Can we add variables and create new functions?

I did not know how to formulate the title, but I try to phrase the question like this:

Say I have a vector $\mathbf{a}=(a_1,a_2)$ and a function $f(\mathbf{a})=f(a_1,a_2)$, such that $\mathbb{R}^2\rightarrow\mathbb{R}$.

Next, say I also have a vector $\mathbf{b}=(b_1,b_2)$ and a function $g(\mathbf{b})=g(b_1,b_2)$, such that $\mathbb{R}^2\rightarrow\mathbb{R}$.

And now I add the vectors, $\mathbf{c}=\mathbf{a}+\mathbf{b}=(a_1+b_1,a_2+b_2)$ and I have a function $h(\mathbf{c})=\dots???$ But hey, what happens here?

Is $h$ now a function of four variables, i.e. $h(\mathbf{c})=h(a_1,a_2,b_1,b_2)$? Or what does $h(\mathbf{c})=h(a_1+b_1,a_2+b_2)$ mean?

• It the domain of $h$ is $\mathbb R^2$ (as it seems here) then it is a function on $2$ real variables, period. The fact that its argument can be written as a sum (actually sums, because for a fixed $c$ there are more possibilities to write $c=a+b$) of two does not affect that. – drhab Jun 24 '17 at 9:22
• How could something like, say, $h\bigl((7,3)\bigr)$ be determined? – Hagen von Eitzen Jun 24 '17 at 9:22

$\require{begingroup}\begingroup\newcommand{\Reals}{\mathbf{R}}\newcommand{\Vec}{\mathbf{#1}}$tl; dr: A function $y = f(x)$ is not its output value $y$, but a relationship between two variables, $x$ and $y$. You're implicitly using the same letter $h$ to denote two different functions, one of two variables and one of four variables.

To a modern mathematician, if $X$ and $Y$ are sets, then a function $f:X \to Y$ is a subset $\Gamma$ of the Cartesian product $X \times Y$ such that for every $x$ in $X$, there exists a unique $y$ in $Y$ such that $(x, y) \in \Gamma$.

Less formally, a function is a rule associating a unique value $y = f(x)$ to each $x$ in $X$.

Even less specifically (but getting at what seems to be the crux of the question), a function is a relationship between two quantities, where the term "quantity" should be interpreted liberally as "an element of some set" (not necessarily a set of real numbers).

(Caution: Loose historical claims.) This understanding of functions is relatively recent: The stringent legel definition above comes from the 20th century, while the viewpoint that a function is a single-valued relation emerged in the 19th century.

Mathematicians of prior eras, as well as physicists and engineers and calculus textbooks of today that use Leibniz notation, often focus on the variables or quantities themselves, as in, "Let $x = x(t)$ be the position of a particle at time $t$", or "Let $A = A(x, y)$ be the area of a rectangle of width $x$ and height $y$." (There are good Leibniz-notational reasons for this practice, but the notation comes at a price, particularly the large number of spurious "paradoxes" and other confusions regarding partial derivatives and the chain rule. But I digress.)

Your question is predicated on a "variables-based" viewpoint regarding functions. When you write "$\Vec{c} = \Vec{a} + \Vec{b}$", and "$h(\Vec{c}) = h(a_{1} + b_{1}, a_{2} + b_{2})$", and "Is $h$ now a function of four variables, i.e. $h(\Vec{c}) = h(a_{1}, a_{2}, b_{1}, b_{2})$? Or what does $h(\Vec{c}) = h(a_{1} + b_{1}, a_{2} + b_{2})$ mean?" there is (as I read it) an implicit assumption that "a function is its value".

Explicitly, one might be tempted to write $\Vec{y} = h(\Vec{c}) = h(a_{1} + b_{1}, a_{2} + b_{2}) = h(a_{1}, a_{2}, b_{1}, b_{2})$, and then ask whether $\Vec{y}$ is a function of two variables or of four.

The mathematical problem is, such an equation uses the symbol "$h$" to stand for two (or three, if you distinguish $\Reals^{2}$ and $\Reals \times \Reals$) distinct functions (relationships between variables). To disambiguate, a modern mathematician might write $$\Vec{y} = h(a_{1} + b_{1}, a_{2} + b_{2}) = H(a_{1}, a_{2}, b_{1}, b_{2}),$$ with $H:\Reals^{4} \to \Reals$ defined by the second equality. The resolution to your question would then be, "$h$ is a function of two variables and $H$ is a function of four variables, and these functions are distinct", though they are themselves closely related: $h(\Vec{a} + \Vec{b}) = H(\Vec{a}, \Vec{b})$. (In the spirit of hair-splitting, we're now identifying $\Reals^{4}$ and $\Reals^{2} \times \Reals^{2}$.)$\endgroup$