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Square roots of j and ε

Original question on math.stackexchange by A.J.

Answer by Ivo Terek

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.

If Δ<0, then C_α,β is a field.
If Δ=0 , the zero divisors of C_α,β are precisely the elements a+bu with a+βb/2=0 , while the others have inverses.
If Δ>0 , the zero divisors are precisely the elements a+bu such that a+(β+√Δ)b/2=0 or a+(β−√Δ)b/2=0, while the others have inverses.

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!