1. ## Proof, linear independence

Let $f,g,\in{F(R,R)}$ be the functions defined by $f(t)=e^{rt}$and $g(t)=e^{st}$, where $r\neq{s}$. Prove that $f$and $g$are linearly independent in $F(R,R)$.

Assume $ae^{rt}+ be^{st}=0$
$ae^{rt} = -be^{st}$
$ln(a)+rt = ln(-b) +st$
$ln(\frac{-a}{b})=t(s-r)$

For our assumption to be true, the values of $a$ and $b$ are dependent on the value of $t$, therefore there is no universal value of $a$ or $b$, except $0$, for which $af+bg=0$
Thus, these functions are linearly independent

This proof is correct?

2. You don't know if $a$ or $b$ are positive. But you can take $t=0$ in the equation, go back to the equation and take the derivative on both side, then take $t=0$.

3. Actually, we do know that one of the two constants is negative and the other is positive. $t\in{R}$ and $e^x$ is never negative, so for the equation to be true, one of the constants must be negative and the other positive.

4. Yes, it would be perfect if you add "We assume $a>0$, if it's not the case we switch $r$ and $s$.

5. $ae^{rt}+be^{st}$ is the 0 function, not just the number 0. this means this equation holds for any and all values of t.

in particular, it holds for t = 0, whence b = -a.

thus we have $a(e^{rt} - e^{st}) = 0$ for all t. if a is not 0, then

$e^{rt} = e^{st}$ for all real t, taking logs of both sides gives r = s, a contradiction, so a = 0 and b = -a = 0.

EDIT: why do i do it this way? because in the original poster's proof, he takes the log of a number which he is trying to prove is 0.

now, by the previous two comments, you can actually assume that a > 0, and b < 0, and then the original poster's argument works,

but it is not easy to see that ln(a/-b) = t(s - r) immediately implies a contradiction. i feel it is more straight-forward to use the fact that R

is a field (and thus has no zero divisors).