# how to prove $(-1)\cdot(-1)=1$ based only on the field axioms?

How do I prove $(-1)\cdot(-1)=1$ (or in other words that $-1$ is the multiplicative inverse of itself), based only on the field axioms?

Thanks

This is to get the result of the multiplication $\cdot$ of $- 1$ by $-1$. The number $-1$ is the symmetric of $1$ with respect to addition $+$. The axioms that relate addition $+$ to multiplication $\cdot$ are distributivity to the left and right distributivity. Any way to prove this result has to pass through the axiom distributive property which connects the multiplication operation with the addition operation. These equality is a relatively easy exercise.

\begin{align} (-1)\cdot (-1)= & (-1)\cdot(-1)+\color{red}{0}\\ = & (-1)\cdot(-1)+\color{red}{[(-1)+1]}\\ = & (-1)(-1)+\color{red}{\Big(1\cdot(-1)+1\cdot 1\Big)}\\ = & \color{blue}{\Big((-1)(-1)+1\cdot(-1)\Big)}+\color{red}{1\cdot 1} \\ = & \color{blue}{\Big([(-1)+1]\cdot(-1)\Big)}+\color{red}{1} \\ = & \color{blue}{\Big(0\cdot(-1)\Big)}+\color{red}{1} \\ = &\color{blue}{0}+ \color{red}{1}\\ =&1 \end{align}

• wow...thanks!!!
– MCL
Oct 26, 2013 at 8:03

$\textbf{Hint:}$ Think about $(1-1)(1-1)$.

Principles of Mathematical Analysis - Walter Rudin, 3rd Ed, p7: • Regarding the last equality, -(-(xy))=xy, here is the proof: By addition inverses axiom, -x+x=0. Replace x by -x. This gives -(-(x))=x Oct 25, 2013 at 15:18
• By the way you will still have to prove $0 \cdot y = 0$, this shouldn't be difficult though. Oct 25, 2013 at 15:24
• Also still need to prove that ax+b=0 has one and only solution but I know how to prove these two, so that's really good enough for me.
– MCL
Oct 25, 2013 at 15:28

$(-1) \cdot (-1)=(-1) \cdot (-1)+(-1) \cdot (1)-(-1) \cdot 1=(-1) \cdot (-1+1)-(-1)=1.$

$-1\cdot\text{sth.}=\text{additive inverse of sth.}$,thus $-1\cdot(-1)=1$, as $1$ is the additive inverse of $-1$.

• You have to prove that $-1 \cdot x = -x$... Oct 25, 2013 at 14:55
• $0 = 0\cdot x = (1+(-1))\cdot x = 1\cdot x + (-1)\cdot x = x + (-1)\cdot x \implies (-1)\cdot x = -x$. Oct 30, 2013 at 11:51