$$f(x,y,z) = \frac{1}{x-z}$$
I see two possible domains:
$$D = \{x,y,z \in \mathbb{R}\mid x \neq z\}$$ or $$D = \{x,z \in \mathbb{R}\mid x \neq z\}$$
Is it equally valid to say the function has either domain?
$$f(x,y,z) = \frac{1}{x-z}$$
I see two possible domains:
$$D = \{x,y,z \in \mathbb{R}\mid x \neq z\}$$ or $$D = \{x,z \in \mathbb{R}\mid x \neq z\}$$
Is it equally valid to say the function has either domain?
The domain of the function is the set of inputs that are allowed into the function (i.e., the inputs that, when input into the function, make sense).
You are right that we cannot have $x = z$. But look at how the function is defined. It is $f(x,y,z)$. It takes as input a vector with $3$ components, and it outputs a number. So since the domain is the set of inputs that have an output, that means our domain must be a set of vectors with 3 components.
Here is how I would write the domain:
$D = \{ (x,y,z) \mid x,y,z \in \mathbb{R}, x \neq z \}$. Here, $D$ is a subset of $\mathbb{R}^{3}$, which the space of vectors with 3 components and real numbers as components.
You can't just say the domain is $D = \{ x,y,z \in \mathbb{R} \mid x \neq z \}$ because this set is a subset of $\mathbb{R}$, not $\mathbb{R}^{3}$. The input for the function $f(x,y,z)$ are vectors of the form $(a,b,c)$ with $a,b,c \in \mathbb{R}$ (i.e., $(a,b,c) \in \mathbb{R}^{3})$.