I have a question that says...
THEOREM: The function $f: \mathbb{R}_{\geq 0}\rightarrow \mathbb{R}$ given by $f(x) = ln(x)$ is onto. If you were going to prove this statement, what is the first sentence. After this sentence what is the new goal?
So I known this is a proof involving functions, but I'm unsure how to start a proof of this. Is the first sentence just suppose to be $f: \mathbb{R}_{\geq 0}\rightarrow \mathbb{R}$? In general how to you start proof involing functions.
The first sentence will be a definition of "surjective" in this particular context.
For example:
We need to show that for every $y \in \mathbb{R}$, there is $x \in \mathbb{R}^{>0}$ such that $\ln(x) = y$.
So, you need to show $f(x)=\ln(x)$ for $(x>0)$ is onto. That means, you need to show that for every $y\in \mathbb{R}$, you can find an $x\in \mathbb{R}_{>0}$ such that $f(x)=\ln x=y$. This tells me $x=e^{\ln x}=e^{y}$. Here's how I'd write the proof.
Fix $y\in\mathbb{R}$. Let $x=e^{y}$. Since $e^{y}>0$ for all $y\in\mathbb{R}$, it follows that $x=e^{y}\in\mathbb{R}_{>0}$. And, $f(x)=\ln(x)=\ln(e^{y})=y$. Hence, since $y$ was arbitrary, $\ln:\mathbb{R}_{>0}\to\mathbb{R}$ is onto.
The notation is already defined in your "theorem". What you start with would thus be unmangling the definitions used in the theorem. It says
The function $\ln : \mathbb R_{>0} \to \mathbb R$ is surjective (=onto).
(note $\mathbb R_{\ge 0}$ is not the domain of $\ln$ since $\ln 0$ is undefined)
Thus you need to find two definitions:
And then you try to prove surjectivity, i.e. $$\forall y\in\mathbb R: \exists x\in \mathbb R_0 \text{ s.t. } \ln x = y$$
