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Original question on math.stackexchange by A.J.
There are no square roots for j. Say that (a+bj)^2=j. Then
a^2+2abj+b^2=j
leads to a^2+b^2=0 and 2ab=1, which has no solution. Similarly, if (a+bϵ)2=ϵ, then
a^2+2abϵ=ϵ
leads to a^2=0 and 2ab=1, which has no solution.
This probably owes to the fact that the split-complex algebra and the dual numbers algebra are not fields. In general, if α and β are real numbers, the behavior of the set
C_α,β={a+bu∣a,b∈R and u^2=α+βu},
equipped with the obvious operations, can be controled by the discriminant Δ=β^2+4α.
Clearly the last two cases are, in reality, a single one, but I think it is easier to see what happens if you state it like this. Proof of these facts? Exercise!