Is the infinite decimal fraction $1.23456...n$ irrational?

Question

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.

Answer 1

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.