Is a number with two consecutive sets of differing infinite digits a real number? Suppose I construct a number starting 0. (followed by an infinite number of digits that are 3) (followed by an infinite number of digits that are 1), like $0.\bar{3}\bar{1}$.
Is this a real number?
Can I do calculations with it?
Is it real if I simply follow it by a single digit as in $0.\bar{3}1$
What are the limitations here in combining sets of overbar statements to create numbers?
 A: There is no object matching your description. In fact, the notation "$0.\bar{3}\bar{1}$" is not defined. More specifically, the phrasing "(followed by an infinite number of digits that are 3)(followed by an infinite number of digits that are 1)" does not make sense in any formalism I've heard of. Similarly, $0.\bar{3}1$ is undefined (and again, there is not really any sensible definition for it).
For reference, we define: $$0.a_{1}a_{2}\ldots a_{n}\bar{b} = 0.a_{1}a_{2}\ldots a_{n}+\sum_{k=n}^{\infty}\frac{b}{10^{k+1}},$$ and we may give a similar definition for when $b$ is replaced by a string of digits $b_{1},b_{2},\ldots, b_{m}$: $$0.a_{1}a_{2}\ldots a_{n}\overline{b_{1}b_{2}\ldots b_{m}} = 0.a_{1}a_{2}\ldots a_{n}+\sum_{k=n}^{\infty}\frac{\sum_{i=1}^{m}10^{m-i}b_{i}}{10^{k+m}}.$$ We simply neglect to define $0.\bar{3}1$ because the concept itself makes no sense.
A: If your number existed,
it could be defined as
$\begin{array}\\
\lim_{n \to \infty}
\left(
\sum_{k=0}^n 3\cdot 10^{-k}
+
\sum_{k=0}^n 10^{-k-n-1}
\right)
&=\lim_{n \to \infty}
\left(
3\sum_{k=0}^n \cdot 10^{-k}
+
10^{-n-1}\sum_{k=0}^n 10^{-k}
\right)\\
&=\lim_{n \to \infty}
(3+10^{-n-1})\sum_{k=0}^n \cdot 10^{-k}\\
&=\lim_{n \to \infty}
(3+10^{-n-1})\frac{1-10^{-n-1}}{9}\\
&=\lim_{n \to \infty}
(\frac13 -\frac29 10^{-n-1}+\frac19 10^{-2n-2})\\
&= \frac13\
\end{array}
$
which is the same as
$\lim_{n \to \infty}
\sum_{k=0}^n 3\cdot 10^{-k}
= \frac13
$.
