3 ms·
Seems pretty true with 1 and -1. Map R with f(x) = -x and f(x) = |x - 1| for x in your new mapping is indistinguishable from f(x) = |x + 1| in R. In any case I
by cgriswald 2y ago
Seems pretty true with 1 and -1. Map R with f(x) = -x and f(x) = |x - 1| for x in your new mapping is indistinguishable from f(x) = |x + 1| in R.
In any case I’d say this is arbitrary like using + for addition and - for subtraction. It seems like you’re just talking about the symbols themselves. I’m not sure how you get to half plane from there.
- xigoi 2y ago1 and -1 are distinguishable: one of them equals its square, the other does not.
- cgriswald 2y agoSure, but I'm not sure I'm understanding the argument. I don't understand how a function like f(z) = e^z has to be symmetric about the real number line or how i and -i aren't distinguishable with something like Im(z) > 0. Is there a proof somewhere I can read?
- bandrami 2y agoIt falls out of complex numbers satisfying the conditions of a field though I don't know of a specific "proof" of that (you generally don't "prove" definitions). You could equally say "i is indistinguishable from 1/i" or "i's additive inverse is its multiplicative inverse"; in either case it's an arbitrary choice which of the conjugates is positive and which is negative. The key being that you cannot say "i > -i" because of that.
- cgriswald 2y agoAh, gotcha. I think I wasn’t understanding exactly what was being said. Thanks for the explanation.
- red75prime 2y agoI interpreted your words as "the complex plane has topology of a plane where conjugates are glued together".
- bandrami 2y agoAh, sorry, no; it is a half plane in the sense that which way is "up" is completely arbitrary