# Conclusion about the vectors from linear independence

Suppose $S=\{1,s_1, s_2,...,s_n\}\subset\Bbb R$, show that if $S$ is linearly independent over $\Bbb Q$ then $s_1,s_2,...,s_n$ are irrational.

I can see that if $S$ is linearly independent then $\lambda_1s_1 + \lambda_2s_2 + ... + \lambda_ns_n=0 \Rightarrow \lambda_1=\lambda_2=...=\lambda_n=0.$ I don't see how you can conclude from this that each of the $s_i$'s are irrational though.

• $\;\{1\}\subset\Bbb R\;$ is independent over the rationals. – DonAntonio Feb 16 '16 at 21:42
• Oops, I missed the 1. I've made a change to this now. – user313163 Feb 16 '16 at 21:44

Suppose that some $s_i$ is rational, say $s_i = p/q$. Then $-p\cdot 1 + qs_i = 0$, meaning that $\{1,s_1,s_2,\ldots,s_n\}$ is linearly dependent over $\Bbb{Q}$.