I'm practicing my proof-writing and was hoping you could let me know if this proof looks good. I would like to know if the proof is incorrect, if there are parts that are overly wordy/complicated, or if I'm missing some element of a proof that is helpful to see, if not strictly necessary.
The Prompt:
Show that $f: x → e^{−x}$ is a bijection from $R_+$ onto $(0, 1]$.
My Proof:
Define $x,y∈R_+$ and suppose $e^{-x} = e^{-y}$. Then $-x = -y$ and $x = y$. If $f(x)=f(y)$ then $x=y$, so $f$ is injective.
Now suppose $a ∈ (0,1]$ and let $x=-ln(a)$, which will be an element of $R_+$ for any $a ∈ (0,1]$.
$e^{-x} = e^{-(-ln(a))} = a$
Thus every $a$ element of the co-domain has some $x∈R_+$ s.t. $f(x) = a$, so the mapping is surjective. Thus, f is a bijection.
*Is the latter half of the proof acceptable? Am I allowed to use the properties of the natural log to prove that $e^{-x}$ is surjective? I'm not sure how I could prove surjectiveness without using that information.
You've pretty much nailed it. However, a problem I notice is that while the statements you make are correct, you aren't showing why they are correct.
For example, the first I notice "suppose $e^{-x}=e^{-y}$. Then $-x=-y$".... "So $f$ is injective".
You need to show how $e^{-x}=e^{-y}\iff -x=-y$.
I would do this by taking the derivative $(e^{-x})'=-e^{-x}$. This is continuous and non-zero over our range, and thusly $f$ is monotone, which is sufficient for injectivity.
The same issue lies in the surjectivity part of your proof. See if you can add the necessary clarity there.