Is 1/x a surjective function when the domain and the range are all real numbers? SO the question is, is $f(x)=1/x$ an injective, surjective, bijective or none of the above function? The domain is all real numbers except 0 and the range is all real numbers.
It's injective because 
$f(x)=f(y)$ 
$1/x=1/y$ 
$y=x$ 
I also think that it's surjective in the given domain, as 
$x=1/y$
$f(y)=1/1/y$
$f(y)=y$
Thus it's surjective and by addition it's also bijective. Am I correct?
 A: The first part showing that it is injective is good, though you could be a little more descriptive. Like:
Let $x,y \in \mathbb{R} - \{0\}$ such that $f(x) = f(y)$. This implies that $1/x = 1/y$. Rearranging, we have that $x = y$. Thus, $f$ is injective.
The part showing that $f$ is surjective is a little jumbled. For one, it doesn't make sense to say that $f(y) = (1/y)^{-1}$. Because by definition $f(y) = 1/y$ and these two expressions are generally not the same. 
In general, to show that a function is surjective, you pick an element $y$ in the codomain and show that the function maps some element in the domain to $y$. But there is an issue when you try to do this for the function $f$ in this case. Can you spot a real number $y$ that is not in the range of $f(x) = 1/x$? If you are having trouble, just look at the graph of $f(x)$.
A: It is indeed injective but not surjective.
Because the range is $\mathbb{R} $ not $\mathbb{R} *$
$$f: \mathbb{R}*  \to \mathbb{R} \\ \quad \ x\mapsto \frac 1 x$$
$f$ is not surjective because $\forall x \in \mathbb{R}*, \  f(x)\neq 0$ 
$$g: \mathbb{R}*  \to \mathbb{R} *\\ \quad \ x\mapsto \frac 1 x$$
$g$ however is bijective.
