# Reduce function produces result as False for long inequalities

    Reduce[(2 (-1 + w + G) (4 (-1 + w) w + (4 + w (4 + 3 w)) G) -  w (4 - 8 G + w (-14 + 6 G +  w (10 + 3 G))) Ep + (-2 + w) w (2 +  3 w) Ep^2)/(4 (-2 + w) (-1 + w) w (2 + w) (2 G - Ep)) < 0 &&
-(((-1 + G + w (1 + G - Ep)) (2 - 2 G + w (-2 + Ep)))/(2 (-1 + w) w (2 + w) (2 G - Ep))) < 0 &&
0 < 2 + (2 (-1 + w) (-1 + G))/( w (-2 G + Ep)) < 1 &&
0 < w < 1 &&
Ep > 0 &&
G > 0, {w, Ep, G}]


The code above results False in Mathematica. Could someone tell me what excatly False means here?

• Your code can't execute as it has syntax errors. Please re-check it. Sep 16, 2014 at 16:29
• Don't use capital letters as symbols. There are quite a few already used by the language (for example E ,I) Sep 16, 2014 at 16:30
• After a probable syntax correction (removing a parasitic ]) and replacing E for an unasigned symbol it seems there are no solutions Sep 16, 2014 at 16:37
• @belisarius I used epsilon instead of E in my actual code. I just wanted to make it simple here. You are right E might have different meaning. But thanks, it's good to know that False means it doesn't have any solutions.
– UTK
Sep 16, 2014 at 16:49

{eq1, eq2,  eq3} = {(2 (-1 + w + G) (4 (-1 + w) w + (4 + w (4 + 3 w)) G) -
w (4 - 8 G + w (-14 + 6 G + w (10 + 3 G))) Ep + (-2 + w) w (2
+ 3 w) Ep^2)/(4 (-2 + w) (-1 + w) w (2 + w) (2 G - Ep)),
-(((-1 + G + w (1 + G - Ep)) (2 - 2 G + w (-2 + Ep)))/(2 (-1 +
w) w (2 + w) (2 G - Ep))),
2 + (2 (-1 + w) (-1 + G))/(w (-2 G + Ep))};


You can check that each condition intersects the other two somewhere:

a = Reduce[eq2 < 0 && 0 < eq3 < 1 && 0 < w < 1 && Ep > 0 && G > 0, {w, Ep, G}];
c = Reduce[eq1 < 0 && 0 < eq3 < 1 && 0 < w < 1 && Ep > 0 && G > 0, {w, Ep, G}];
b = Reduce[eq2 < 0 && 0 < eq1     && 0 < w < 1 && Ep > 0 && G > 0, {w, Ep, G}];

If[! #, False, "Not false", "Not false"] & /@ {a, b, c}
(* {"Not false", "Not false", "Not false"} *)


However those regions don't intersect among them:

Reduce[a && b && c]
(* False *)