Calculating the width of the interval defined by an inequality












1














I am looking for a Mathematica function that takes an inequality as the input and gives back the width defined by upper bound - lower bound:



Example:



Fn[1 <= x <= 2.5]



1.5




If the inequality is evaluated to False (e.g., 2 <= x <= 1), then I need the function to return 0.



I truly appreciate your help.










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    1














    I am looking for a Mathematica function that takes an inequality as the input and gives back the width defined by upper bound - lower bound:



    Example:



    Fn[1 <= x <= 2.5]



    1.5




    If the inequality is evaluated to False (e.g., 2 <= x <= 1), then I need the function to return 0.



    I truly appreciate your help.










    share|improve this question









    New contributor




    Monire Jalili is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.























      1












      1








      1







      I am looking for a Mathematica function that takes an inequality as the input and gives back the width defined by upper bound - lower bound:



      Example:



      Fn[1 <= x <= 2.5]



      1.5




      If the inequality is evaluated to False (e.g., 2 <= x <= 1), then I need the function to return 0.



      I truly appreciate your help.










      share|improve this question









      New contributor




      Monire Jalili is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











      I am looking for a Mathematica function that takes an inequality as the input and gives back the width defined by upper bound - lower bound:



      Example:



      Fn[1 <= x <= 2.5]



      1.5




      If the inequality is evaluated to False (e.g., 2 <= x <= 1), then I need the function to return 0.



      I truly appreciate your help.







      function-construction inequalities






      share|improve this question









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      Monire Jalili is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











      share|improve this question









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      Monire Jalili is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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      share|improve this question




      share|improve this question








      edited 1 hour ago









      m_goldberg

      84.3k872195




      84.3k872195






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      asked 4 hours ago









      Monire Jalili

      61




      61




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      New contributor





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          2 Answers
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          f[ineq_, var_] := RegionMeasure[ImplicitRegion[ineq, var]]

          f[1 <= x <= 2.5, x]



          1.5




          This works also for some systems of inequality in several variables:



          f[{1 <= x <= 2.5, 0 <= y <= x}, {x, y}]



          2.625







          share|improve this answer





























            2














            fn[expr_] := Module[{},
            If[! expr, Return [0]];
            Return[Abs[expr[[3]] - expr[[1]]]]
            ]

            fn[2 <= x <= 1]
            (*0*)

            fn[1 <= x <= 2.5]
            (*1.5*)

            fn[2.5 > x > 1]
            (*1.5*)


            Don't know if this works in all cases, but works in the simple cases you provide.






            share|improve this answer





















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              2 Answers
              2






              active

              oldest

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              2 Answers
              2






              active

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              active

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              active

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              3














              f[ineq_, var_] := RegionMeasure[ImplicitRegion[ineq, var]]

              f[1 <= x <= 2.5, x]



              1.5




              This works also for some systems of inequality in several variables:



              f[{1 <= x <= 2.5, 0 <= y <= x}, {x, y}]



              2.625







              share|improve this answer


























                3














                f[ineq_, var_] := RegionMeasure[ImplicitRegion[ineq, var]]

                f[1 <= x <= 2.5, x]



                1.5




                This works also for some systems of inequality in several variables:



                f[{1 <= x <= 2.5, 0 <= y <= x}, {x, y}]



                2.625







                share|improve this answer
























                  3












                  3








                  3






                  f[ineq_, var_] := RegionMeasure[ImplicitRegion[ineq, var]]

                  f[1 <= x <= 2.5, x]



                  1.5




                  This works also for some systems of inequality in several variables:



                  f[{1 <= x <= 2.5, 0 <= y <= x}, {x, y}]



                  2.625







                  share|improve this answer












                  f[ineq_, var_] := RegionMeasure[ImplicitRegion[ineq, var]]

                  f[1 <= x <= 2.5, x]



                  1.5




                  This works also for some systems of inequality in several variables:



                  f[{1 <= x <= 2.5, 0 <= y <= x}, {x, y}]



                  2.625








                  share|improve this answer












                  share|improve this answer



                  share|improve this answer










                  answered 3 hours ago









                  Henrik Schumacher

                  49.6k468140




                  49.6k468140























                      2














                      fn[expr_] := Module[{},
                      If[! expr, Return [0]];
                      Return[Abs[expr[[3]] - expr[[1]]]]
                      ]

                      fn[2 <= x <= 1]
                      (*0*)

                      fn[1 <= x <= 2.5]
                      (*1.5*)

                      fn[2.5 > x > 1]
                      (*1.5*)


                      Don't know if this works in all cases, but works in the simple cases you provide.






                      share|improve this answer


























                        2














                        fn[expr_] := Module[{},
                        If[! expr, Return [0]];
                        Return[Abs[expr[[3]] - expr[[1]]]]
                        ]

                        fn[2 <= x <= 1]
                        (*0*)

                        fn[1 <= x <= 2.5]
                        (*1.5*)

                        fn[2.5 > x > 1]
                        (*1.5*)


                        Don't know if this works in all cases, but works in the simple cases you provide.






                        share|improve this answer
























                          2












                          2








                          2






                          fn[expr_] := Module[{},
                          If[! expr, Return [0]];
                          Return[Abs[expr[[3]] - expr[[1]]]]
                          ]

                          fn[2 <= x <= 1]
                          (*0*)

                          fn[1 <= x <= 2.5]
                          (*1.5*)

                          fn[2.5 > x > 1]
                          (*1.5*)


                          Don't know if this works in all cases, but works in the simple cases you provide.






                          share|improve this answer












                          fn[expr_] := Module[{},
                          If[! expr, Return [0]];
                          Return[Abs[expr[[3]] - expr[[1]]]]
                          ]

                          fn[2 <= x <= 1]
                          (*0*)

                          fn[1 <= x <= 2.5]
                          (*1.5*)

                          fn[2.5 > x > 1]
                          (*1.5*)


                          Don't know if this works in all cases, but works in the simple cases you provide.







                          share|improve this answer












                          share|improve this answer



                          share|improve this answer










                          answered 3 hours ago









                          Bill Watts

                          2,8231516




                          2,8231516






















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