Explicitly give an infinity of rational numbers who come from below the number $9.123412341234 ...$, Explicitly give an infinity of rational numbers who come from below the number $9.123412341234 ...$, and an infinity who come from above.
I know this number is $\frac{91225}{9999}$, but how can I give that set of numbers?
Also I have to give an infinity of irational numbers who come from below the number $1.989898989898 ...$, and countless who come from above.
Any ideas or hint?
 A: For
$9.123412341234...$,
you can have,
for each integer $k$,
a lower fraction of
$9+\sum_{i=1}^k \frac{1234}{10^{4i}}
$
which is just $k$ copies
of $1234$ past the decimal point.
The sum is
$9+1234\frac{10^{-4}-10^{-4(k+1)}}{1-10^{-4}}
=9+1234\frac{10^{4k}-1}{10^{4(k+1)}-10^{4k}}
=9+\frac{1234(10^{4k}-1)}{9999\ 10^{4k}}
$.
To do it from above,
do the same but add 1 to the last digit.
This gives
$9+\sum_{i=1}^k \frac{1234}{10^{4i}}+\frac1{10^{4k}}
=9+\frac{1234(10^{4k}-1)}{9999\ 10^{4k}}+\frac1{10^{4k}}
=9+\frac{1234(10^{4k}-1)+9999}{9999\ 10^{4k}}
$.
For
$1.989898989898...
$,
use
$1+\sum_{i=1}^k \frac{98}{10^{2i}}
$
(which is just $k$ copies
of $98$ past the decimal point)
for the lower fraction and
$1+\sum_{i=1}^k \frac{98}{10^{2i}}+\frac1{10^{2k}}
$
for the upper fraction.
A: The set of rationals $\frac{n}{9999}$ where $n < 91225$ and $n$ is an integer is countably infinite with a cardinality of $ \aleph_0$. Likewise, the set of rationals $\frac{n}{9999}$ where $n > 91225$ and $n$ is an integer is countably infinite with a cardinality of $ \aleph_0$. In general, if you have a rational number $\frac{a}{b}$ there are countably infinite rational numbers $\frac{a-n}{b}$ and $\frac{a+n}{b}$ less than and greater than $\frac{a}{b}$ respectively, for rational $n$.
The second decimal expansion you give is equal to $\frac{197}{99}$. I leave it as an exercise to you to follow the procedure outlined above to produce your two infinite sets.
A: The phrasing “come from below/above” suggests to me that the intent is a monotone sequence converging toward the given number.  How about $\frac{91225n\pm 1}{9999n}$?  For the irrationals, similarly, $\frac{197x \pm 1}{99x}$, where $x$ belongs to any increasing sequence of positive irrationals, such as natural logarithms of integers starting with 2.
