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What does the graph for $\left|x\right|-x=\left|y\right|-y$ look like for $x, y \ge0$

When considering absolute values of $x$ and $y$ there are 4 cases:

  1. $x\lt 0 \; y\lt 0$
  2. $x\ge 0 \; y\lt 0$
  3. $x \lt 0 \; y\ge 0$
  4. $x\ge0, \; y\ge 0$

Solutions for the first three ones are:

  1. $x = y; \; x,y < 0$
  2. $y = 0; \; x\ge0, y<0$
  3. $x = 0; \; x<0, y\ge0$

But for $x, y \ge 0$ the equation holds for every single value of $x$ and $y$. Does that mean that the whole plane $x, y \ge 0$ is part of the graph? If so then what tools could I use to see that? Desmos and W|A are not showing that the plane is part of the graph.

Here is how Mathematica plots the function:enter image description here

roman
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    See https://math.stackexchange.com/questions/2642860/plotting-a-graph-of-a-equation/2642868#2642868 for a similar plot generated with Mathematica's RegionPlot. – Jalex Stark Jul 09 '18 at 10:40
  • Are you asking how to plot it correctly in Mathematica or how to plot it in general? – Rory Daulton Jul 09 '18 at 10:42
  • @RoryDaulton I was a bit confused that mathematica didn't fill the first quarter, and based on link provided by Jalex Stark i should have used RegionPlot instead. – roman Jul 09 '18 at 10:51
  • In Desmos you can get a good idea of the region by plotting $$\left|(|x|-x)-(|y|-y)\right| < k$$ for a small positive $k$ (say $0.01$). The idea is that as $k \to 0$, the solution will approach $\left|(|x|-x)-(|y|-y)\right| = 0$, which is equivalent to your equation. For implicit plots, it's easier for Desmos (and most other software) to plot a region when there's a bit of leeway. – Théophile Jul 09 '18 at 11:21

2 Answers2

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$$|x|-x=|y|-y\tag 1$$

Case $x> 0$ and $y>0$ : $$|x|-x=0=|y|-y=0 \qquad\text{Eq.}(1)\text{ is true any }x>0,y>0.$$

Case $x<0$ and $y<0$ : $$|x|-x=-x-x=-2x=|y|-y=-y-y=-2y$$ $$y(x)=x\qquad\text{ any }x<0.$$

Case $x>0$ and $y<0$ : $$|x|-x=0=-y-y=-2y \qquad\text{no solution since }y<0\neq 0.$$

Case $x<0$ and $y>0$ : $$|x|-x=-x-x=-2x=y-y=0\qquad\text{no solution since }x<0\neq 0.$$

Case $x=0$ and $y>0$ : $$|x|-x=0=|y|-y=0 \qquad\text{Eq.}(1)\text{ is true any }y>0.$$

Case $x=0$ and $y<0$ : $$|x|-x=0=|y|-y=-2y \qquad\text{no solution since }y<0\neq 0.$$

Case $x>0$ and $y=0$ : $$|x|-x=0=|y|-y=0 \qquad\text{Eq.}(1)\text{ is true any }x>0.$$

Case $x<0$ and $y=0$ : $$|x|-x=-2x=|y|-y=0 \qquad\text{no solution since }x<0\neq 0.$$

Case $x=0$ and $y=0$ : Eq.$(1)$ is true.

On figure below, the red points satisfy Eq.$(1)$.

enter image description here

JJacquelin
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The program Graph found here is somewhat out of date, but it does plot your question correctly.

enter image description here

Rory Daulton
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