Take the 2-minute tour ×
Mathematica Stack Exchange is a question and answer site for users of Mathematica. It's 100% free, no registration required.

I have two intervals $A = (-3, m^2- 2]$ and $B=[m, +\infty)$. How to find the values of $m$ so that $A \cap B = \emptyset$ and $A \cup B = (-3,+\infty)$? I tried

{a = x > -3 && x < m^2 - 2, b = x >= m} {Reduce[a || b]}

and

{a = x > -3 && x < m^2 - 2, b = x >= m}
{Reduce[a && b]}

We have $A \cap B = \emptyset$ when and only when $$\begin{cases} m^2 - 2 > -3&\\ m^2 - 2 <m. \end{cases}$$ and $A \cup B = (-3,+\infty)$ when and only when $$\begin{cases} m^2 - 2 > -3&\\ m^2 - 2 \geqslant m. \end{cases}$$ Therefore we need to solve two systems of inequalities.

share|improve this question
1  
Two systems of inequalities to be solved at the same time are just one system consisting of all the inequalities from the two systems. –  celtschk Oct 20 '12 at 7:32
    
Because each of $A$ and $B$ is relatively closed in $A\cup B$ and $A\cup B$ is connected, the only possible solutions would be when either $A$ or $B$ is empty. (It would be interesting to find ways to get Mathematica to perform such forms of topological reasoning.) –  whuber Oct 20 '12 at 17:21
    
Why do you wrap your commands in { }? –  Sjoerd C. de Vries Oct 20 '12 at 19:25

1 Answer 1

Just reduce the four inequalities you gave:

Reduce[m^2 - 2 > -3 && m^2 - 2 < m && m^2 - 2 > -3 && m^2 - 2 >= m, m]

(* False *)

Your second and fourth inequalities contradict each other, so there is no solution.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.