4 ms·
A=Aa+e A-Aa=e A(1-a)=e A=e/(1-a) not e/(e-a)
by emilfihlman 8y ago
A=Aa+e
A-Aa=e
A(1-a)=e
A=e/(1-a) not e/(e-a)
- dataflow 8y agoNope, I'm pretty sure it's not wrong. I believe ϵ is the multiplicative identity ("1") here. Note that ϵ is the empty string, and concatenating it with anything results in the same thing, so Aϵ = A. Hence we have Aϵ - Aa = A(ϵ - a) = ϵ.