Dual division rules

[juliadiff]

Dual division rules

[goretkin]

I understand the temptation of seeing ε/ε and calling it 1, however it's not valid. It would imply that 1/ε exists, which means that there is dual number x such that xε=1. x can't be purely real, because you'd end up with a number with an epsilon part. xcan't be purely epsilon, because then you just end up with zero. And combination in between is a combination of zero and epsilon. You can't make 1 out of that.

In short, ε does not have an inverse. (At the very least, the above shows it doesn't have an inverse that is itself a dual number.)

(Note also, that you can start with a false statement and get to a true one by multiplying both sides by ε. For example 3ε=4ε, so you must take care with any argument that multiples or divides by ε both sides of an equation.)

[KristofferC]

I realize it makes no sense for an operation on 2 infinitesimal numbers to result in a real,

I agree with this :)

From what I have seen division between Duals is typically only defined when the real part of the denominator is non zero.

[safiire]

In short, ε does not have an inverse.

Yes, your explanation convinces me. :) That makes sense.