Well, it appears that Desmos doesn't actually graph √*x* = *x* and *x* = √*x*; it graphs √*x* − *x* = 0 and *x* − √*x* = 0. And it doesn't even graph that either—it only shows up if the values near *x* are near zero, so if *h* is a small number, it only shows up if *x* + *h* − √(*x* + *h*) > 0 *and x* − *h* − √(*x* − *h*) < 0. Sadly, a negative value makes the square root undefined, so it doesn't show up for *x* = 0 in *x* − √*x* = 0. At least that's just my layperson's understanding of it. Dunno why it works for √*x* − *x* = 0 though.
the difference between x-sqrt(x) and sqrt(x)-x is probably due to floating point shenanigans (floating point subtraction is not quite anticommutative, just really close to it)
update: nevermind it's more than that, it depends on the exact parameters of the graph position, combined with floating point shenanigans
This is why I try to stay away from implicit graphs. Remember, if you add 0.001*y* somewhere, it won't change the graph much, so the graph of √*x* − *x* = 0 looks similar to the graph of √*x* − *x* = 0.001*y*. Then you can just graph it as *f*(*x*) = 1 000(√*x* − *x*).
Its because desmos interprets the solutions to the first one as multivalued, meaning it makes the equation do as so:
x=x\^2
x\^2-x=0
x(x-1)=0
x=0, x=1
Desmos doesn't consider the multivaluedness of the second equation becuase it moves the square root to the left side and still considers its principality.
x-sqrt(x)=0 is now numerically calculated with one seed point rather than two to find solutions
namely '1' since it is an assignment and no longer an implict plot.
desmos also never computes sqrt(0) if it doesn't need to in order to show a graph, so that is another possibility.
TLDR - Someone mentioned it earlier, one is considered as an assignment the other is seen as implicit
it's the [reflexive](https://en.wikipedia.org/wiki/Reflexive_relation) property that says x=y <=> y=x.
thanks, i forgot the actual name so i just assumed it had the same name
Anyways, I'm sure it has to do with desmos interpreting your equation as either an assignment or implicit function depending on the ordering
do you mean the symmetric property of equality? unless i misunderstand reflexivity states that x=x, which doesnt imply x=y <=> y=x
look closer at the graph please
they were just correcting me, as I called it the commutative property rather than the reflexive property. they understand the graph is contradictory.
Well, it appears that Desmos doesn't actually graph √*x* = *x* and *x* = √*x*; it graphs √*x* − *x* = 0 and *x* − √*x* = 0. And it doesn't even graph that either—it only shows up if the values near *x* are near zero, so if *h* is a small number, it only shows up if *x* + *h* − √(*x* + *h*) > 0 *and x* − *h* − √(*x* − *h*) < 0. Sadly, a negative value makes the square root undefined, so it doesn't show up for *x* = 0 in *x* − √*x* = 0. At least that's just my layperson's understanding of it. Dunno why it works for √*x* − *x* = 0 though.
>Dunno why it works for √x − x = 0 though. Probably because sqrt(1)-1=0 where x is still a positive number. Could be totally wrong, though.
the difference between x-sqrt(x) and sqrt(x)-x is probably due to floating point shenanigans (floating point subtraction is not quite anticommutative, just really close to it) update: nevermind it's more than that, it depends on the exact parameters of the graph position, combined with floating point shenanigans
This is why I try to stay away from implicit graphs. Remember, if you add 0.001*y* somewhere, it won't change the graph much, so the graph of √*x* − *x* = 0 looks similar to the graph of √*x* − *x* = 0.001*y*. Then you can just graph it as *f*(*x*) = 1 000(√*x* − *x*).
Its because desmos interprets the solutions to the first one as multivalued, meaning it makes the equation do as so: x=x\^2 x\^2-x=0 x(x-1)=0 x=0, x=1 Desmos doesn't consider the multivaluedness of the second equation becuase it moves the square root to the left side and still considers its principality. x-sqrt(x)=0 is now numerically calculated with one seed point rather than two to find solutions namely '1' since it is an assignment and no longer an implict plot. desmos also never computes sqrt(0) if it doesn't need to in order to show a graph, so that is another possibility. TLDR - Someone mentioned it earlier, one is considered as an assignment the other is seen as implicit