Formal proof for $(-1) \times (-1) = 1$ Is there a formal proof for $(-1) \times (-1) = 1$? It's a fundamental formula not only in arithmetic but also in the whole of math. Is there a proof for it or is it just assumed?
We use only the usual field axioms for the real numbers. First we prove an intermediate result.

$0\times 0$

$=0\times(0+0)$

$=0\times 0+0\times 0$

Subtract $0\times 0$ from each side to get $0=0\times 0$. Now we are ready for the final kill.

$0$

$=0\times 0$

$=(1-1)\times(1-1)$

$=1\times 1+1\times (-1)+(-1)\times 1+(-1)\times (-1)$

$=1+(-1)+(-1)+(-1)\times (-1)$

$=(-1)+(-1)\times (-1)$

Add $1$ to each side to get $1=(-1)\times (-1)$.
Another possibility using Euler's identity:

$$\begin{align}
(-1)\times (-1) &amp; = e^{\imath \pi} \times e^{\imath\pi}\\
&amp;=e^{\imath 2\pi} = 1
\end{align}
$$