When and why was $-1 \times -1$ defined to equal $1$?

Question

When and why was $-1 \times -1$ defined to equal $1$?

My thoughts was that since the rational and natural numbers shared similar algebraic properties, maybe they wanted distributivity to hold for the negative numbers also. I have seen an algebraic derivation of the formula in question, but it relies on distributivity of negative multiplication.

Any thoughts or opinions appreciated. Thank you

Answer 1

I really doubt that $-1 \cdot -1$ was defined to be $1$. However, you can derive that $-1 \cdot -1 = 1$ once you extend the Distributive property from natural numbers to all integers … though you also need to extend various other elementary properties regarding addition and multiplication from the natural numbers to the integers:

First, $-1$ is defined as that number such that, when you add $1$ to it, you get $0$.

So, if for now we refer to that number as $a$, we have

$$a+1=0$$

Multiplying both sides by $a$, we get:

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

So if we extend the Distributive property:

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

So, also extending the normal multiplicative properties involving $1$ and $0$:

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

Adding $1$ to each side

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

So, using the Associative property:

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

Since $a+1 =0$ and $0+1=1$, we get:

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

And so:

$$a \cdot a = 1$$

And there you have it:

$$-1 \cdot -1 =1$$

Answer 2

Just providing a faster variant of @Bram28's derivation: in any ring, $$(-x)y=-(xy)$$ since $$(-x)y+xy=(-x+x)y=0y=0.$$ Therefore (letting $x=1$ and $y=-1$) $$(-1)^2=-(1\cdot(-1))=-(-1)=1.$$