everything else cannot make sense!). And then this can be proved by mimicing the proof that $ e$ is irrational.
ali666
15.12.2007 00:27
Peter wrote: I just wonder how this can be correct... if all are equal to $ i - 1$ gives rational number $ q\in\mathbb{Q}$, then putting all equal to $ \frac {i - 1}{2}$ instead simply gives you $ \frac {1}2 q\in\mathbb{Q}$... $ \frac {i - 1}{2}$ is not integer for even $ i$ therefore we can not put all $ a_{i} = \frac {i - 1}{2}$ sorry when I begin writing this post,zetaX dosen't post...
Peter
15.12.2007 00:35
ZetaX wrote: The $ a_i$ have to be integers (everything else cannot make sense!). That makes sense, thanks!
t0rajir0u
15.12.2007 02:17
Actually, what you probably want to do first is to prove that every rational number has a unique "factorial base" expansion. The rest of the proof should then proceed analogously to the proof for rational numbers in base $ b$.