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Does this mean that x and y can only be square integers? 1, 4, 9 etc. ?

The problem I am trying to understand is R = {(x,y) ∈ Z^2 : ∃k ∈ Z x=ky}

Matthew
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  • Note that $\mathbb{Z^2} = \mathbb{Z}\times \mathbb{Z}$. Does this help? Your initial idea is wrong. – xyz Jul 21 '21 at 19:27
  • $\Bbb{Z^2}$ refers to $\Bbb{Z} \times \Bbb{Z}$ which represents ${(x,y) ~: ~x \in \Bbb{Z} ~\text{and} ~y \in \Bbb{Z}}$. – user2661923 Jul 21 '21 at 19:28
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    To be fair, occasionally someone might use $\Bbb Z^2$ as a notation for the set of perfect squares in their context. But the default understanding in set theory of $S^2$ should indeed be $S\times S$. – Greg Martin Jul 21 '21 at 19:30
  • @GregMartin I wouldn't use $\mathbb{Z}^2$ to mean the set of perfect squares, due to the risk of confusion. I would use a different symbol instead. – David Lui Jul 21 '21 at 19:31
  • Welcome to Mathematics Stack Exchange. It means that $(x,y)$ is an ordered pair of integers – J. W. Tanner Jul 21 '21 at 19:40
  • When asking about the meaning of notation, it is good to site where you found the notation. In math one is allowed to give definitions for notation convenient to the exposition at hand, especially if there is no standard notation suitable. – hardmath Jul 21 '21 at 20:32

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No, the exponent is being applied to the set $\mathbb{Z}$, so it's saying that $(x, y)$ is an element of the set $\mathbb{Z}^2$.

$\mathbb{Z}^2$ is the set of all ordered pairs of integers. For example, $(1, 2) \in \mathbb{Z}^2$, and $(12345, 23456) \in \mathbb{Z}^2$. But : $(1.234, 2) \notin \mathbb{Z}^2$ and $3 \notin \mathbb{Z}^2$.

David Lui
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