∀x,n(x^n∉ℝ) ⇔ x≤0,n∈ℝ,n∉ℚ

I asked this question already, but this is, I think, a more correct mathematical statement?

Question

∀x,n(x^n∉ℝ) ⇔ x≤0,n∈ℝ,n∉ℚ

I asked this question already, but this is, I think, a more correct mathematical statement?

anonymous · Answer

this is finite math -.-

anonymous · Answer

right?

anonymous · Answer

This is innaccurate. I should say

∀x,n(x^n∉ℝ) <- x<0,n∈ℝ,n∉ℚ

Finite math?

anonymous · Answer

I learned this notation in logic theory.

anonymous · Answer

okies i failed that class

anonymous · Answer

sorries cant help you there either

anonymous · Answer

:c it's k

anonymous · Answer

:P

anonymous · Answer

$$\forall x,n(x <0,n \in \mathbb{R}, n \notin \mathbb{Q} \rightarrow x^{n} \notin \mathbb{R} )$$
is that it? what is it you want to know?

anonymous · Answer

well, basically x^n, where x<0, and n is a transcendental, x^n MUST be in the complex plane, right? how can we prove this?

anonymous · Answer

Proof by contrapositive assume $$x \in \mathbb{R} \rightarrow x \ge 0 \cup n \notin \mathbb{R} \cup n \in \mathbb{Q}$$

sasogeek · Answer

anonymous · Answer

Oh ok. Thanks.