onto functions disproving method 
Let $f: \mathbb R^* \to \mathbb R$ with $f(x) = \frac{x+1}x,$ where $\mathbb R^*$ is the set of all real numbers different from zero. 
Determine whether or not $f$ is an onto function.

I know that this is not onto. But how do I go about disproving it in a formal way. I know the range does not contain the element $1$ whilst the codomain does. And since the codomain is not equal to the range the function is not onto. But like I said how do I formally disprove it??
 A: Hint: Put $f(x)=y.$ Then $(x+1)/x=y.$ From this one obtains $x=\frac{-1}{1-y}.$
Can you see what happens when $y=1?$
A: Let's recall the following definiton.
A function $f \colon X \to Y$ is called onto (i.e. surjective) if for each $y \in Y$, there exists a $x \in X$ such that $f(x) = y$, which means that the range is exactly the codomain $f(X) = \{f(x) \mid x \in X\} = Y$.
Therefore, a function $f \colon X \to Y$ is not onto if and only if there exists a $y \in Y$ such that $f(x) \neq y$ for all $x \in X$.
For the given function $f \colon \mathbb{R}^{*} \ni x \mapsto 1 + \frac{1}{x} \in \mathbb{R}$, we know that there exists a $y = 1 \in \mathbb{R}$ and $f(x) = 1 + \frac{1}{x} \neq 1 = y$ for all $x \in \mathbb{R}^{*}$, which completes the proof.
Your approach works as well and gives a finer consequence. In fact, you have shown that the restriction function $g \colon \mathbb{R}^{*} \to \mathbb{R} \setminus \{1\} = \{y \in \mathbb{R} \mid y \neq 1\}$ is a one-to-one and onto function (i.e. bijection).
However, in order to give a formal proof, it is better to avoid using the imprecise phrase "the inverse of the function" if you haven't proved that the function is one-to-one and onto. Because a function is invertible if and only if it is one-to-one and onto.
