# Universal Quantification and Sets $\forall x \in Z \exists y \in Z(x *y = x + y)$

$\forall x \in Z \exists y \in Z(x *y = x + y)$

If I'm understanding this correctly this is FALSE.

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I think you are right, for example, if $x=3$, there doesn't exist a $y$ such that $3y=3+y$, since this implies $y=3/2$, which is not in $\mathbb{Z}$. – yunone Feb 16 '11 at 6:52
Note that you haven't asked a question yet...(If your question is "Am I understanding this correctly?" the answer is "Yes, it seems that you are." Maybe it would make for a more useful question if you described what doubts you have that your reasoning is correct.) – Pete L. Clark Feb 16 '11 at 8:26

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.

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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.

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