# How to "read" the solution of an inequality?

The solution of fun[x,y]>0 leads to the following output

 fun[x_, y_] = (1/16)*(1 - 6*x*y + (-1 + x)*x*y)*(1 + x*(-1 + 6*y) +
x*(-1 + x + y - x*y))*(-1 + 2*x -
x^2 + (x + (-1 + x)*x*(-1 + y) - 6*x*y)*(6*x*y + (x - x^2)*y));
Reduce[fun[x, y] > 0 && 0 < x < 1 && 0 < y < 1, y]


$$\color{blue}{\left(\frac{1}{2} \left(7-3 \sqrt{5}\right)

The way I understand this is:

1. blue and red are two solutions
2. when $$x$$ lies in range $$\frac{1}{2} \left(7-3 \sqrt{5}\right), $$y$$ can take values in the range $$\color{green}{-\frac{1}{x^2-7 x}
3. when $$x$$ lies in range $$2-\sqrt{2}, $$y$$ can take values in the range $$\frac{x-2}{2 (x-7)}-\frac{1}{2} \sqrt{\frac{x^4-4 x^3+8 x-4}{(x-7)^2 x^2}} $$\textbf{or}$$ $$\color{green}{-\frac{1}{x^2-7 x}

Is Mathematica telling me that $$y$$ exists in range $$-\frac{1}{x^2-7 x} for two different ranges of $$x$$?

Doesn't 2 and 3 imply $$-\frac{1}{x^2-7 x} for $$\frac{1}{2} \left(7-3 \sqrt{5}\right)?

• Your interpretation seems correct to me. You could plot the solution to visualize it. Sep 30 at 12:42

In Make Reduce produce nicer output @chuy presented a comand line to make a nice looking output ot results of Reduce  . I here give the form that also worked in version 8.0.

fun[x_, y_] = (1/16)*(1 - 6*x*y + (-1 + x)*x*y)*(1 + x*(-1 + 6*y) +
x*(-1 + x + y - x*y))*(-1 + 2*x -
x^2 + (x + (-1 + x)*x*(-1 + y) - 6*x*y)*(6*x*y + (x - x^2)*y));

red = Reduce[fun[x, y] > 0 && 0 < x < 1 && 0 < y < 1, y];

red //. Or ->
Composition[(Column[#, Right, Background -> {{White, LightGray}},
Frame -> All] &), List]]


TraditionalForm[
N[red] //.
Or -> Composition[(Column[#, Right,
Background -> {{White, LightGray}}, Frame -> All] &), List]]


FullSimplify[(1/2 (7 - 3 Sqrt[5]) < x <=
2 - Sqrt[2] && -(1/(-7 x + x^2)) < y < 1) || (2 - Sqrt[2] < x <
1 && ((-2 + x)/(2 (-7 + x)) -
1/2 Sqrt[(-4 + 8 x - 4 x^3 + x^4)/((-7 + x)^2 x^2)] <
y < (-2 + x)/(2 (-7 + x)) +
1/2 Sqrt[(-4 + 8 x - 4 x^3 +
x^4)/((-7 + x)^2 x^2)] || -(1/(-7 x + x^2)) < y < 1)),
Assumptions -> 0 < x < 1 && 0 < y < 1]


1 + (-7 + x) x y < 0 || (-2 + x + Sqrt[-4 + 8 x - 4 x^3 + x^4]/x)/( 2 (-7 + x)) < y < (-2 + x - Sqrt[-4 + 8 x - 4 x^3 + x^4]/x)/( 2 (-7 + x))

RegionPlot[1 + (-7 + x) x y < 0 || (-2 + x + Sqrt[-4 + 8 x - 4 x^3 + x^4]/x)/

RegionPlot[1 + (-7 + x) x y < 0, {x, 0, 1}, {y, 0, 1},PlotPoints -> 50]

RegionPlot[(-2 + x + Sqrt[-4 + 8 x - 4 x^3 + x^4]/x)/(2 (-7 + x)) < y < (-2 + x - Sqrt[-4 + 8 x - 4 x^3 + x^4]/x)/(2 (-7 + x)), {x, 0, 1}, {y, 0, 1}, PlotPoints -> 50]

• It should be noticed that RegionPlot[1 + (-7 + x) x y < 0 || (-2 + x + Sqrt[-4 + 8 x - 4 x^3 + x^4]/x)/ (2 (-7 + x)) < y < (-2 + x - Sqrt[-4 + 8 x - 4 x^3 + x^4]/x)/(2 (-7 + x)), {x, 0, 1}, {y, 0, 1}, PlotPoints -> 400] shows this is the union of the two nonoverlapping domains. Sep 30 at 15:05