I need to represent the following statements using quantifiers in this task:
(a) For any real number $x$, if $x$ is rational, then also $\sqrt{x}$.
(b) For some natural numbers $n \geq 3$ and integers $x, y$ and $z$, $x^{n} \cdot y^{n} = z^{n}$ holds.
Are these correct:
(a) $\quad \forall x \in \mathbb{R} : x\in\mathbb{Q}\Rightarrow\sqrt{x} \in \mathbb{Q}$
(b) $\quad \exists n \in \mathbb{N} \quad \land \quad \exists x,y,z \in \mathbb{Z} \quad , n \geq 3 :\quad x^{n} \cdot y^{n} = z^{n}$
And in the second step we have to negate the above statements. Are these correct:
(a) $\quad \lnot(\forall x \in \mathbb{R}: x\in\mathbb{Q}\Rightarrow\sqrt{x} \in \mathbb{Q}) \Leftrightarrow \exists x \in \mathbb{R} : x\in\mathbb{Q} \quad \land \quad \sqrt{x} \notin \mathbb{Q}$
(b) $ \quad \lnot(\exists n \in \mathbb{N} \quad \land \quad \exists x,y,z \in \mathbb{Z} \quad , n \geq 3 :\quad x^{n} \cdot y^{n} = z^{n}) \Leftrightarrow \forall n \in \mathbb{N} \quad \land \quad \forall x,y,z \in \mathbb{Z}\quad , n \geq 3 : \quad x^{n} \cdot y^{n} \neq z^{n}$
Some touchups and minor correction:
$\forall x {\in} \mathbb{R} \; \big( x{\in}\mathbb{Q}\Rightarrow\sqrt{x} {\in}\mathbb{Q}\big)$
‘∧’ operates on sentences, which $\:∃n{∈}N\:$ is not.
$\exists n {\in} \mathbb{N}\;\, \exists x,y,z {\in} \mathbb{Z} \;\big(n \geq 3 \;\land\; x^{n} {\cdot} y^{n} = z^{n}\big)$
Neither look correct; the correct negations are
a) $\exists x {\in} \mathbb{R} \; \big( x{\in}\mathbb{Q}\;\land\; \sqrt{x} {\notin}\mathbb{Q}\big)$
b) $\forall n {\in} \mathbb{N}\;\, \forall x,y,z {\in} \mathbb{Z} \;\big(n < 3 \;\lor\; x^{n} {\cdot} y^{n} \ne z^{n}\big)$