# Is this a proper proof of (-1)(-1) = 1?

I am a novice in proof writing and have just started a book on analysis. I have no other pure math experience or knowledge of abstract algebra.

I am trying to prove that $(-1)*(-1) = 1$. I will first show my attempt and follow it with the standard technique that I have found online. I would like to understand why the second method is preferred (or perhaps the only valid one).

Method 1 (my attempt): We are given the nine field axioms for $\Bbb R$. Using these, it has been shown (in the text) that $(-1)x = -x$ for all $x \in \Bbb R$. Therefore, $(-1)(-1) = -(-1) = 1$ because we have the identity element of multiplication $1 \in \Bbb R$ and there exists $-1 \in \Bbb R$ such that $-1$ is the additive inverse of $1$.

Method 2 (found by Google search): We know that $(0)(0) = 0$ because it was shown (in the text) that $0x = 0$ for all $x \in \Bbb R$. Then $$(0)(0) = ((-1) + 1)*((-1) + 1) = (-1)(-1) + (-1)(1) + (1)(-1) + (1)(1) \\= (-1)(-1) - 1 - 1 + 1 = (-1)(-1) - 1 = 0$$ Therefore $(-1)(-1) = 1$ because we see that $(-1)(-1)$ is the additive inverse of $-1$.

Is the second method simply a more precise version of the first method, while the first is overly verbal? Or is the first somehow using circular reasoning?

• Nothing's wrong with your method (I assume that $-(-x)=x$ is known). – Hagen von Eitzen May 6 '14 at 17:09
• $-(-x)$ isn't just notation for the additive inverse of $-x$? – Vale132 May 6 '14 at 17:17
• $-(-x)$ is notation for the additive inverse of $-x$, but it still needs to be shown (this is admittedly very quick) that this is equal to $x$. Hagen is just saying that he presumes that's already been done; otherwise, you need to do that as well. – Steven Stadnicki May 6 '14 at 18:04

$$(-1)(-1)+(-1)\stackrel{Dist.}=(-1)\left[(-1)+1\right]\stackrel{\text{additive inv.}}=(-1)\cdot 0\stackrel{text}=0$$
Thus, $\;-1\;$ is the additive inverse of $\;(-1)(-1)\;$ , but also $\;-(-1)=1\;$ is the additive inverse of $\;(-1)\;$, so by uniqueness of the inverses we're done.
• @Vale132, we showed it is an additive inverse, so by the axiom of uniqueness of additive inverse (this very axiom existsin any abelian group, if you mind), we conclude it has to be equal to $\;1\;$ ... – DonAntonio May 6 '14 at 17:15