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If we have two sets, the set of natural numbers and the set of integers and we relate each member of $\Bbb N $ to its squared value in $\Bbb Z$ then $f(x) = x^2$ and $\Bbb N $ is the domain of function and $\Bbb Z$ is the codomain. Is it okay to define the square function in this manner?

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    $\begingroup$ What you have written is a valid function from $\mathbb{N}$ to $\mathbb{Z}$. If you want to call it a square function, that would also be fine, thought it would beg the question "why "a" and not "the"", ie this is an example, not a general definition. $\endgroup$ – Tobias Kildetoft Apr 1 '13 at 15:34
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    $\begingroup$ Yes. This defines a function with no ambiguity. $\endgroup$ – Julien Apr 1 '13 at 15:34
  • $\begingroup$ @TobiasKildetoft I'm sorry Tobias, I'm not a native speaker, I may make mistakes in grammar. $\endgroup$ – Samama Fahim Apr 1 '13 at 15:55
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Yes, your definition is fine: $$f:\mathbb N \to \mathbb Z$$ $$f(x) = x^2\quad\text{i.e.,}\quad x\in \mathbb N \overset{\large f}{\longmapsto} x^2 \in \mathbb Z$$ is defined everywhere on $\mathbb N$, and is a well-defined function, pure and simple.

Note that the image of $f$ is $$\{x^2\mid x\in \mathbb N\} \subset \mathbb N \subset \mathbb Z$$ That is, the image of $f$ is a proper subset of the codomain you've defined for $f$, since $x^2 \geq 0$ for all $x \in \mathbb N$ (and for all $x \in \mathbb Z$, if the domain were $\mathbb Z$). That's not a problem with your definition. That's just an observation.

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    $\begingroup$ I should sue you for plagiarism. But I'll upvote you instead. $\endgroup$ – Julien Apr 1 '13 at 15:37
  • $\begingroup$ @julien Oh!...I just saw your "no ambiguity" Yikes! $\endgroup$ – Namaste Apr 1 '13 at 15:38
  • $\begingroup$ @julien I think you have a case. $\endgroup$ – Pedro Tamaroff Apr 1 '13 at 15:48

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