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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.

user10360304
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1 Answers1

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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.

Rhys Hughes
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  • How about this for the surjectivity part of the proof:

    Define $f: x \rightarrow e^{-x}$. Suppose $a∈(0,1]$ and $x∈R_+$ s.t. $e^{-x}=a$. Since $(e^{-x})' = -e^{-x}$, which is continuous and strictly negative on $R_+$, the function is continuous and strictly decreasing. We also know that $e^0=1$ and $\lim_{n\to∞} e^{-n} = 0$. We then know that $f(0)=1$ and as x decreases $f(x)$ converges to zero in a continuous manner across $R_+$, and so since $R_+$ is complete, $Ran(f) = (0,1]$ and so $f$ is surjective.

     – user10360304 Apr 20 '20 at 20:37