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When I type the following implicit equation into Mathematica I only get 'one quarter' of what the resulting plot should look like, i.e., I only get the upper-right quadrant.

ContourPlot[x^(2/3) + y^(2/3) == 1, {x, -1, 1}, {y, -1, 1}]

If you type $x^{\frac{2}{3}} + y^{\frac{2}{3}} = 1$ into this graphing calculator you will see the correct output which has a curve in each of the four quadrants.

So why is Mathematica giving an incorrect result, and is it possible to make it plot this implicit function correctly?


marked as duplicate by rhermans, Bob Hanlon, Daniel Lichtblau, Henrik Schumacher, m_goldberg plotting Aug 8 '18 at 23:35

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  • 5
    $\begingroup$ Try ContourPlot[Abs[x]^(2/3) + Abs[y]^(2/3) == 1, {x, -1, 1}, {y, -1, 1}]. Mathematica is giving the correct result; for instance, how to plot (-0.5)^(2/3) ? $\endgroup$ – b.gates.you.know.what Aug 8 '18 at 5:30
  • 3
    $\begingroup$ ContourPlot[CubeRoot[x^2]+CubeRoot[y^2]==1,{x,-1,1},{y,-1,1}] or ContourPlot[(x^2)^(1/3)+(y^2)^(1/3)==1,{x,-1,1},{y,-1,1}] $\endgroup$ – mathe Aug 8 '18 at 6:08
  • 2
    $\begingroup$ The calculator is apparently interpreting x^(a/b) as (x^a)^(1/b) whereas in Mathematica this is treated as (x^(1/b))^a. The two are of course not the same. The Mathematica treatment is a consequence of defining x^y as (in more customary notation) exp(y*log(x)) $\endgroup$ – Daniel Lichtblau Aug 8 '18 at 14:28
  • $\begingroup$ ContourPlot[(x^2)^(1/3) + (y^2)^(1/3) == 1, {x, -1, 1}, {y, -1, 1}] $\endgroup$ – Bob Hanlon Aug 8 '18 at 23:39

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