Does $x>0$ imply that $x\in\mathbb{R}$?

Question

Let’s assume I input

Assuming[x > 0, expression]

Is it assumed by Mathematica that $x$ is a real number? Or that the real part of $x$ is positive? Something else?

A simple Mathematica illustration would be welcome.

Answer 1

The most direct way to test this is probably the following:

$Assumptions = x > 0;
Element[x, Reals] // Simplify
(* Out[1]= True *)

$Assumptions = True;
Element[x, Reals] // Simplify
(* Out[4]= x ∈ Reals *)

So $x>0$ seems to imply that $x$ is real.

Answer 2

It is assumed that $x$ is a real number. Everything else would mathematically not make sense because on complex numbers there does not exist an ordering relation.

An example would be to take the expression $\sqrt{x^2}$ and to imagine that this is not equal $x$ for $x=-\mathbb{i}$. Therefore the expression is in a general form not simplified

In[37]:= Sqrt[x^2]

(* Out[37]= Sqrt[x^2] *)

If you now say that $x \geq 0$ should hold you get

In[33]:= Assuming[x >= 0, Refine[Sqrt[x^2]]]

(* Out[33]= x *)

Note that if $x \geq 0$ would mean the real part is non-negative, the value $x=-\mathbb{i}$ would still be possible. Therefore, it can be assumed, that using an ordering does automatically force the variable to be real.

Answer 3

In general the situation is much more subtle than the other answers suggest. For example this issue is present in version 8 while not in version 7 :

Integrate[ Exp[-a^2] Sin[2 t] (a^2 + b^2 + b*Cos[t] + a*Sin[t]), {t, 0, 2 Pi}]
$Assumptions = {x > 0};
Integrate[ Exp[-a^2] Sin[2 t] (a^2 + b^2 + b*Cos[t] + a*Sin[t]), {t, 0, 2 Pi}]

0
8/3 Sqrt[a^2 + b^2] E^-a^2

The identical integrand (not depending on x) yields different results if we assume x > 0. This bug may appear in different cases when we deal with complex variables.

One may encounter certain inconsequences working with these examples :

Assuming[ y > 0 && x > 0, 
         Integrate[1/Sqrt[z^2 + y^2], {z, -x, x}]]                 (* I *)

Assuming[ y > 0 && Element[x, Complexes] && Re[x] > 0, 
          Integrate[1/Sqrt[z^2 + y^2], {z, -x, x}]]                (* II *)

Assuming[ y > 0 && Element[x, Complexes], 
          Integrate[1/Sqrt[z^2 + y^2], {z, -x, x}]]                (* III *)

Assuming[ Element[y, Reals] && Element[x, Complexes], 
          Integrate[1/Sqrt[z^2 + y^2], {z, -x, x}]]                 (* IV *)

2 Log[(x + Sqrt[x^2 + y^2])/y]
2 Log[(x + Sqrt[x^2 + y^2])/y]
ConditionalExpression[2 Log[(x + Sqrt[x^2 + y^2])/y], x > 0]
ConditionalExpression[2 ArcSinh[x/Abs[y]], y != 0 && x >= 0]

For example assuming in (III) weaker conditions we get ConditionalExpression[..., x > 0] while in (II) under a more restrictive condition we get a more general expression.

FunctionExpand[2 ArcSinh[x/Abs[y]], x > 0 && y > 0] // TrigToExp

2 Log[Sqrt[1 + x^2/y^2] + x/y]

Answer 4

Mathematica will always assume that all the arguments of an inequality relation are real but there are situations the presence of an inequality will lead to a stronger assumption. This is the case with Reduce. If you evaluate:

Reduce[x^2 + y^2 <= 1, {x, y}]

Mathematica will assume that both x and y are real. If you do not want this assumption you need to tell Reduce explicitly:

Reduce[x^2 + y^2 <= 1, {x, y},Complexes]

Answer 5

The most direct hint that x>0 implies Element[x,Reals] is the following:

Reduce[Element[x, Reals] && x > 0]
(*
==> x > 0
*)