Friends,
Do you know of a neat proof of the linear independence (over $\mathbb{C}$) of the functions $f(t) = e^{at}$ and $g(t)=e^{bt}$ when $a$ and $b$ are linearly independent over $\mathbb{Q}$?
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This is a consequence of the Lindemann-Weierstrass-theorem, but you surely want an easier solution. – Peter Nov 16 '16 at 17:42
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You are right... – Jamai-Con Nov 16 '16 at 17:43
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Can't you do with $$\alpha e^{at}+\beta e^{bt}=0\iff e^{(a-b)t}=-\frac\beta\alpha=Cst$$ What am I missing ? – Nov 16 '16 at 17:47
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1Rolling back the edit as the intended version was asked anew. Reclosing, because the older version is a duplicate. The edited version is here. – Jyrki Lahtonen Nov 16 '16 at 19:49
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$f$ and $g$ are linearly dependent over $\mathbb{C}$ iff $f=\alpha g$, for some $\alpha \in \mathbb{C}$.
This implies $\alpha=1$ by evaluating at $t=0$.
But clearly $f=g$ iff $a=b$.
lhf
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