# Weird function or not

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?

• Yes, this is a function. In fact, $f = \emptyset,$ and $\emptyset$ is a function. – Dave L. Renfro Nov 19 '14 at 22:01
• So f: ∅ → Y where Y is any set including the non empty set is always a function – Namch96 Nov 19 '14 at 22:03
• @Namch96, it may help to read the accepted answer at math.stackexchange.com/questions/60365/… – Barry Cipra Nov 19 '14 at 22:06
• 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. – user192614 Nov 19 '14 at 22:06
• Ahhh ok i understand – Namch96 Nov 19 '14 at 22:13

• It's a bad definition anyway, because $(-1)^{1/2}$ is undefined. – egreg Nov 19 '14 at 23:04
• 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 0:23
• 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 0:30