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.


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.

  • $\begingroup$ A linear combination with which coeffients? Over $\mathbb R$ you just need 1. With coefficients in $\mathbb Q$ or $\mathbb N$ you can't have a finite basis! $\endgroup$
    – Maffred
    Jul 16, 2016 at 4:52
  • $\begingroup$ Over what ring do you want your basis? Obviously $\mathbb{R}$ is one-dimensional over itself. And it is infinite-dimensional over $\mathbb{Q}$. $\endgroup$
    – ಠ_ಠ
    Jul 16, 2016 at 4:54
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    $\begingroup$ When talking about a basis, you usually want every element (i.e. real number) to be representable using a finite linear combination of basis elements. If your basis is countable, and your coefficient set is counrable, then the set of possible finite linear combinations is countable. $\endgroup$
    – Arthur
    Jul 16, 2016 at 5:04
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    $\begingroup$ @Ovi: the answer to the question is no: the number of linear combinations of a countably infinite set with integer coefficients is countable, because the integers are countable. $\endgroup$ Jul 16, 2016 at 5:07
  • 1
    $\begingroup$ A basis spans a vector space. A vectors space is taken over a field, Integers aren't a field. You shouldn't call it a basis then. Maybe I make mistake if so I apologize. $\endgroup$ Jul 16, 2016 at 7:20

2 Answers 2


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 $LinComb(A,C)=\mathbb{R}$.

  • $\begingroup$ Isn't your third bullet missing exactly the most important "piece", namely the linear combinations which, for every $n\in\mathbb{N}$, there exists a non-zero coefficient $c_m$ with $m>n$? $\endgroup$ Nov 29, 2022 at 20:59
  • $\begingroup$ @RonKaminsky No, it's not missing anything. Infinite sums are not linear combinations (cf. the difference between "Hamel basis" and "Schauder basis"). $\endgroup$ Nov 29, 2022 at 21:04
  • $\begingroup$ Thanks, terminology always gets me! I have to wonder, then, why this is tagged "calculus". $\endgroup$ Nov 29, 2022 at 21:38

$1$ is a basis which spans the real numbers. Also a vectors space to have the said basis requires coefficients from a field.


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