This is a formal justification, from the ring axioms, of the formula (−a)(−b) = ab. As the article mentions, this is often phrased as "the product of two negatives is positive," but, of course, the presence of a minus sign in front of a variable does not indicate a negative number (for example, if a = −3, then −a is positive); and the formula makes sense even in a ring with no notion of positive and negative numbers.
I am not sure how you can prove this more "quickly". Trying to do it any more quickly involves claiming some result (no matter how trivial) that is not directly present in the ring axioms. But the whole point of this post is to derive everything strictly from first principles, using nothing beyond the ring axioms themselves.
Here is your argument elaborated step by step.
STEP 1: First we want to show that ab is the additive inverse of (-a)b. This is Theorem 3 of the post.
STEP 2: Next we want to show that (-a)(-b) is the additive inverse of (-a)b. This follows similarly to the proof of Theorem 3: (-a)(-b) + (-a)(b) = (-a)(-b + b) = (-a)(0) and (-a)(0) = 0 by Theorem 2 of the post.
But nothing in the ring axioms directly says that the above results mean ab and (-a)(-b) must be equal. How do we know for sure that ab and (-a)(-b) are not two distinct additive inverses of (-a)b?
THEOREM 5: We now prove the uniqueness of additive inverse of an element from the ring axioms. Let b and c both be additive inverses of a. Therefore b = b + 0 = b + (a + c) = (b + a) + c = 0 + c = c.
Now from Steps 1 and 2, and Theorem 5, it follows that ab = (-a)(-b).
So what did we save in terms of intermediate theorems? Nothing! We no longer need Theorem 1 (inverse of inverse) of the post. But now we introduced Theorem 5 (uniqueness of additive inverse). We have exactly the same number of intermediate theorems with your approach.
This is a formal justification, from the ring axioms, of the formula (−a)(−b) = ab. As the article mentions, this is often phrased as "the product of two negatives is positive," but, of course, the presence of a minus sign in front of a variable does not indicate a negative number (for example, if a = −3, then −a is positive); and the formula makes sense even in a ring with no notion of positive and negative numbers.
A simple example of how this is true _even if you don't have negative numbers_:
Let's use mod 5 arithmetic. You have 5 elements in the ring -- 0,1,2,3,4
The additive inverses are as follows:
Which is to say that 1 is the additive inverse of 4 and 2 is the additive inverse of 3, and vice versa. 0 is the identity, of course.So what happens if you multiply 2 * -3 (2 times the additive inverse of 3).
The additive inverse of 3 is just 2, so the answer is 2 * -3 = 2 * 2 = 4.
The other way to calculate it is to find the additive inverse of the product:
2 * -3 = -(2 * 3) = -(1) which is the additive inverse of 1: 4 again.
ab and (-a)(-b) can each be quickly proved to be the additive inverse of (-a)b. So they equal each other. No intermediate theorems are really needed.
I am not sure how you can prove this more "quickly". Trying to do it any more quickly involves claiming some result (no matter how trivial) that is not directly present in the ring axioms. But the whole point of this post is to derive everything strictly from first principles, using nothing beyond the ring axioms themselves.
Here is your argument elaborated step by step.
STEP 1: First we want to show that ab is the additive inverse of (-a)b. This is Theorem 3 of the post.
STEP 2: Next we want to show that (-a)(-b) is the additive inverse of (-a)b. This follows similarly to the proof of Theorem 3: (-a)(-b) + (-a)(b) = (-a)(-b + b) = (-a)(0) and (-a)(0) = 0 by Theorem 2 of the post.
But nothing in the ring axioms directly says that the above results mean ab and (-a)(-b) must be equal. How do we know for sure that ab and (-a)(-b) are not two distinct additive inverses of (-a)b?
THEOREM 5: We now prove the uniqueness of additive inverse of an element from the ring axioms. Let b and c both be additive inverses of a. Therefore b = b + 0 = b + (a + c) = (b + a) + c = 0 + c = c.
Now from Steps 1 and 2, and Theorem 5, it follows that ab = (-a)(-b).
So what did we save in terms of intermediate theorems? Nothing! We no longer need Theorem 1 (inverse of inverse) of the post. But now we introduced Theorem 5 (uniqueness of additive inverse). We have exactly the same number of intermediate theorems with your approach.