# Linear combination is not an integer

Let $n$ be a positive integer and $s>0$ a real number. Must there exist positive real numbers $x_1,x_2,\dots,x_n<s$ such that no nontrivial linear combination of $x_1,x_2,\dots,x_n$ (with rational coefficients) is an integer?

This should be possible by taking the $x_i$'s to be "independent" irrational numbers. But I'm not sure how to prove it formally or what theorem to use.

## 3 Answers

Assuming the linear combinations you allow have rationals (or integers) multiplying the $x$s there are independent sets. Start with $x_1$ being some irrational. There are only countably many linear combinations of $x_1$, so pick $x_2$ to be some irrational that is not equal to any of them. Now there are only countably many linear combinations of $x_1,x_2$, so pick $x_3$ to be some irrational not equal to any of them. As there are uncountably many irrationals less than $s$ you can find one. Keep going. You can allow the list of $x$s to be countably infinite and the argument still holds.

You are totally right. Formally, we can go like this:

1. $\mathbb R$ is a vector space over $\mathbb Q$.
2. Using cardinality argument, $\dim_{\mathbb Q} \mathbb R = \infty$. That is, any finite subspace of $\mathbb R$ is a proper subspace.
3. Starting with the subspace $V_0 = \mathbb Q = \langle 1\rangle$

a. there's an $x_1\in \mathbb R \backslash V_0$, let $V_1 = \langle 1, x_1\rangle$.

b. there's an $x_2\in \mathbb R \backslash V_1$, let $V_2 = \langle 1, x_1, x_2\rangle$.

c. ...

Finally, you would get the desired $x_1,\dots, x_n$, and if any of them is $\ge s$, just divide it by some positive integer so that it becomes less than $s$.

I left it to you to prove that $1, x_1, \dots, x_n$ are linearly independent. But I'm pretty sure that it should be mentioned in most linear algebra course.

The fact that $\mathbb R$ is uncountable guarantees that it is infinite-dimensional as a $\mathbb Q$-vector space. So there exist irrational $y_1,\ldots,y_n$ such that $y_1,\ldots,y_n,1$ are linearly independent. Now take $x_j=|y_j|$, $j=1,\ldots,n$. As $-1\in\mathbb Q$, these are still linearly independent (together with $1$) and so no rational combination is an integer.

Finally, if $s>0$ is fixed, take $m\in\mathbb N$ with $m>\max \{x_1/s,\ldots,x_n/s\}$. Then $$\frac {x_1}{m},\ldots,\frac {x_n}{m}$$ are all smaller than $s$ and still no rational linear combination of them can be an integer.