Let $n$ be a Number with a Decimal Expansion. Must $n$ be Real? I know that every real number has a decimal expansion.
Intuitively, it seems that the converse must also be true; that is, if a number has a decimal expansion then it must be real. However, I have never been able to find a definitive answer to this.
Furthermore, while perusing some old questions some days ago on this site, a user who posted an answer to another question stipulated something to the effect, "and assuming that a number with a decimal expansion is real \ldots"; also, I have not since been able to find the question in which the comment was made.
So, my question is:
If $n$ is a number with a decimal expansion, is $n$ a real number? 
 A: If we have an infinite decimal expansion $.a_1a_2a_3\ldots$, it is defined to be the limit of the sequence of partial expansions, i.e. $.a_1, .a_1a_2, .a_1a_2a_3, \ldots, .a_1a_2a_3\ldots a_n, \ldots$.
It should be clear that this is a Cauchy sequence of real numbers, and it is a defining property of the reals that every Cauchy sequence of real numbers converges to a real. So, our infinite decimal does in fact correspond to a real number.
A: Write your decimal in scientific notation i.e. 3.476E-10. Then your decimal representation can also be represented as the quotient of two integers 3476/10E13 (dividing the decimal representation without a decimal by some power of 10). Then, not only is any number with a finite decimal expansion a real number, but it is also a rational number, as the quotient of two integers.
A: Yes, every decimal expansion corresponds to a real number.
Specifically, if we have


*

*a finite sequence $a_1 a_2 a_3 {\tiny \ldots} a_n$ of decimal digits, and

*an infinite sequence $b_1 b_2 b_3 {\tiny \ldots}$ of decimal digits,


then the decimal expansion $a_1 a_2 a_3 {\tiny \ldots} a_n . b_1 b_2 b_3 {\tiny \ldots}$ is the decimal expansion of a real number. Specifically, the decimal expansion $a_1 a_2 a_3 {\tiny \ldots} a_n . b_1 b_2 b_3 {\tiny \ldots}$ is defined as the real number
$$\left (\sum_{i=1}^n a_i 10^{n - i} \right) + \left (\sum_{i=1}^\infty b_i 10^{- i} \right).$$
The series on the right must converge because its terms are bounded by the exponentially decreasing sequence $0.9, 0.99, 0.999, \ldots$.
If you have a number with a finite sequence of digits after the decimal point, then you can turn that into an infinite sequence by appending infinitely many $0$s on the end.
Note that it's not permissible to have infinitely many digits before the decimal point. It's also not permissible to have digits that are "infinitely far beyond the decimal point"—in other words, there may be infinitely many digits after the decimal point, but each individual digit has only finitely many digits before it. (Hence why there is no "last digit of $\pi$", or "infinitieth digit of $\pi$", and why there's no such number as $0.999\ldots9$, with infinitely many $9$s between the first $9$ and the last one.)
