Significance of multiplying $-1$ by $-1$

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 $$\bigg(ab + (-a)b\bigg) + (-a)(-b) = \bigg(ab + (-a)b\bigg) + (-a)(-b)$$ $$\bigg(ab + (-a)b\bigg) + (-a)(-b) = ab + \bigg((-a)b + (-a)(-b)\bigg)$$ $$\bigg(a+(-a)\bigg)b+(-a)(-b) =ab +\bigg(b+(-b)\bigg)(-a)$$ $$0b+(-a)(-b) =ab +0(-a)$$ $$0+(-a)(-b) =ab +0$$ $$(-a)(-b) =ab$$