Discrete math question - nested quantifiers

Question

question regarding nested quantifiers.

$$\forall x \forall y\big((x < y)\to (x^2 < y^2)\big)$$ Determine the truth value for this question. I think this is false because if $x$ is $4$ and $y$ is $2$, the statement is false. but then it could be true if $x$ is $2$ and $y$ is $4$ ... I'm so confused at the moment...

Answer 1

The statement $\forall x \forall y\big((x < y)\to (x^2 < y^2)\big)$ is true iff

$$(x < y)\to (x^2 < y^2)$$

is true for all $x, y$. But the above fails for $x = -5, y = 0$, so $\forall x \forall y\big((x < y)\to (x^2 < y^2)\big)$ is false.

Answer 2

If you want to test the truth, you can see if the question can be falsified; we negate the statement:

~$\forall x \forall y((x < y) \rightarrow(x^2 < y^2))$

Then we have to follow these rules: i) We start with the outermost quantifiers ii) ~ $\forall x$ becomes $\thereexists x $ iii) Negate the relation/predicate. We then have:

$1$) ~$\forall x \forall y((x < y)\rightarrow (x^2 < y^2))$

$2$)(There is x) ~$ \forall y((x < y) \rightarrow (x^2 < y^2))$

$3$)(There is x )(There is y)~$(x<y)\rightarrow(x^2 <y^2)$

$4$)(There is x )(There is y) $(x\geq y)\rightarrow (x^2<y^2) $

And now you test whether there is a pair $(x,y)$ satisfying $4$