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just wondering if this statement is true or false? And can anyone give an example // Counter Example if the statement is true or false respectively

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  • $\begingroup$ What kind of the function did you mean? $\endgroup$ – Mikasa Sep 24 '17 at 13:29
  • $\begingroup$ If $A$ and $B$ are sets, then there is a function $f$ with domain $A$ and codomain $B$. $\endgroup$ – user296602 Sep 24 '17 at 13:29
  • $\begingroup$ x^2; 3x+2; |y|=x $\endgroup$ – user451844 Sep 24 '17 at 13:36
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    $\begingroup$ @user296602 Unless $B$ is empty and $A$ is not. $\endgroup$ – Theo Bendit Sep 24 '17 at 13:37
  • $\begingroup$ I expect that $R$ stands for the set of real numbers. The answer to your question is "no". Functions do not have necessarily real numbers as input. $\endgroup$ – drhab Sep 24 '17 at 13:37
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I can just define a function that takes ordered pairs as input by $f((1,2))=3$. Its domain is $\{(1,2)\}$, which is not a subset of $\Bbb R$

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  • $\begingroup$ at least you didn't do one between two Cartesian products. $\endgroup$ – user451844 Sep 24 '17 at 14:19
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from my comments if you start with the complex numbers their distance to the origin in the complex plane is a real number so we then have a domain of the complex numbers and a codomain of the reals.

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It's obviously false.

Suppose a function is defined on a plane. Then its domain is a plane, which certainly is not a subset of reals. As a result the domain of any function of a complex agrument complex number is not a subset of reals.

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  • $\begingroup$ mathworld.wolfram.com/ComplexArgument.html shows that complex argument may not be the best use of words here. $\endgroup$ – user451844 Sep 24 '17 at 13:58
  • $\begingroup$ @RoddyMacPhee Right. Fixed. Thank you. $\endgroup$ – CiaPan Sep 24 '17 at 14:04

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