Simple proof using only field axioms: $(-x)(-y) = xy$

Question

I need to prove that $(-x)(-y) = xy$ using only the field axioms. I tried starting with $ since$ $(-(-x)(-y)) + (-x)(-y) = 0$ by A6, or the additive inverse. And then adding $xy$ to both sides by A2 or transitivity. But I'm sure that to say $(-(-x)(-y))=-(xy)$ is outside the field axioms. So I'm stuck and would appreciate some help.

Answer 1

First, we need a lemma $\forall y, 0y = 0$.

$$ \begin{array}{rll} 0y &= 0y + 0 &0 \text{ is additive identity}\\ &= 0y + (0y + (-0y))& -0y \text{ is additive inverse of } 0y\\ &= (0y + 0y) + (-0y)& \text{addition is associative}\\ &= (0+0)y + (-0y) &\text{multiplication is distributive}\\ &= 0y + (-0y)&0 \text{ is additive identity}\\ &= 0 & -0y \text{ is additive inverse of } 0y \end{array} $$

With this lemma, we have $$ \begin{array}{rll} (-x)(-y) &= (-x)(-y) + 0 & 0 \text{ is additive identity}\\ &= (-x)(-y) + 0 y & \text{ lemma }\\ &= (-x)(-y) + (x + (-x)) y & -x \text{ is additive inverse of } x\\ &= (-x)(-y) + (xy + (-x)y) & \text{ multiplication is distributive}\\ &= (-x)(-y) + ((-x)y + xy) & \text{ addition is commutative} \\ &= ((-x)(-y) + (-x)y) + xy & \text{ addition is associative} \\ &= (-x)((-y) + y) + xy & \text{ multiplication is distributive }\\ &= (-x)(y + (-y)) + xy & \text{ addition is commutative}\\ &= (-x)0 + xy & -y \text{ is additive inverse of } y\\ &= 0(-x) + xy & \text{ multiplication is commutative}\\ &= 0 + xy & \text{ lemma }\\ &= xy & 0 \text{ is additive indentity } \end{array} $$

Answer 2

It depends how picky you are, and how minimal your axioms are. The below is middle of the road.

1. We have $(-x)y+xy=(-x+x)y$ (distributive law) $=0y$ (defn of additive inverse)

2. We want $0y=0$, but this takes a surprising amount of effort. We have $1y=y$ (defn of mult identity) so $0y+1y=(0+1)y$ (distrib law) $=1y$ (defn additive identity). We are still not quite home (unless your axioms give that the additive identity is unique). So let $z$ be additive inverse of $1y$. Then $(0y+1y)+z=0y+(1y+z)=0y+0$ (defn additive inv) $=0y$ (defn of additive inv). But $0y+1y=1y$, so $(0y+1y)+z=1y+z$, so $0y=1y+z=0$ (defn additive inv).

3. So we have established that $(-x)y+xy=0$. In other words $xy$ is the additive inverse of $(-x)y$.

4. But $(-x)y=y(-x)$ (mult commutative) and since $x$ was any element of the field, this is the additive inverse of $(-y)(-x)=(-x)(-y)$. Thus we have established that $xy$ and $(-x)(-y)$ are both additive inverses of $(-x)y$.

5. If your axioms require inverses to be unique you are done. If not you have to prove it. So assume $b,c$ are both additive inverses of $a$. Then $c=0+c=(b+a)+c=b+(a+c)=b+0=b$.

Answer 3

We are so used to the operators in arithmetic that we don't realise when they are being used in a way that has yet to be defined.

In the field axioms for the additive operator there is an identification of -x with the inverse of x, and the minus sign is blithely tossed around and assigned elsewhere as we know it can be from the arithmetic we have been doing for years. But there is nothing in the axioms that says that you can divorce the minus sign from the real number in this way: the -x is an indivisible whole just as the x alone is.

However, we can divorce the sign from its number and toss it about as we know we can if we can prove that -1.x = -x; and, in general, that -x.y = x.-y:

     (x + -x).y = x.y + -x.y = 0

hence

     -x.y = -(x.y) ;

also

     x.(y + -y) = x.y + x.-y = 0

hence

     x.-y = -(x.y) ;

thus

     -x.y = x.-y

Answer 4

If we think in terms of number lines. Travel to the left is negative. Going backwards is negative. Going backwards to the left goes positive.

But some of us, and probably birds, think spatially. In a sphere, if you drop down and pass the origin, you start to go up. A child knows digging a ditch ends up in China, or England if you're in China.

Spherical movements in opposite directions cancel. If force 1 spins the sphere 5 times faster in one direction, and force 2 spins it 5 times faster in the opposite direction, they cancel. Two negative spherical vectors multiplied increase their negativity.

If spheres are embedded in 2D space, they're just local oddities. But reality is different. Our 2D space is embedded in 3D. I think this is what makes mathematics of negatives unintuitive to children.