# Using only the field axioms of real numbers prove that $(-1)(-1) = 1$

Using only the field axioms of real numbers prove that $(-1)(-1) = 1$
(1) I start with an obvious fact:$$0 = 0$$ (2) Add $(-1)$ to both sides of the equation: $$0 + (-1) = 0+ (-1)$$ (3) Zero is the neutral element of addition $$(-1) = (-1)$$ (4) One is the neutral element of multiplication $$(-1)(1) = (-1)$$ $$(-1)(1+(-1)+1)=(-1)$$ (5) Multiplication is distributive under addition $$(-1)(1)+(-1)(-1)+(-1)(1) = (-1)$$ (6) One is the neutral element of multiplication $$(-1)+(-1)(-1)+(-1)=(-1)$$ (7) Add $1$ to both sides $$(-1)+(-1)(-1)+(-1)+1=(-1)+1$$ (8) Negative one is the additive inverse of one $$(-1)+(-1)(-1) +0 = 0$$ (9) Add 1 to both sides $$1 + (-1) + (-1)(-1)+0 = 0 + 1$$ $$0 + (-1)(-1) + 0 = 0 + 1$$ (10) Zero is the neutral element of addition $$(-1)(-1) = 1$$ Is my proof good? Should I change something?

• Looks good to me. – aleden Nov 3 '17 at 19:30
• Is there any reason to start with $0=0$ and not with $(-1)=(-1)$? – G Tony Jacobs Nov 3 '17 at 19:30
• This all looks right to me. – Santana Afton Nov 3 '17 at 19:31
• @GTonyJacobs, Actually, I thought that it would be more natural. – Aemilius Nov 3 '17 at 19:31
• @Aemilius No, I mean adding the same quantity in right and left sides – Raffaele Nov 3 '17 at 19:35

What do you think of this?

Lemma. In a field $a\cdot 0=0\cdot a=0$, for any $a\in\mathbb{F}$

proof

$a\cdot 0=a\cdot (1+(-1))=a+(-a)=0$

$0\cdot a=a\cdot 0=0$ for the commutativity of product

end proof

end lemma

main proof

$1=1\\ 1+0\cdot 0=1\\ 1+(1-1)(1-1)=1\\ 1+1\cdot 1+1(-1)-1(1)+(-1)(-1)\\ 1+1-1-1+(-1)(-1)=1\\ 0+0+(-1)(-1)=1\\ (-1)(-1)=1$

• Ridiculous though it may sound, how do you know that $0 \cdot 0 = 0$? – Aemilius Nov 3 '17 at 19:47
• @Aemilius it's an issue of the other answerer, too :) – Raffaele Nov 3 '17 at 19:48
• @Aemilius should work now. Have a look. Disclaimer: I am not saying your proof is bad. I just wanted to show mine. – Raffaele Nov 3 '17 at 19:54
• Yeah, now this works just fine! – Aemilius Nov 3 '17 at 19:56

Some people like to condense these things into one string of equalities:

\begin{align}(-1)(-1)=(-1)(-1)+0 &= (-1)(-1)+(-1)+1\\ &= (-1)(-1)+(-1)(1)+1\\ &=(-1)(-1+1) + 1\\ &=(-1)(0) + 1\\ &=0+1\\ &=1 \end{align}

This is basically just your argument, rearranged. It begins with $(-1)(-1)$ and ends with $1$, and we can justify each equal sign along the way: i) $0$ is the additive identity, ii)$-1$ and $1$ are opposites, iii) $1$ is the multiplicative identity, iv) distribution, v) $-1$ and $1$ are opposites, vi) $0$ is an annihilator for multiplication, vii) $0$ is the additive identity.

The only step here that isn't an axiom is that $0\cdot a=0$ for all $a$, but this is usually one of the first things you prove.