Simple proof that $i \neq 0$

Question

This is an obvious fact, but I'm working from first principles, and it seems that rigorously formalizing such a fact is necessary.

Theorem: $i \neq 0$

Proof. Consider the inclusion map of $\mathbb{R}$ in $\mathbb{C}$, \begin{align*} \iota : & \mathbb{R} \hookrightarrow \mathbb{C} \\ & x \mapsto x + 0i. \end{align*} For $x, y \in \mathbb{R}$, $\iota(x) = \iota(y)$ implies that $x + 0i + y + 0i$. Since $0i = 0 + 0i$ is the additive identity in $\mathbb{C}$, $x + 0i = x$ and $y + 0i = y$. Hence, $x = y$, meaning that $\iota$ is injective, so $\iota$ embeds $\mathbb{R}$ in $\mathbb{C}$. Hence, $0$ is the unique preimage in $\mathbb{R}$ of the element $0 + 0i$ in $\mathbb{C}$. Furthermore, since $\iota$ is well-defined, $\iota$ maps $0$ to some unique $z$. Since $i = 0 + 1i \neq 0 + 0i$, $i$ is not the image of $0$ under $\iota$, so $i \neq 0$.

Remarks. I feel like I wrote a lot more than I needed to. The fact that $\iota$ is well-defined (which I admit that I did not prove, though I can't even think of a proof of this fact other than to say that it's rather obvious) seems sufficient to write the proof.

How does this look? I'd appreciate any critiques, whether it's of my reasoning, my writing style, etc.

Answer 1

Your proof is incorrect. Near the end you write $i=0+1i \neq 0$. But here you assume that $i \neq 0$, which is what you want to prove. Also, note that your proof doesn't 'do' much. You mearly look at the identity map, and don't use any specific property of $i$, which should arise suspicion.

A very simple correct proof is the one given above by PrudiiArca.

Answer 2

$i^4 = 1$, so $i$ is a unit in the ring/field $\mathbb C$ and not a zero divisor. In particular $i\neq 0$.

Another perspective is that $\mathbb R$ has no root for $-1$ and $\mathbb C$ is obtained by adjoining an element $i$ not in $\mathbb R$, ie. $\mathbb C = \mathbb R(i)$. In particular $i \neq 0$.