Is there a possibility that we graph a sub-plane in xy plane using an equation based on 2 variables?

An equation with 2 variables (x and y) is observed to be giving a graph of a point or a curve at most, in xy-plane. But since an Equation with 2 variables can give a graph of multiple points (a curve), ain't it make it possible for such an equation to give a graph of a sub-plane (a plane smaller than xy-plane) that lies anywhere in xy-plane ?

I mean we may yet haven't found such a 2 variables-based equation but there is always a possibilty that such entity exists, right?

In case you haven't got me:

If you haven't got me yet here I have something for you that might help. Suppose we have a random equation in 2 variables xy-3x=6, we all know that when we plot these kind of equations in xy-plane, we get a curve. So my question is, can there be a similar equation (equation in 2 variables; x and y) which actually gives us a graph of a plane (surface) when we plot it in xy-plane? I mean, can it (equation in 2 variables) give us the coordinates (x, y) of a rectanglular region or any 2D-region upon solving?

• Your assumptions are wrong, even for the one-variable case. $x=|x|$ gives the entire non-negative real line, not a point. – Parcly Taxel Mar 11 '18 at 6:12
• But when you are not in xy-plane i.e you have only x axis, x=x defines a point only but I got you. I should have made it clear. @ParclyTaxel – Anonymous Mar 11 '18 at 7:17
• graph of a sub-plane (a plane smaller than xy-plane) What does that even mean? – dxiv Mar 11 '18 at 7:41
• can it (equation in 2 variables) give us the coordinates of a rectangle (x, y) For example $\,(x^2-1)^2+(y^2-1)^2=0\,$ has the vertices of the unit square $\,(\pm1, \pm1)\,$ as solutions, if that's what the question is about. – dxiv Mar 11 '18 at 7:50
• a simple plane drawn in xy-plane There is no such thing math-wise. A rectangle is not a plane, and the interior of a rectangle is not a plane, either. It would help your question if you clarified what you are asking in terms others can understand. – dxiv Mar 11 '18 at 8:06

Using that $\,a+|a|=0 \iff a \le 0\,$:

• $\,y-|y|=0\,$ represents the upper half-plane $\,y \ge 0\,$;

• $\,x^2+y^2-1+|x^2+y^2-1| = 0\,$ represents the unit disc $\,x^2+y^2 \le 1\,$;

• $\,\big(x^2-1+|x^2-1|\big)^2+\big(y^2-1+|y^2-1|\big)^2=0\,$ represents the square $\,x,y \in [-1,1]\,$.

[ EDIT ]   If the equations are restricted to polynomial equations, however, then the answer is negative:  if the zero set of a bivariate polynomial includes an open disk then the polynomial is identically zero on the entire plane, in other words it is the zero polynomial (see e.g. this). Therefore it is not possible to write a polynomial equation whose solution set is the interior of a 2D curve.

I don't quite understand what you want but maybe it is something like this,

$$x+y < 5$$

This when graphed looks like what you are calling a "sub-plane"

Is this what you wanted?

Note that you can easily get the whole $XY$ plane by writing something like this,

$$x^2 + y^2 \geq0$$

If you only want a equation and not an inequality then a example would be,

$$x+y = x+y$$

This also maps the whole $XY$ plane.

• x+y=x+y becomes 0=0. And I can't really imagine it being plotted. Any additions? – Anonymous Mar 11 '18 at 7:36
• @Anonymous $0=0$ is another way of saying that no restrictions on the points you can choose. Any point will satisfy the self identity $x+y=x+y$ hence all the points on the $XY$ plane will be plotted. – Agile_Eagle Mar 11 '18 at 8:55