Why is $1 \over x$ a function?

Question

I have just learned that every relation which is both left-total and right-unique is called map or function.

Why is $f: \mathbb{R} \rightarrow \mathbb{R}, x \mapsto {1 \over x}$ or as relation $G = \{(x,y) \in \mathbb{R} \times \mathbb{R} \mid y = {1 \over x}\}$ a function?

To be left-total, $\forall x \in \mathbb{R} \exists y \in \mathbb{R}: (x,y) \in G$ must be true. But that is not true for $x=0$, since ${1 \over 0} \notin \mathbb{R}$.

Is that special case simply ignored since it is not "allowed" to divide by zero or am I missing something?

Answer 1

There is no function $f: \mathbb{R} \to \mathbb{R}$ defined by $x \mapsto \frac{1}{x}$, for the reason you state: it doesn't assign a value to the input $0$.

The function $x \mapsto \frac{1}{x}$ is usually given with domain $\mathbb{R} \setminus \{0\}$.

Answer 2

Sometimes functions are lazily defined with implicit domains which you take to be the largest set of $\mathbb R$ that the function makes sense. This occurs frequently in first-year calculus courses, but should be more explicit if you are operating at your level of detail.

Your relation as defined is not a function over $\mathbb R$, but rather $\mathbb R \setminus \{0\}$. However, you can make it a function from $\mathbb R$ to $\mathbb R$ by explicitly defining a value for $f(0)$ to make it left-total.