× denotes the multiplication operation.
$ (\forall x)(\forall z)(\exists y)(x × y = z) $
Domain: $\mathbb{Q}$
(Q.1) The answer I have says this formula would evaluate to true for this domain, which I don't understand.
If x was 0 ($\frac{0}{1}$) and z was 5 ($\frac{5}{1}$), then there exists no rational number y, such that 0 × y = 5
$\therefore$ $ (\forall x)(\forall z)(\exists y)(x × y = z) $ is false under this interpretation.
Am I correct or am I missing something here?
Domain: $\mathbb{R}$
(Q.2) Essentially my argument is the same for why this formula is false for this domain, but for brevity:
If x was 0.0 and z was 5.0, then there exists no real number y, such that 0.0 × y = 5.0
$\therefore$ $ (\forall x)(\forall z)(\exists y)(x × y = z) $ is false, under this interpretation.
Am I correct or am I missing something here?
With the help of the users in the comment section, I can conclude the arguments made are correct.
The formula was changed to:
$ (\forall x) (\forall z) (x \ne 0 \rightarrow (\exists y)(x \times y = z))$