0
$\begingroup$

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?

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

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

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

$\endgroup$
  • $\begingroup$ Ridiculous though it may sound, how do you know that $0 \cdot 0 = 0$? $\endgroup$ – Aemilius Nov 3 '17 at 19:47
  • $\begingroup$ @Aemilius it's an issue of the other answerer, too :) $\endgroup$ – Raffaele Nov 3 '17 at 19:48
  • $\begingroup$ @Aemilius should work now. Have a look. Disclaimer: I am not saying your proof is bad. I just wanted to show mine. $\endgroup$ – Raffaele Nov 3 '17 at 19:54
  • $\begingroup$ Yeah, now this works just fine! $\endgroup$ – Aemilius Nov 3 '17 at 19:56
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.