Negative multiplied by negative is positive.

This is usually taught to us in school after our knowledge about numbers expands to include the negative numbers. Most of us take this fact for granted. I took it on faith as well.

One of my pastimes is to ponder on these facts that I once learnt by faith and observable evidence rather than reason and logic. I see whether I can see these facts in a new light. A few years ago, I thought over this one as well and proving this fact was easy.

First let me provide an intuitive understanding of why it should be so. In the discussion to follow, we will assume that we already know some basic properties like distributive property, a × (−b) = −(a × b), etc. because I don't want to prove each and every thing involved here right from the Dedekind–Peano axioms.

We know that 7 × 8 = 56. Let us express 7 as (10 − 3) and 8 as (10 − 2).

So, the following must be true.

(10 − 3) × (10 − 2) = 56

Using the distributive property of multiplication over addition, we get,

10 × 10 + (−3) × 10 + 10 × (−2) + (−3) × (−2) = 56

We already know that −3 × 10 = −30 and 10 × −2 = −20. But we do not know what (−3) × (−2) is. So, let us see what we get by using what we know.

10 × 10 + (−3) × 10 + 10 × (−2) + (−3) × (−2) = 56

⇔ 100 + (−30) + (−20) + (−2) × (−3) = 56

⇔ 50 + (−2) × (−3) = 56

⇔ (−2) × (−3) = 6

We see that (10 − 3) × (10 − 2) = 56 is true if and only if (−2) × (−3) is considered as 6.

However this is not a proper proof. One may ask whether this rule is true for all such calculations even when the numbers are different. Can we prove that −a × −b = a × b is absolutely necessary for all real numbers a and b? So, let me try a proper algebraic proof.

Let a and b be two positive real numbers. We know that a − a = 0.

Multiplying both sides by −b, we get,

(a − a) × (−b) = 0 × (−b)

Again, using the distributive property of multiplication over addition, we get,

a × (−b) + (−a) × (−b) = 0.

We know that a × (−b) = −(a × b), but we do not know what (−a) × (−b) is. Using what we know, we get,

−(a × b) + (−a) * (−b) = 0.

Adding (a × b) to both sides, we get,

(−a) × (−b) = a × b

4 comments

Suni said:

hmmm... this time i really understood something. :)

Prunthaban said:

Nice proof. So now we should be able to talk in a very generic sense. -a is the inverse element of addition. So from your proof it looks like in a field the multiplication of 2 values is equal to the multiplication of their inverse elements under addition. That makes your proof independent of anything and assumes only that the system in question is a field (assuming your proof is not using anything other than the 9 field axioms).

Susam Pal said:

Prunthaban,

That is an interesting way to look at it. We can generalize it further. We neither need commutativity of multiplication in our proof nor do we need existence of multiplicative inverse. So, our proof holds good for elements of any ring as well.

Anirudh said:

Nice Blog! You might also like my post: http://hypocritecat.blogspot.com/2010/05/on2-to-on-sum-of-squares-of-differences_6569.html

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