I want to solve a complicated equation with the sign > instead of ==. However, mathematica Solve function is not defined for >. Can somebody tell me how to do it?
Code example:
Solve[a^2-b^2 > 0,a] Expected Result:
a > b and a > -b. $\begingroup$ $\endgroup$
7 
1 Answer
$\begingroup$ $\endgroup$
It is not true that Solve can not deal with inequalities.
This problem has not one or a finite number of solutions. The solution are separated region in the {a,b}-plane.
Reduce[a^2 - b^2 > 0, a, Reals] a < -Sqrt[b^2] || a > Sqrt[b^2] Even better seems:
RegionPlot[a^2 - b^2 > 0, {a, -2, 2}, {b, -2, 2}] 
This is a correct input for the question:
Solve[a^2 - b^2 > 0, {a, b}, Reals] 
{{}} The Mathematica documentation for this result for the built-in Solve is
"Solve uses {} to represent the empty solution or no solution:"
What does that mean?
Wolfram Inc gives the example
Solve[x == x, x] This is a tautology. The intend is to state this is a correct input, but it has no finite set of solutions. Only complete Reals is the solution.
Same is true for Your input. The same is programmed in Mathematica if the input is valid but overdetermined or already the solution.
The most appropriate input is
Reduce[a^2 - b^2 > 0, {a,b}, Reals] (a < 0 && -Sqrt[a^2] < b < Sqrt[a^2]) || (a > 0 && -Sqrt[a^2] < b < Sqrt[a^2]) This is the symbolic representation of the RegionPlot visualization.
is specialized for such problems. It. contains Solve and many more algorithms. Reduce in a generalization of Solve for more complex infinite solutions and inequalities. The proper question would be, why is not Solve redundant to Reduce. Solve may rise less costs in use.
There is indeed a workaround for the question to get meaningful results with Solve.
sol = Solve[x^2 - y^2 == 0, {x, y}] Plot[Evaluate[Re[y /. sol]], {x, -2, 2}] 
Mathematicians are usually both interested in the region and the boundary of the region.
The same result comes with
sol = Solve[x^2 - y^2 == 0, {x, y}, Reals] {{y -> -Sqrt[x^2]}, {y -> Sqrt[x^2]}} Both are different for Mathematica.
Plot[Evaluate[Re[y /. sol]], {x, -2, 2}] 
sol = Solve[x^2 - y^2 > 0, {x, y}, Reals] 
An example for finishing this:
sol = Solve[x > 0 && y > 0, {x, y}, Reals] 
Solve seems not to work with Assumptions like Simplify does. It knows logical connectives.
For deeper insight than Mathematica documentation does:
What is the difference between Reduce and Solve?
Nice is
Solve[a*x == 0, x] (* Out: {{x -> 0}} *) Reduce[a*x == 0, x] (* Out: a == 0 || x == 0 *) So the implicit assumption already made if using Solve are much wider than using Reduce for two independent variables. That should answer the question and give advise how to solve the question.
Reduce[a^2 - b^2 > 0, a, Reals]$\endgroup$PowerExpandto the result will replaceSqrt[b^2]withb, so if you know that $b \geq 0$, you can make the result look even nicer. $\endgroup$PowerExpand. I would recommend usingSimplifyso that the assumptions are explicit, e.g.,Simplify[#, b >= 0] &$\endgroup$