Let $x\in \mathbb{R}$ and $z\in\mathbb{C}$. And define equality ($x=z$) iff $x=(0,x)$. Is this equality well defined?

Question

Let $x\in \mathbb{R}$ and $z\in\mathbb{C}$. And define equality ($x=z$) iff $x=(0,x)$. Is this equality well defined ?

Okay. It is easily shown that something goes wrong if you define equality in this way.

$$1=1\cdot 1=(0,1)\cdot(0,1)=i^2=-1$$

But how do I call this what goes wrong here? In some sense I would think that my definition is a well defined equivalence relation. But I'm certainly doing something wrong here.

I would like to understand this better then just finding with counterexamples that I can't define something this way. I would like to understand the "root" of my wrong definition.

Note: I know that the usual definition is that $x=(x,0)$, I'm just playing around. I think my head has a hard time understanding the logic behind the equality $x=(x,0)$. My head says something, can't I define such a equality just like that. What do I need to show to say that this equality is well defined ?

Answer 1

You assume that $\,x \equiv ix\,$ for all $\,x\in\Bbb R.\,$ As you show, this implies $\,1 \equiv -1,\,$ so $\,2 \equiv 0,\,$ so $\,1 \equiv 0,\,$ so the quotient ring generated by this congruence relation is the trivial one-element ring (or, you have reached a contradiction if you are working only with fields, where $\,1\ne 0\,$ by hypothesis).

Answer 2

Your "equality" is actually an equivalence relation. So we have x is equivalent to z iff x=(0,x). But in C x=(x,0). Thus, x is equivalent to z iff x=0, and the relation is not reflexive. Okay, so maybe the second equality was also supposed to mean your equivalence relation? Then you are defining your relation in terms of itself. In your example it seems like you are saying (a,o) is equivalent to (o,a) {That is, your equivalence relation is only defined for elements in R and elements in the imaginaries, not their sums). To preserve addition, you must have (a,a) is equivalent to (2a,0) for all a. With a little fiddling, you should have (a,b) is equalivent to (c,d) if (a+b=c+d). This is already going beyond your definition though! This leads to 1=-1. What the moral of the story should be is that equivalence relations do not necassarily preserve a particular operation. You can always extend your equivalence relation to accomodate the operations but then you might get a trivial algebraic structure in the end. You have to pick your equivalence relation carefully (such as in quotienting by an ideal)