Is there a basis which spans the real numbers? 
Is there a finite set of real numbers $S=\{a_1, a_2, ..., a_n \}$ such that every real number can be written as a linear combination (with integer coefficients) of the elements of $S$? If no, is there a set $M= \{a_1, a_2, ... \}$ with a countably infinite number of elements, such that every real number can be written as a linear combination (with integer coefficients) of the elements of $M?$

I suspect the answer is no for both. However, if $S$ and/or $M$ exist, they must contain at least one transcendental number; a linear combination of only algebraic numbers should still be algebraic.
Motivation:
First, I solved a simpler problem. I wanted to see if we can span the integers with a finite basis containing any integers except $1$. The answer is yes. One such basis is $\{2, 3 \}$, because for any integer $n$, we have $n=n \cdot (3-2) = 3n - 2n=3n+2(-n).$
Next, I want to know if this can be done with rationals. I haven't worked on it, but I think I might have a shot at discovering the answer myself. However, I'm pretty sure that I can't solve the case for real numbers. 
 A: This is essentially addressed in the comments, but let me flesh it out (if only to move this question off the "unanswered" queue):
Your suspicion that this isn't possible is correct, and the easiest way to see this is via cardinality.
If $A$ is any countable set of reals and $C$ is any set of "allowed coefficients" (you use $\mathbb{Z}$, but $\mathbb{Q}$ would give the same answer, as would the set of algebraic numbers, or so on and so forth), then the set $$LinComb(A,C)=\{x: \exists c_1,..., c_n\in C, a_1,..., a_n\in A(x=c_1a_1+...+c_na_n)\}$$ is countable. Showing this is a good exercise; the key points are:


*

*The union of countably many countable sets is countable.

*The Cartesian product of two (hence, of finitely many) countable sets is countable.

*We can break $LinComb(A,C)$ into countably many "pieces" - namely, for each $n\in\mathbb{N}$ we have the set $LC_n(A,C)$ of reals which can be written as a linear combination of elements of $A$ using coefficients from $C$ of length $n$. 
Now, what can you say about the cardinality of $LC_n(A,C)$, and how that determines the cardinality of $LinComb(A,C)$?
So $LinComb(A,C)$ is countable as long as $A$ and $C$ are. But the reals are uncountable, so we'll never have $LinComb(A,C)=\mathbb{R}$.
A: $1$ is a basis which spans the real numbers. Also a vectors space to have the said basis requires coefficients from a field.
