Embedding $\mathbb{R}$ in $\mathbb{C}$: Why is $\iota \colon \mathbb{R} \rightarrow \mathbb{C}$ injective? Problem: When embedding $\mathbb{R}$ in $\mathbb{C}$ with a function $\iota \colon \mathbb{R} \rightarrow \mathbb{C},x\mapsto(x,0)$ , it seems obvious that $\iota$ is injective. But how do I show that?
My attempt/understanding: For showing injectiveness I need to show that $\forall x ,y\in \mathbb{R}: \iota(x):=(x,0)=(y,0)=:\iota(y) \implies x=y$ is true. For me this is pretty obvious, but I don't know how to show this formally. 
The reasons given for injectiveness in my reading are: 


*

*$(x,0)+(y,0)=(x+y,0)$

*$(x,0)\cdot (y,0)=(x\cdot y,0)$


I understand that this shows that the operations of $\mathbb{R}$ are analogous to $\mathbb{C}$, but I do not understand how this contributes to proving injectiveness. 
 A: As a set, $\mathbb{C}$ is the set of ordered pairs $(a,b)$ of real numbers $a,b$.
Now, by definition of equality for ordered pairs, we have  $(a,b)=(a',b')$ iff $a=a', b=b'$.
Therefore, $\iota(x)=\iota(y) \implies (x,0)=(y,0) \implies x=y,0=0$.
A: One feature about monic maps is that if $g \circ f$ is monic, then $f$ is monic too.
Here is an outline of a method to show $\iota$ is monic:


*

*Construct the "real part" map $\mathrm{Re} : \mathbb{C} \to \mathbb{R}$ 

*Show that $\mathrm{Re} \circ \iota$ is the identity map on $\mathbb{R}$


Since the identity map is monic, you can conclude $\iota$ is as well.
Since you're expressiong $\mathbb{C}$ as tuples, the real part function is pretty simple: it's $(x,y) \mapsto x$. The only reason I bothered to call it $\mathrm{Re}$ is that I wanted to state the sketch in a way that might be a bit easier to generalize to other situations.

The mechanics of following this proof sketch aren't very much different from the other suggestions involving tuples, but you may find the organization of the proof more intuitive since it boils down to a formal calculation.
