TITLE: Why does the imaginary number $i$ satisfy $i\times 0=0$?
QUESTION [9 upvotes]: Why does the imaginary number $i$ satisfy $i\times 0=0$?
I mean, we don't really know what $i$ is. How could we be sure about that? I think there's a reason behind why mathematicians decided that.

REPLY [5 votes]: Technically, $0\,i\not\in\mathbb{R}$; instead, $0\,i\in\mathbb{C}$. Thus, being (overly?) pedantic, $0\times i\ne0$; instead $0\times i=0+0\,i$. However, being the zero element of $\mathbb{C}$, we usually abbreviate $0+0\,i$ as $0$, not meaning an element of $\mathbb{R}$ but an element of the subset of $\mathbb{C}$ that is homeomorphic to $\mathbb{R}$, that is $\{x+0\,i:x\in\mathbb{R}\}$. With this abbreviation, we do have $0\times i=0$

This comment by AlexB says

The thing is that when one defines the integers, they come with a canonical embedding of $\mathbb{N}$. Similarly, the rationals come with a canonical embedding of $\mathbb{Z}$ and so on. So it does make sense to speak of subsets. Indeed, it becomes so cumbersome to distinguish between genuine subsets and images under an embedding that one fairly quickly drops this distinction in practice, unless one really has to thing about the foundations.

The question "Why does the imaginary number $i$ satisfy $i\times0=0$?" seems to me to be fairly foundational, so I think that my approach above seems appropriate.
Once we have established why $i\times0=0$, then we can move on as Arturo Magidin says in this answer:

So even though they are actually very different sets, we have copies of each sitting inside the "next one", copies that respect all the structures we are interested in, so we can still think of them as being "subsets".