I would like to ask the following. If we have the proposition $$\forall x\in\mathbb{R}^{+}(x^2>0)$$ and we wish to use as a domain the $$\mathbb{R}$$ instead. Is it correct that it will translate to: $$\forall x\in\mathbb{R}(x>0\land x^2>0)$$?
Similarly am I correct when from $$\exists x\in\mathbb{Z}^{+}(x^2=\sqrt2)$$ go to $$\exists x\in\mathbb{Z}(x>0\rightarrow x^2=\sqrt2)$$ ?
On
Two translatings are incorrect. First, $\forall x \in \mathbb{R}^+ (x^2 >0)$ is true, but $\forall x\in \mathbb{R} (x > 0 \wedge x^2 >0)$ is false. It might be corrected as $\forall x\in \mathbb{R}(x>0 \to x^2 > 0)$. Second, $0,-1,-2,-3,\cdots$ show that $\exists x\in \mathbb{Z}(x>0\to x^2=\sqrt{2})$ is true, whereas there is no $x$ to show $\exists x\in\mathbb{Z}^+(x^2=\sqrt2)$.
For second example, $\exists x\in\mathbb{Z}^{+}(x^2=\sqrt2)$ means "There is an element $x$ such that $x\in \mathbb{Z}^+$ and $x^2=\sqrt{2}$". Then we can conclude cardboard_box's answer is correct.
No, you have your implications and conjunctions swapped.
$$ \forall x \in \mathbb{R} (x \gt 0 \land x^2 > 0) $$
Means "All real numbers are positive and have positive squares".
You're looking for
$$ \forall x \in \mathbb{R} (x \gt 0 \implies x^2 > 0) $$
For the 2nd one, it's also the wrong way around.
$$ \exists x \in \mathbb{Z}(x \gt 0 \implies x^2 = \sqrt{2}) $$
Means "There is an integer which, if it is positive then its square is $ \sqrt{2} $".
i.e. "There is an integer which is either non-positive or its square is $\sqrt{2}$".
You're looking for
$$ \exists x \in \mathbb{Z}(x \gt 0 \land x^2 = \sqrt{2}) $$