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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2Yes, this is a function. In fact, $f = \emptyset,$ and $\emptyset$ is a function. – Dave L. Renfro Nov 19 '14 at 22:01
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So f: ∅ → Y where Y is any set including the non empty set is always a function – Namch96 Nov 19 '14 at 22:03
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@Namch96, it may help to read the accepted answer at http://math.stackexchange.com/questions/60365/what-is-the-set-theoretic-definition-of-a-function – Barry Cipra Nov 19 '14 at 22:06
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1A 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. – Nov 19 '14 at 22:06
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Ahhh ok i understand – 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?
Asaf Karagila
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The definition should be formalizable in some language; if the language is unspecified, the definition is ill posed. – egreg Nov 20 '14 at 00:15
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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! – Asaf Karagila Nov 20 '14 at 00:23
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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. – egreg Nov 20 '14 at 00:30
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Yes, I agree that there is some inherent ambiguity. But the nice thing about oblivious freshmen is that you can lie to their face and use what "they think they know and don't know it's wrong yet". Of course I rarely ever do it, but I also hate numerical examples. – Asaf Karagila Nov 20 '14 at 00:35