$\forall x \in Z \exists y \in Z(x *y = x + y)$
If I'm understanding this correctly this is FALSE.
$\forall x \in Z \exists y \in Z(x *y = x + y)$
If I'm understanding this correctly this is FALSE.
To prove it false, you just have to exhibit a single x for which there is no y ... This is because the negation of
$\forall x \in Z \exists y \in Z(x *y = x + y)$
is
$\exists x \in Z \forall y \in Z(x *y \ne x + y)$
yunone has done just that.
Let $x\in\mathbb{Z}$. Firstly, if $x\in\{0,2\}$ then $y=x$. Suppose if so $x\notin\{0,2\}$.
We have that $x=\frac{x+y}{y}=\frac{x}{y}+1$. Since both sides are integers we have $\frac{x}{y}\in\mathbb{Z}$. Let $x$ be some prime number which is not $2$, then $y=\{\pm 1,\pm x\}$. Since $\frac{x}{y} = x-1 \neq \pm 1$ (because $x\neq 0,2$) we have that $y\neq\pm x$ therefore $y=\pm 1$ but then we have $x-1 = \pm x$ which is a contradiction.