Proof that $i$ is not a real number This will not be a very rigorous proof of that $i=\sqrt{-1}$ is not a real number.
Using the properties of the $\mathbf{R}$:

*

*The real numbers are closed under multiplication

*The real numbers can negative, zero or positive

Assuming that $i\in \mathbf{R}$,
Therefore working on three cases:
Case 1: $i=0$
If $i=0 \Rightarrow i\cdot i=0 \cdot i\Rightarrow -1=0$ which is clearly wrong.
Hence $i\not=0$
Case 2: $i>0$
If $i>0 \Rightarrow i\cdot i>0 \cdot i\Rightarrow -1>0$ which is once again clearly wrong
Hence $i\not>0$
Case 3: $i<0$
If $i<0 \Rightarrow i\cdot i>0 \cdot i\Rightarrow -1>0$
Hence $i\not<0$
Henceforth $i$ is not positive, neither is it negative nor is it zero.
This can only be possible if $i\notin \mathbf{R}$
I'd like to know if this is a good proof if it was backed up group theory.
 A: This is correct, but there is no need to consider $3$ different cases. You could just say that if $i\in\Bbb R$, then $i^2\geqslant0$. But this is false, since $i^2=-1$.
A: Alternative approach
By definition, $i^2 = -1.$  A property of the real numbers is that for all $x \in \Bbb{R}, x^2 \geq 0.$
Therefore, $i$ can't be a real number.
A: I'll address a point of confusion the OP has had with existing answers. Let's begin by proving the given properties of $\Bbb R$, together with rules such as $a,\,b>0\implies ab>0$, imply $x^2\ge0$ for $x\in\Bbb R$: if $x\in\Bbb R$ either $x\ge0$ so $x^2\ge0$, or $x<0$ so $x^2>0$. This is a consequence of our assumptions, not a further assumption beyond our allowed scope. So when we note this contradicts $i^2=-1<0$ if $i\in\Bbb R$, we have proven $i\notin\Bbb R$; we don't need to worry that maybe we've discovered a counterexample to a conjecture.
A: Of course, this answer is based on the fact that $ i ^ 2 = -1$ anyway.  But for me it's a different beauty of mathematics.
If, $i\in\mathbb R$, then $e^{i \pi}>0$. But, this contradicts with Euler's well-known identity.
$$e^{i \pi}=-1$$
