Hello I’m new here and firstly sorry for my English.

Following is a question from a discrete mathematics textbook.

Define a relation C from R to R as follows:

For any $(x,y) \in$ R $\times$ R

$(x,y) \in$ C means that $x^2+y^2=1$

Question: What are the domain and codomain of C

And the answer is:

The domain and co-domain of C are both R, the set of all real numbers.

Obviously we can draw the relation $x^2+y^2=1$ as a circle in the Cartesian plane whose radius is 1 and center is $(0,0)$.

I don’t understand why the domain is R rather than $[-1,1]$.

And I’ve learned that codomain is different from range that it can be large than range. But I still wonder could the the codomain of this relation be $[-1,1]$? Or any other interval which is larger than $[-1,1]$?

Really appreciated for you help!

  • $\begingroup$ the domain is $R$ because it holds $\forall x, y \in R$ $\endgroup$ – The Integrator Apr 21 '18 at 7:00

The domain and codomain are both $\Bbb R$, because checking whether $(x,y)\in C$ makes sense for all $(x,y)\in\Bbb R\times\Bbb R$. They even explicitly said so:

For any $(x,y) \in$ R $\times$ R

Now, whether "domain" means this, or it means the more restrictive "The set of all $x$-values which appears in pairs of $C$" is really up to your textbook's author. It really ought to be defined somewhere in your book, so you should look for it. Regarding the other side, "codomain" is less restrictive, but "range" more restrictive. So the codomain can, technically, be any set which contains $[-1,1]$, but again the quote above hints that they want $\Bbb R$ as the answer.

  • 1
    $\begingroup$ Thank you! Anyway I’m feeling strange that the definition of “domain” can be up to the author. Sorry but this is completely new idea to me because I thought the “domain” always means “the set of all $x$-values”. Again, really appreciate it! $\endgroup$ – Alxt Apr 21 '18 at 7:19
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    $\begingroup$ @Alxt I also think "domain" should have the stricter definition. But you should consult your book to be certain that you get the right one. $\endgroup$ – Arthur Apr 21 '18 at 7:47

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