Square root of dual number

Question

We know the ring of dual numbers over the real field $\mathbb{R}$, is defined as $\mathbb{R}[\epsilon] = \{a+b\epsilon~|~a,b\in\mathbb{R} ~\text{and}~ \epsilon^2 = 0\}.$ Is the square root of $\epsilon,~i.e.~\sqrt{\epsilon}$ defined? If yes, what is its value? If no, how to show its non-existence? To make it more general, does $\epsilon^\alpha$ for $0<|\alpha|<1$ exist?

Answer 1

Suppose $\sqrt{\epsilon}$ exists. Then $\epsilon = (a+b\epsilon)^2$ for some $a,b \in \mathbb{R}, ~\text{with}~ a \neq 0.$ Then $a^2+2ab\epsilon = \epsilon$ which implies $a=0,$ contradiction.