Showing that $(-1)\cdot(-1) = 1$ in Field theory.

Question

I know that

$(-a) = (-1)\cdot a$ and $-(-a) = a$

So that I could say $a=1$ and use this in:

$-(-a) = - [(-1)\cdot a]$

such that $1 = -[(-1)\cdot 1]$.

But I don't know how to proceed from here, how can I turn $-[(-1)\cdot 1]$ into a simple $(-1)\cdot (-1)$?

For some basic information about writing mathematics at this site see, e.g., here, here, here and here. — Another User, Aug 26 '22 at 18:32
choose $a=1$ for $-(-a) = a$ but choose $a=-1$ for $(-a) = (-1)*a$ — 311411, Aug 26 '22 at 18:36
$$\begin{align} (-1)(-1)&=(-1)(-1)+((-1)+1)\&=(-1)(-1)+1(-1)+1\&=((-1)+1)(-1)+1\&=0\cdot (-1)+1\&=0+1=1. \end{align}$$ — Thomas Andrews, Aug 26 '22 at 18:39

score 2 · Answer 1 · answered Aug 26 '22 at 18:43

2

If you know that

Note that

$1=_2-(-1)=_1(-1)\cdot (-1)$

answered Aug 26 '22 at 18:43

This is perfect, I am so dumb, could not see it for the life of me. – Denis Dinand Aug 26 '22 at 18:53

score 0 · Answer 2 · answered Aug 26 '22 at 19:56

0

You can also see that $x^2=1$ has two solutions $1$ and $-1$ since a field is necessarily an integral domain.

answered Aug 26 '22 at 19:56

Lelouch

2 Answers2