Maybe this is a weird question but it's been bugging me.
In the childhood we were taught that $4 \times 3$ means $4+4+4$ i.e. adding 4, 3 times.
My question is then how would you explain $-1 \times -1$ using some kind of mathematical logic?
I want to know the significance in real life.
It doesn't have a meaning when I say adding $-1$, $-1$ times.
Look at it this way,
$3 \times 4$ , you have a defined credit of $3$ and you gain $4$ times as much as that credit:
$$(+3+3+3+3) = (+12)$$
$3 \times -4$ , here you have a defined credit but you lose $4$ times as much of that credit:
$$(-3-3-3-3) = (-12)$$
$-3 \times 4$ , you have a debt of $-3$ and you gain 4 times as much debt as that debt:
$$[+(-3)+(-3)+(-3)+(-3)] = (-3-3-3-3) = (-12)$$
$-3 \times -4$ , define a debt of $-3$ and lose 4 times of that debt, which means you gain credit by losing the debt:
$$[-(-3)-(-3)-(-3)-(-3)] = (+3+3+3+3) = (+12)$$
Here is a nice little video about this topic:
At $11$:$38$ He explains the same thing I've just written.
We can also reason from the ring axioms. Not quite so intuitive, but this is the core reason that $-1\cdot -1=1$, so I feel like for completeness' sake, it should be included. We'll quickly do a couple of lemma's that it follows from:
Lemma. We have $-(-a)=a$.
Proof. We know \begin{align} a&=a+0\\ &=a+((-a)+-(-a))\\ &=(a+(-a))+-(-a)\\ &=0+-(-a)\\ &=-(-a) \end{align}
Another lemma. We have $0\cdot a=0$.
Proof. We know \begin{align} a\cdot 0&=a\cdot 0+0\\ &=a\cdot 0+((a\cdot0+-(a\cdot 0))\\ &=((a\cdot 0)+(a\cdot0))+-(a\cdot 0)\\ &=a\cdot(0+0)+-(a\cdot 0)\\ &=a\cdot0+-(a\cdot 0)\\ &=0 \end{align} (This proof can also be tweaked to have $a\cdot0=0=0\cdot a$ in non-commutative rings)
Last lemma. We have $-a=a\cdot -1$.
Proof. We know \begin{align} -a&=-a+0\\ &=-a+(a\cdot 0)\\ &=-a+(a\cdot (1+-1))\\ &=-a+((a\cdot1)+(a\cdot-1))\\ &=(-a+(a\cdot1))+(a\cdot-1)\\ &=(-a+a)+(a\cdot-1)\\ &=0+(a\cdot-1)\\ &=a\cdot-1 \end{align} Now it follows that $-1\cdot -1=-(-1)=1$.
I've often seen this proof/explaination $$\bigg(ab + (-a)b\bigg) + (-a)(-b) = \bigg(ab + (-a)b\bigg) + (-a)(-b)$$ $$\bigg(ab + (-a)b\bigg) + (-a)(-b) = ab + \bigg((-a)b + (-a)(-b)\bigg)$$ $$\bigg(a+(-a)\bigg)b+(-a)(-b) =ab +\bigg(b+(-b)\bigg)(-a)$$ $$0b+(-a)(-b) =ab +0(-a)$$ $$0+(-a)(-b) =ab +0$$ $$(-a)(-b) =ab$$
It is a very interesting question and much deeper than it seems to be when it occurs in a non-commutative setting. But let us come back to children. One possible way to justify the product by $-1$ is to rely on distributivity. Actually, saying that $4 \times 3$ means $4 + 4 + 4$ amounts to define $3$ as $1 + 1 + 1$ and then use $$ 4 \times (1 + 1 + 1) = (4 \times 1) + (4 \times 1) + (4 \times 1). $$ Then of course one also needs $4 \times 1 = 4$ to fully justify this approach. Now, suppose you want to give a meaning to $4 \times (-2)$ while keeping distributivity. You may observe that $$ (4 \times 2) + (4 \times (-2)) = 4 \times (2 + (-2)) = 4 \times 0 = 0. $$ Again, one needs to know that $4 \times 0 = 0$, but children should buy that easily. Anyway, now, the only possibility is to have $4 \times (-2) = - (4 \times 2)$. Last step, $(-4) \times (-2)$. Requiring distributivity forces the equality $$ ((-4) \times (-2)) + 4 \times (-2) = (4 + (-4)) \times (-2) = 0 \times (-2) = - (0 \times 2) = - 0 = 0 $$ And finally, yes, $(-1) \times (-1) = 1$ !
In the XVIII century, an outstanding scientist, mathematician and engineer Leonhard Euler explained generally multiplying negative numbers like this. Clearly,$-5 \cdot 4 =-20.$ Therefore, the product $-5 \cdot (-4)$ can not be equal $-20$, but the product has to be somehow related with the number $20$. There remains only one possibility: $-5 \cdot (-4)= 20$
Therefore $-1 \cdot (-1)=1$
So you start with nothing, and you take away the value $-1$, you are left with $1$.
– 5xum Mar 03 '16 at 10:59