What is undefined in Mathematical Logic? A question that arises from negating a nested predicate. Negate the following expression and indicate whether the negated statement is true.
$$\forall x \in \mathbb{R}, \exists n \in \mathbb{Z}, x^n > 0 $$


*Relevant equations


De Morgan's Law for negating quantifiers.


*The attempt at a solution


$$
\neg(\forall x \in \mathbb{R}, \exists n \in \mathbb{Z}, x^n > 0)
\\\neg\forall x \in \mathbb{R}, \exists n \in \mathbb{Z}, x^n > 0
\\\exists x \in \mathbb{R}, \neg\exists n \in \mathbb{Z}, x^n > 0
\\\exists x \in \mathbb{R}, \forall n \in \mathbb{Z}, \neg(x^n > 0)
\\\exists x \in \mathbb{R}, \forall n \in \mathbb{Z}, x^n \leq 0
$$
The above negated expression states that for some value of x in the domain of Real numbers, and for all values of n in the domain of Integers, x to the power of n is less than or equal to zero.
Assuming that I did the above negation correctly here's what is really really driving me crazy.
First of all I know that the original equation $\forall x \in \mathbb{R}, \exists n \in \mathbb{Z}, x^n > 0$ is false. This is shown for
when x = 0 and n is a positive integer, then x^n is 0 and not greater than 0,
when x= 0 and n is also 0, x^n is undefined,
when x = 0, and n a negative integer, x^n is also undefined.
So $\forall x \in \mathbb{R}, \exists n \in \mathbb{Z}, x^n > 0$ is false. So I would definitely expect the negation $\exists x \in \mathbb{R}, \forall n \in \mathbb{Z}, x^n \leq 0$ to be true right?
Here's what I find:
when x is some value greater than 0
if n > 0, x^n > 0
if n = 0, x^n = 1
if n < 0, x^n > 0
So no luck finding some x to satsify x^n <= 0 here.
When x = 0
if n > 0, x^n = 0
if n = 0, x^n is undefined
if n < 0, x^n is undefined
So it works when n is a positive integer but is undefined for all else, which means that the negated statement has still not been found to be completely true. But since undefined is neither true nor false can I assume that the instance x = 0 does prove the negated statement to be true? What does undefined mean in mathematics?
When x is some value less than 0
if n > 0, if n is odd then x^n < 0, if n is even then x^n > 0
if n = 0, x^n = 1
if n < 0, if n is odd then x^n < 0, if n is even then x^n > 0
So if n is a non zero odd integer than x^n <= 0 but is false for all other integers, which means that the negated statement still has not to be found true....
So we conclude that $ \exists x \in \mathbb{R}, \forall n \in \mathbb{Z}, x^n \leq 0$ is false...
But I know the the original statement was false so the negation has to be true... What did I do wrong or how should I think differently?
 A: $0^0$ as well as $\frac {0} {0}$ are undefined.
Thus, in order to correctly "formalize" the statement, we have to take into account this condition :
$∀x \in \mathbb R,∃n \in \mathbb Z(defined(x^n) \land x^n > 0)$ --- (#).
Thus, the negation of (#) is :

$∃x \in \mathbb R,∀n \in \mathbb Z(\lnot defined(x^n) \lor x^n \le 0)$.

Now we have to express the "defining condition" which is :
$\lnot (x=0 \land n \le 0)$.
With it, we can rewrite the above formula as :


$∃x \in \mathbb R,∀n \in \mathbb Z,((x=0 \land n \le 0) \lor x^n \le 0)$ --- (§).


Is this formula true ?
For $x \in \mathbb R$ such that $x \ne 0$ it is not true that $x^n \le 0$ for all $n$.
Thus, we are left with $x=0$.
If we substitute $0$ in place of $x$ in the formula above, we have :
$∀n \in \mathbb Z,((0=0 \land n \le 0) \lor 0^n \le 0)$ --- (*)
We have two cases to consider :
i) For $n=0$, we have that the left disjuct is : $(0=0 \land 0 \le 0)$, which is obviously true.
ii) For $n \ne 0$, we have that the right disjunct is $0^n=0 \le 0$, which is true (because $n \ne 0$).
In both case, the disjunction $(0=0 \land n \le 0) \lor 0^n \le 0$ is true.
Thus, we can conlude that it is true for all $n$, i.e. the formula (*) is true.
Having found a value of $x$ such that : $∀n \in \mathbb Z,((x=0 \land n \le 0) \lor x^n \le 0)$ is true, we can conclude that the formula (§) is true.
Conclusion


It is correct to say that the formula (#) is false.



Note
This issue arises in first-order logic, due to the fact that the standard syntax of the language does not allow undefined predicates or functions.
Thus, with the function "exponentiation" defined as : 

$y=e(x,n)$ iff $y=x^n$, 

whe have to be careful to avoid the case in which it is undefined, i.e. when $x=0$ and $n \le 0$.
