I have a question pertaining to the nature of graphs in 3 dimensions.
Consider Fermat's Last Theorem:
$$x^n + y^n = z^n$$
When $n$ is varied and the solution is graphed, the function is in $\mathbb R^3$ (real coordinate space, 3 dimensions) rather than $\mathbb R^2$, since it has a $z$ variable. For some reason, I have trouble wrapping my head around why, precisely, the introduction of a third variable increases the number of dimensions in a solution. Fundamentally, I mean. It's a factor in vector algebra, for example, and I know how to visualize it and do maths with it in that zone, but that's because I'm accepting the fact that it is in $\mathbb R^3$ space because it's modelling our physical world and we're considering functions with unit vectors $i$, $j$, and $k$ which all project perpendicularly to eachother. Oddly enough, that works for me, but not necessarily $x$, $y$, and $z$ instead.
Essentially, am I right in assuming adding a new variable to a function creates $\mathbb R^{n+1}$ (another "dimension", per se)? Like, if I were to take $x + y = b$ where b is some constant, this is in $\mathbb R^2$. However, if I add $z$ to it, so it's say, $x + y + z = c$ where $c$ is also a constant, is this not now in $\mathbb R^3$? If so, why?
If you have an equation in $n$ real variables $x_1,\ldots,x_n$, then solutions can be viewed ordered sequences of $n$ real numbers $r_1,\ldots,r_n$, so they are vectors $(r_1,\dots,r_n)\in\Bbb{R}^n$. So adding a new variable adds another dimension to the space in which solutions live.