Prove no real number satisfies $x^{2} = -1$ I ran a search, but, oddly enough, I can't to find a similar question on here. (If so, kindly point me in that direction, and I'll take this one down.) It seems like a pretty basic question in real analysis, but I'm struggling to come up with a suitable proof that no real number satisfies $x^{2} = -1$. I assume it's a proof by contradiction, but I'm just not seeing it.
 A: Can you prove:
Proposition $(-1)(-1)=1$
Proposition If $\ a<0\ $ then $\ a=(-1)|a|$
Assuming these, if $a\ge 0$ then it is clear $a^2\ge 0$. If $a<0$ then 
$$a^2=a\cdot a = (-1)|a|\cdot (-1)|a| = (-1)(-1)|a|^2 = |a|^2 > 0$$
A: Here's a proof that uses calculus.
We show the function $f(x) = x^2 + 1$ has no real roots. Its derivative is $f'(x) = 2x$, so its only critical point occurs when $x=0$. Since $f''(x) = 2$, by the second derivative test, $x=0$ is a global minimum. Since $f(0) = 1$, for all $x\in\mathbb{R}$, $f(x)\geq 1 > 0$, so no real number satisfies $f(x) = 0$.
A: If possible suppose that there is a non zero real number, say $r$, with $r^2=-1$
Now if $r>0$ then $-1=r^2>0$ (since multiplying both side by $r$ does not change the inequality) which is a contradiction, 
and if  $r<0$ then multiplying by a negative number reverse the inequality and as in the first case you will get a contradiction.  
A: Assume there actually is a real number satisfying $x^2=-1$. Since $0^2=0$, it follows that $x \neq 0$. Assume $x > 0$. From
$$x \cdot x= x^2 =-1$$
We obtain (Dividing by $x$):
$$0<x=-\frac{1}{x}<0$$
Contradiction.
Assume $x < 0$. Again we obtain:
$$0>x=-\frac{1}{x}>0$$
Contradiction.  
The conclusion (The only thing that would make the above implication be true as a whole) is that there does not exist a real number $x$ for which $x^2=-1$.
A: If $x>0$, then $x\cdot x>0$ since the product of two positive numbers is positive.
If $x=0$, $\;x\cdot x=0\cdot0=0\cdot0+(0\cdot0-(0\cdot0))=(0+0)\cdot0-(0\cdot0)=0\cdot0-(0\cdot0)=0$.
If $x<0$, then $-x>0$ and \begin{align}
&(-x)(-x)\\=&(-x)(-x)+0\\=&(-x)(-x)+(0\cdot x-(0\cdot x))
\\=&(-x)(-x)+((0+0)x-(0\cdot x))\\=&(-x)(-x)+((0\cdot x+0\cdot x-(0\cdot x))
\\=&(-x)(-x)+0\cdot x
\\=&(-x)(-x)+(-x+x)\cdot x
\\=&(-x)(-x)+(-x)\cdot x+x\cdot x
\\=&(-x)(-x+x)+x\cdot x
\\=&(-x)\cdot0+x\cdot x
\\=&(-x)\cdot0+((-x)\cdot0-((-x)\cdot0))+x\cdot x
\\=&(-x)\cdot(0+0)-((-x)\cdot0)+x\cdot x
\\=&((-x)\cdot0-((-x)\cdot0))+x\cdot x
\\=&x\cdot x>0
\end{align}
since $(-x)(-x)>0$.
Finally,  $1=1\cdot1>0$ and so, $-1<0$ which implies $-1\not\geq0$ by trichotomy.
Hopefully, the axioms that I've used are clear enough.
A: Direct proof: 
Assume $z \in \mathbb{C}$ obeys $z^2=-1$. So:
$$z(z-\overline{z})=z^2-z \overline{z}=-1-|z|^2 \neq 0 \implies z-\overline{z} \neq 0 \iff \Im{(z)} = \frac{z-\overline{z}}{2i} \neq 0 \iff z \notin \mathbb{R}$$ 
A: This might be cheating a bit, but you could also give reference to the Fundamental theorem of Algebra: Any $n$ degree polynomial in $\mathbb{C}[x]$ has exactly $n$ roots counted with multiplicity. If you know that $i$ and $-i$ are the two roots of $x^2 +1 = 0$ and you know that $i$ and $-i$ are not real, then there are no real roots.
