we want to show that :
$ f(x) = e^{-x} -x -1 $ is strictly decreasing.
For all $x\in \mathbb{R}$ with :
$ \left.\begin{matrix} x<y \Leftrightarrow e^{-x}> e^{-y} \\ x<y \Leftrightarrow -x-1>-y-1 \\ \end{matrix}\right\} \Leftrightarrow f(x)>f(y) $ therefore, f is strictly decreasing.
I know that the definition only requires $\Rightarrow$ and its not necessary for the proof, but my question is : can we use $\Leftrightarrow$ too?
Is there any point in the proof above that makes the use of this symbol $\Leftrightarrow$ incorrect? (For example when i sum the 2 expressions).
Thanks in advance.
You can't here, the reverse implication is false, e.g. you can have $f$ and $g$ two functions such that $f+g$ is increasing but not $f$ nor $g$:
Take for $x<0$, $f(x) = x$, $g(x) = -x/2$ and for $x\geq 0$ $f(x) = -x/2$, $g(x) = x$.
$(f+g)(x) = x/2$ (increasing) and $f$ and $g$ are clearly not increasing.
So in summary, you get $f,g$ increasing $\implies$ $f+g$ increasing.
but not the converse.
You have the same thing with inequalities.