Is the following (decimal) number irrational?
0.10100100010000100000100000010000000100... etc.
My intuition tells me it is irrational. My informal "proof" is simply that it doesn't contain a repeating set of digits.
For irrationality, is it both a necessary and sufficient condition that the digits never revert to a repeating sequence?
Is there a more formal proof for this case?
On
Yes, a number is rational if and only if its decimal representation is eventually periodic (including the possibility of a period $\overline 0$)
A formal proof for your specific number requires a formal definition. I assume that your number is $$\sum_{n=1}^\infty 10^{-\frac{n^2+n}2}.$$ Any decimal representation that has infinitely non-zero digits (which is the case for your number) and has blocks of zeroes of arbitrary size (which is also the case for your number) cannot be eventually periodic: Some late period must lie completely in a sufficiently big block of zeroes, hence the period must be all zeroes, contradicting the fact that some non-zero digit occurs further to the right.
A number with a similar expression can even be shown to be not only irrational, but in fact transcendent: $$\sum_{n=1}^\infty 10^{-n!}.$$
On
What you have there is a trancedental number; a number for which there is no variable polynomial equation with rational coefficients that has this number as a root. Trancedental numbers are always irrational (but not all irrational numbers are trancedental). Therefore, yes, your number is irrational.
The proof is in the construction; like the well-known trancedental numbers $\pi$ and $e$, your number is the asymptotic limit of the sum of an infinite series; in this case, the sum of a reducing fraction:
$$\sum_{n=1}^{\infty} \dfrac{1}{10^{\dfrac{n(n+1)}{2}}}$$
This is similar to the construction of the Champernowne constant which is proven transcedental. The sum is constructed such that no 2 terms ever modify the value of a decimal place of the same order of magnitude, very much like $C_{10}$, and so the number constantly increases but can never reach a rational sum, unlike the infinite sum of $\frac{1}{2^n}$.
A formal justification of your informal proof can be achieved by noting that in the process of long division, the fact that you have a finite number of possible remainders guarantees that eventually a remainder will be repeated. That is, for any rational number its decimal expansion becomes periodic.