Is $\sqrt{-(x^2)}$ discontinous on the reals?

Question

I know this function does not exist on the reals, but does that mean it is discontinuous on the reals?

Answer 1

The domain is $\{0 \}$, and the function is indeed continuous there. For any fixed $\epsilon>0$, choose $\delta = \epsilon$. Then it is true that $\forall x \in (-\delta, \delta): |f(x) - 0| < \epsilon$, since the only $x$ there is to check is $x=0$, and at that point we have $|f(x)-0|=|0-0|=0<\epsilon$.

Answer 2

The question cannot be answered. This is would be like saying "If an object doesn't have a color, is it's color a shade of blue?".

The definition of "discontinuous on the reals" assumes that we have a function on the reals. If we do not have such a function, the definition doesn't make sense.