One part of math that was drilled into my head when learning negative numbers was that a negative times a negative equals a positive. Somewhat astonishingly, I never learned the actual proof of this until college. The proof requires only a few basic principles of arithmetic.
First, we know that \( 0 \) times any other number is \( 0 \). This is true even when multiplying \( 0 \) and \( -1 \):
$$ 0 * -1 = 0 $$
Next, we know that \( 1 + (-1) \) is \( 0 \). Or if you like the jargon, \( 1 \) is the arithmetic inverse of \( -1 \). Either way, the substitution \( 0 = 1 + (-1) \) can be made in the equation:
$$ (1 + (-1)) * -1 = 0 $$
The distributive law allows this to be rewritten as
$$ (1)*(-1) + (-1)*(-1) = 0 $$
Now, \( 1 \) times any other number is that other number (jargon: \( 1 \) is the multiplicative identity). We can simplify \( (1)*(-1) \) to \( -1 \):
$$ (-1) + (-1)*(-1) = 0 $$
All that remains is to add \( 1 \) to both sides of the equation. This will cancel with the \( -1 \) on the left side of the equation:
$$ 1 + (-1) + (-1)*(-1) = 1 $$
$$ (-1)*(-1) = 1 $$
Therefore, \( -1 \) times \( -1 \) is \( 1 \).
This proof used the following principles:
So if you think \( -1 \) times \( -1 \) is not \( 1 \) but some other value, you must disagree with at least one of the following above principles. It might be a fun thought experiment to think of an arithmetic where some of these rules don't hold true.
I found this proof was powerful to demonstrate how math builds on simple principles to show something else must unquestionably be true.