Let $S$ be the set of real numbers which can be written in the form $ \sum_{n\geq0}{ \frac{\epsilon_{n}}{n!}}$ ,where ${\epsilon_n}^2=\epsilon_n$ and let $K$ be the field generated by $S$ , help me to prove or disprove that $K=\mathbb{R}$ where $\mathbb{R}$ is the set of real numbers. Thanks
field generated by a set
18
$\begingroup$
abstract-algebra
field-theory
-
0Can you prove that $a=\sum\frac{[\log n]}{n!}$ is in $K$? I have trouble decomposing $a$ into elements of $S$. Also, is the question $R=ℝ$ trivial? (where $R$ is the ring generated by $S$) – 2012-05-03
-
0basicly if you know the factorial base , every real $x$ such that $0
, there is a sequence $d_n$ of integeres such that$ 0 \leq d_n < n$ and $ x= \sum{ \frac{d_{n}}{n!}}$ – 2012-05-03 -
0yes, but is it in $K$ if $d_n→∞$? – 2012-05-03
-
0dear jmad that is why I'm asking help , actually if $ d_n < M$ for some $M$ than we can easily prove that x is in $K$ , regards – 2012-05-03
-
0This is why I provided $a$ with an unbounded yet small representation in factorial base: it is more likely to have a unique canoncical way of getting it with elements of $S$. Now have you an answer to the second part of my comment? ($R≠ℝ$? .. or even $G≠ℝ$?). – 2012-05-03
-
0no I don't have an answer – 2012-05-03
-
0Not sure if "$\sum_{n \ge 0}$" is referring to a finite or infinite sum, but I'm going to assume infinite. Obviously $\epsilon_{n} \in \{0,1\}$, so $\displaystyle \sum_{n=0}^{\infty} \frac{\epsilon_{n}}{n!} \in \{0, \sum_{n=0}^{\infty} \frac{1}{n!} \} = \{0,e \}$. $e$ is nonzero, so $\{0,\frac{1}{e},1,e\} \subseteq \langle S \rangle$. Since 1 is in there, $\mathbb{Q} \subseteq \langle S \rangle$. So now we know that $\mathbb{Q} \subset \langle S \rangle \subseteq \mathbb{R}$. Not sure where to go from here. – 2012-06-10
-
0@JacksonWalters: I think you've misinterpeted the question. By your interpretation, $K$ would be $\mathbb{Q}(e)$, which is countable and thus not $\mathbb{R}$. Notice that it's $\varepsilon_n$, not $\varepsilon$; each one is zero or one, but they don't all have to be the same. – 2012-09-16
-
5Is there a reason to write '$\epsilon_n^2=\epsilon_n$' rather than '$\epsilon_n\in\{0,1\}$'? The latter seems substantially clearer and it's not taking up a whole lot more space... – 2012-09-16