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is there some reason why the graph of $\sqrt{(x-y)} = 0$ wouldn't be equal to the graph of $x = y$?

The reason why i'm asking this is that in GeoGebra, when i plot $\sqrt{(x-y)} = 0$, i don't get any points.

Thanks!

Ulivai
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    Being unfamiliar with Geogebra, my blind guess is rounding error. – user2661923 Nov 14 '20 at 23:52
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    Funny. Seems to stump Wolfram Alpha too. On the other hand, WA can handle $\sqrt {x-y}=1$ . – lulu Nov 14 '20 at 23:53
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    Desmos reacts the same. – PinkyWay Nov 14 '20 at 23:55
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    From this post "Most numerical implicit function plotters actually depend on recognizing changes of sign. Basically, if you want to plot f(x,y)=0, you start by sampling a bunch of points f(xi,yj), and you know that if this has opposite signs at two neighbouring points, there should be a piece of the curve somewhere between them." (@RobertIsrael) This seems to be the same case here, we can never have difference of sign because $\sqrt{x}\geq 0$ for all valid $x$. – JMoravitz Nov 15 '20 at 00:02
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    It also explains why it can handle $\sqrt{x-y}=1$ as there are values of $x,y$ such that it is less than $1$ and other values such that it is more than $1$. – JMoravitz Nov 15 '20 at 00:03

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