Provide a counter-example to demonstrate that the following statement is false.
$\forall x \in \mathbb{R} \forall y \in \mathbb{R} (\sqrt{x^2y^2} \geq 16) \rightarrow (xy \geq 4)$
My attempt:
Statement $P$: $\forall x \in \mathbb{R} \forall y \in \mathbb{R} (\sqrt{x^2y^2} \geq 16) \rightarrow (xy \geq 4)$
For $x = 16$ and $y = -1$: $\sqrt{x^2y^2} = \sqrt{16^2 \times (-1)^2} = \sqrt{16^2 \times 1} = 16 \geq 16$ $xy = 16 \times (-1) = -16 < 4$
Therefore, $P$ is false for $x=16$ and $y=-1$: $P$ is not true for all values of $x$ and $y$ in the set of real numbers.
This counter-example shows that the implication in statement $P$ does not hold for all possible values of $x$ and $y$.
Therefore, the statement $P$ is false.
Am I correct ?
Correction to implement the suggestion from the Answer below:
My attempt:
Statement $P$: $\forall x \in \mathbb{R} \forall y \in \mathbb{R} (\sqrt{x^2y^2} \geq 16) \rightarrow (xy \geq 4)$
This counter-example shows that the implication in statement $P$ does not hold for all possible values of $x$ and $y$.
For $x = 16$ and $y = -1$: $\sqrt{x^2y^2} = \sqrt{16^2 \times (-1)^2} = \sqrt{16^2 \times 1} = 16 \geq 16$ $xy = 16 \times (-1) = -16 < 4$
It is false that every real pair of $x$ and $y$ satisfies $(\sqrt{x^2y^2} \geq 16) \rightarrow (xy \geq 4)$
- i.e., $\Big(∀x{\in}\mathbb R\;∀y{\in}\mathbb R \;S(x,y)\Big)$ is false.
Therefore, the statement $P$ is false.
You’ve given an appropriate counterexample. However, your above sentence makes a logical leap by wrongly claiming that
Correction: