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I was writing an article and I stumbled across a situation where I was doubting how to be rigorous in my writing. I'm going to give an example to illustrate the problem.

Let $X = \{(x, y, z), x, y, z \in \mathbb{R}\}$. Now we want to construct a scalar function, but I was doubting on how to go about it. If one would define the function $f: \mathbb{R}^4 \rightarrow \mathbb{R}^3, f(x, y, z, s) = (x \cdot s, y \cdot s, z \cdot s)$ and would take an arbitrary $x \in X$ where $x = (x_1, x_2, x_3)$, then my paranoid mind would think that $f(x, s) = ((x_1, x_2, x_3) \cdot ?, s \cdot ?, ?, ?)$. I put question marks, as with this reasoning, the third and fourth arguments don't exist.
One could say, define a function $f(x, s) = (x_1 \cdot s, x_2 \cdot s, x_3 \cdot s)$. But how would one define the domain then? We know that $x \in \mathbb{R}^3$ and $s \in \mathbb{R}$. But if we would say $f: \mathbb{R}^3 \times \mathbb{R} \rightarrow \mathbb{R}^3$, then that would be equivalent to saying $f: \mathbb{R}^4 \rightarrow \mathbb{R}^3$. Can someone explain this to me and how math views these cases? Is there a difference in putting a couple of arguments as a coordinate/tuple as opposed to putting the individual elements as parameters? Thanks in advance!

Sakoboy
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    The short answer is that we freely switch between $f(x,y)$ and $f(z)$ where $z=(x,y)$, and similarly we freely switch between $\Bbb R^3\times \Bbb R$ and $\Bbb R^4$. Are they formally different? Yes, but the nuanced differences between them are merely pedantic and there is an obvious morphism (of every kind) between them so they might as well be treated as the same for all intents and purposes. – JMoravitz Nov 26 '22 at 04:02
  • Thanks a lot! That explains a lot. So even if I were to say $\phi = (1, 2)$, and $f(x,y,z) = x + y+ z$, then $f(\phi, 3) = 6$? I read the post you shared, and according to that, it would make sense. Thanks again! – Sakoboy Nov 26 '22 at 04:33
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    Yes, that’s pretty much it. It is also very rare that you want to switch the parentheses all the time in all kinds of ways. Often, there are only two or three options that come up over and over. (Say space-time represented as four components $(x,y,z,t)$ or as location + time $(\vec r, t)$. You wouldn’t run into a situation were you combine time with a single location coordinate.) You will get used to those combinations very quickly. – Eike Schulte Nov 26 '22 at 09:57

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