Is the infinite decimal fraction $1.23456...n$ irrational? How to prove that the number
$ 1.23456\dots n$ is an irrational number?
The number consist, of course, of natural numbers in increasing sequence.
 A: A rational number is a number of the form $\dfrac n m$ where $n$ and $m$ are integers.  A question is: what does the decimal expansion of a rational number look like?  Consider a concrete example: $\dfrac n m = \dfrac{55}{148}$.  To find the decimal expansion, do long division:
$$
\begin{array}{ccccccccccccccc}
& & & 0 & . & 3 & 7 & 1 & 6 & 2 & \ldots  \\[12pt]
148 & ) & 5 & 5 & . & 0 & 0 & 0 & 0 & 0 & \ldots \\
    &   & 4 & 4 &   & 4 \\[12pt]
    &   & 1 & 0 &   & 6 & 0 \\
    &   & 1 & 0 &   & 3 & 6 \\[12pt]
    &   &   &   &   & 2 & 4 & 0 & & & & \longleftarrow & \text{remainder $= 24$} \\
    &   &   &   &   & 1 & 4 & 8 \\[12pt]
    &   &   &   &   &   & 9 & 2 & 0 \\
    &   &   &   &   &   & 8 & 8 & 8 \\[12pt]
    &   &   &   &   &   &   & 3 & 2 & 0 \\
    &   &   &   &   &   &   & 2 & 9 & 6 \\[12pt]
    &   &   &   &   &   &   &   & 2 & 4 & 0 & \longleftarrow & \text{remainder $= 24$}
\end{array}
$$
Once we get the same remainder that we had at an earlier step, we're doing the same problem over again: what is $240$ divided by $148$?  And if we got $1.62$ last time, we still get $1.62$ and we still get the same remainder, $24$, and we're starting over again.
So the thing repears $0.37\ 162\ 162\ 162\ 162\ \ldots$.
But how do we know we must at some point get a remainder that we got before?
The answer is if you divide by $148$, the only possible remainders are $0,1,2,3,\ldots,147$.  The list of possible remainders doesn't go on forever, so at some point you get one you saw before.
Therefore in a rational number, the decimal expansion at some point starts repeating and keeps repeating.
That doesn't happen with your proposed decimal expansion, so it cannot be that of any rational number.
