# TrueQ - Test of inequality [duplicate]

I want to look if a term is <0,>0 or ==0. What is wrong when I write

a < 2 b;
b > 0;
a > 0;
TrueQ[Re[(3 a + 6 b +
Sqrt[9 (-36 + 7 a^2) + 4 a (-27 + 8 a^2)*b +
4 (-45 + 16 a^2) b^2 + 32 ab^3])/4 (a + b)] < 0]


It gives me FALSE every time regardless of whether I say <0,>0 or ==0. But I took the real part, hence it one case has to be TRUE!

For a=2b

Simplify[Re[(-3 a - 6 b +
Sqrt[9 (-36 + 7 a^2) + 4 a (-27 + 8 a^2)*b +
4 (-45 + 16 a^2) b^2 + 32 a b^3])/4 (a + b)] < 0,
Assumptions -> {a > 0, b > 0, a = 2 b}]


I get

Re[Sqrt[-9 - 4 b^2 + 16 b^4]] < True


What does this means?

EDIT*EDIT*

For

Simplify[Re[(-3 a - 6 b +
Sqrt[9 (-36 + 7 a^2) + 4 a (-27 + 8 a^2)*b +
4 (-45 + 16 a^2) b^2 + 32 a b^3])/4 (a + b)] < 0,
Assumptions -> {a > 0, b > 0, a == 2 b}]


I get

Re[Sqrt[-9 - 4 b^2 + 16 b^4]] < 2 b

Does this means that this is only true if Re[Sqrt[-9 - 4 b^2 + 16 b^4]] < 2 b

?

• TrueQ returns False for anything that is not explicitly True. Since you have a symbolic expression the truth can not be determined explicitly. Sep 26, 2013 at 15:09
• What do you think the first three inequalities do here? They do not put any constraints upon a and b in what follows! Sep 26, 2013 at 15:19
• But when a and b >0 and with a=2b you can decide if its larger than 0 or not? Is there a possibility in mathematica to do that? Sep 26, 2013 at 15:20
• the point is your first three lines do not do what you think. In fact they do nothing.. Think about it. Sep 26, 2013 at 16:34
• You seem to think that the line a<2b is an assertion, in which you declare a certain truth, but it is not. It is a Boolean test. You may need $Assumptions and a function using them like Simplify. Sep 26, 2013 at 17:03 ## 3 Answers When you do a<0  you are not affecting a in any way. i.e. you are not saying to Mathematica 'a is less than zero'. I think that what you intend to do is Simplify[Re[(3 a + 6 b + Sqrt[9 (-36 + 7 a^2) + 4 a (-27 + 8 a^2)*b + 4 (-45 + 16 a^2) b^2 + 32 a b^3])/4 (a + b)] < 0, Assumptions -> {a > 0, b > 0, a < 2 b}]  This will simplify your expression and, in this case, it returns False. So, your expression is not less than 0 given the constraints. • Ok that works! Thank you! Sep 26, 2013 at 15:29 May be try Assuming as follows: Assuming[{b > 0, a > 0 && a < 2 b}, Refine[Re[(3 a + 6 b + Sqrt[9 (-36 + 7 a^2) + 4 a (-27 + 8 a^2)*b + 4 (-45 + 16 a^2) b^2 + 32 a b^3])/4 (a + b)] > 0]]  Which gives: True Whereas Assuming[{b > 0, a > 0 && a < 2 b}, Refine[Re[(3 a + 6 b + Sqrt[9 (-36 + 7 a^2) + 4 a (-27 + 8 a^2)*b + 4 (-45 + 16 a^2) b^2 + 32 a b^3])/4 (a + b)] < 0]]  gives False Edit And for the case a == 2b  Assuming[{b > 0, a > 0 && a == 2 b}, Refine[Re[(3 a + 6 b + Sqrt[9 (-36 + 7 a^2) + 4 a (-27 + 8 a^2)*b + 4 (-45 + 16 a^2) b^2 + 32 a b^3])/4 (a + b)] < 0]]  We get: False • I edit my first post. For the case a=2b I get an expression I cant interpret! Sep 26, 2013 at 16:11 • @user2098925. Did you use double equal sign ==? So a == 2 b instead of a = 2 b. See my edit. Sep 26, 2013 at 16:53 You can define global assumptions $Assumptions = {b > 0, a > 0, a < 2 b};

Re[(3 a + 6 b +
Sqrt[9 (-36 + 7 a^2) + 4 a (-27 + 8 a^2)*b +
4 (-45 + 16 a^2) b^2 + 32 a b^3])/4 (a + b)] > 0 // Simplify


True

• True + Re[Sqrt[-9 - 4 b^2 + 16 b^4]] > 0....This I get for a=2b. what does this means? Sep 26, 2013 at 16:46
• I get this for a=2b. what does this means? Sep 26, 2013 at 16:47
• I edited my first post, the same like the previous comment Sep 26, 2013 at 17:45
• @user2098925 It means that result depends on value of b, e.g. b==0 and b==1. Sep 26, 2013 at 17:48