2
$\begingroup$

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

If I'm understanding this correctly this is FALSE.

  • 5
    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 $\mathbb{Z}$.2011-02-16
  • 6
    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.)2011-02-16

2 Answers 2