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Is $f\colon\emptyset \to\mathbb{R}$ with $f(x) = (-1)^{\frac{1}{2}}$ a function where $\emptyset$ is the empty set and $\mathbb{R}$ is the set of real numbers?

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    $\begingroup$ Yes, this is a function. In fact, $f = \emptyset,$ and $\emptyset$ is a function. $\endgroup$ – Dave L. Renfro Nov 19 '14 at 22:01
  • $\begingroup$ So f: ∅ → Y where Y is any set including the non empty set is always a function $\endgroup$ – Namch96 Nov 19 '14 at 22:03
  • $\begingroup$ @Namch96, it may help to read the accepted answer at math.stackexchange.com/questions/60365/… $\endgroup$ – Barry Cipra Nov 19 '14 at 22:06
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    $\begingroup$ A function $f:\emptyset\to A$ is the empty function. It is being described as $\{(x,y)\in \emptyset\times A:\ y=f(x),\text{ and }$f(x)=(-1)^{1/2}$ \}=\emptyset$. The only problem here is whether or not $(-1)^{1/2}$ is a valid symbol in your language. $\endgroup$ – user192614 Nov 19 '14 at 22:06
  • $\begingroup$ Ahhh ok i understand $\endgroup$ – Namch96 Nov 19 '14 at 22:13
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If it wasn't a function into the real numbers there was a witness to this fact. Is there?

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  • $\begingroup$ It's a bad definition anyway, because $(-1)^{1/2}$ is undefined. $\endgroup$ – egreg Nov 19 '14 at 23:04
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    $\begingroup$ That's the point methinks. $\endgroup$ – Asaf Karagila Nov 20 '14 at 0:11
  • $\begingroup$ The definition should be formalizable in some language; if the language is unspecified, the definition is ill posed. $\endgroup$ – egreg Nov 20 '14 at 0:15
  • $\begingroup$ It's a valid definition if you agree that the result should generally be $i$, the question remains whether there is a real number equal to this term. Lucky for us, we don't have to check! $\endgroup$ – Asaf Karagila Nov 20 '14 at 0:23
  • $\begingroup$ I disagree: $(-1)^{1/2}$ is not defined anywhere unless you precisely state what branch of the square root you use. But the question is not really a mathematical one; it's simply a bad question. I'm not blaming @Namch96, of course, but whoever posed it to her/him. $\endgroup$ – egreg Nov 20 '14 at 0:30

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