0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ $\;\{1\}\subset\Bbb R\;$ is independent over the rationals. $\endgroup$ – DonAntonio Feb 16 '16 at 21:42
  • $\begingroup$ Oops, I missed the 1. I've made a change to this now. $\endgroup$ – user313163 Feb 16 '16 at 21:44
2
$\begingroup$

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 $\Bbb{Q}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy