Question

Is \(0i\in \mathbb C \setminus \mathbb R\)?

Answer 1

If you edit it it doesn't render any more, lol.

Answer 2

Is \( \Large 0i\in \mathbb C \setminus \mathbb R\)?

Answer 3

This means is zero times sqrt(-1) an element of the complex numbers and not an element of the reals? Jeeze, I honestly don't know haha. How are the complex numbers defined?

Answer 4

Complex numbers are defined as \(a+bi\) where \(a,b\in \mathbb R\) and \(i=\sqrt{-1}\)

Answer 5

Is 0=0i? Since usually I see it as a+bi, I'm going to have to say that it is an element, since b is a real number times i, so really when we see 0 it is assumed to be 0+0i?

Answer 6

Yeah not very proof like of me. So maybe:

a is an element of R

a and b are an elements of C

then C\R is only b.

Since 0 is a possible b, 0i is contained in C\R

Answer 7

I'm in Real Analysis 1 right now, so give me a break, but I feel like that's pretty much what you're looking for, it needs more formalization going on there though haha.

Answer 8

This is kind of like saying the zero vector is contained in every coordinate system, but if you remove one of the axes, you're only removing the 0 in that direction. Similarly, you don't gain an extra point if you add another dimension either. Sort of weird but makes sense if you think about it.

Answer 9

\[0i=0\in\mathbb{R}\]