Multiplication of two negative numbers

 Multiplication of two negative numbers

Right at the start – a disclaimer! The idea of representing a number belonging to number set  ℤ  (ℤ = {…, -3, -2, -1, 0, 1, 2, 3, …})  as a tuple (a, b) where both a and b belong to Natural numbers  W = {0, 1, 2, 3, 4, …} is not mine! I had read it long ago in some maths related forum and the idea stuck with me. In this notation a number  k ∈ ℤ is represented as (a, b) a,b ∈ W and value of k = b – a.

For this discussion below let’s call this notation P system.  Thus a number in P system

a  = (aL, aR) where aL, aR ∈ W and value of a is (aR – aL).

(For the purpose of readability, I have set font color to RED for the number representation is system P)

The idea is to show that multiplication of two negative numbers is a positive number.

In P system, a positive number is one where aR > aL. The number is zero if aR = aL and number is negative if  aR < aL. Thus aLl numbers in ℤ can be represented by number system P I want to show that multiplication of two negative numbers in ℤ is a positive number in ℤ.

For the purpose of this discussion. I will use additive and multiplicative identity axioms.

Additive identity axiom is:  a + 0 = a.

Multiplicative identity axiom is:  a x 1 = a

Multiplication -1 x -1 = 1 is not an axiom and that is what I want to derive.

Let’s start with properties of number representation in number system P.

 

Property-1: Value of number in P representation does not change if both components of tuple aRe added the same number from ℤ.

That is:   (aL, aR) = (aL + m, aR +m) provided both aL+m and aR+m belong to W i.e. both numbers are positive numbers.  

Addition of two numbers in P:

Let A and B be two numbers in P representation – A is (aL, aR) and B is (bL, bR).

A + B = (aL, aR) + (bL, bR).

A + B = (aL+bL,  aR+bR)

8 + 3 = (2, 10) + (1, 4) = (2+1, 10+4) = (3, 14) = 11

8 + 0 = (2, 10) + (3, 3) = (2+3, 10+3) = (5, 13) = 8

Subtraction of two numbers:

A – B = (aL, aR) – (bL, bR) = (aL – bL, aR – bR).

However, in P representation – the components of the tuple aRe expected to belong to W.  And (aL – bL) or (aR – bR) – can be a number not belonging to W.  

But using the Property-1 – we can rewrite the expression above as:

A – B = (aL, aR) – (bL, bR) = (aL+bR, aR+bL)

(For example: 8 – 3 = (2, 10) – (1, 4)

= (2+4, 10+1)

= (6, 11)

= 5.

Or another 3 – 8 = (1, 4) – (2, 10)

= (1+10, 4+2)

= (11, 6)

= -5

Or another 8 – 0 = (2, 10) – (4, 4)

= (2+4, 10+4)

= (6, 14)

= 14

Or 0 – 8 = (4, 4) – (2, 10)

= (4+10, 4+2)

= (14, 6)

= -8

ALso -8 = (14, 6) = (14-4, 6-4)

= (10, 2)

Thus if A = (aL, aR) then  -A = (aR, aL)

 

Now let’s derive Multiplication:

m . n = m + m .(n - 1)

          = m + m + m .(n -2) Until (n-2) ∈ W

… Thus it is to add number m n times.

Hence, A . B = (aL, aR). (bL, bR)

= (aL(bL, bR), aR(bL, bR

= (aL (bR – bL), aR(bR-bL))

= (aL.bR-aL.bL, aR.bR-aR.bL)

readjusting to ensurebelonginness of both terms of the tuple to  W

= (aL.bR+aR.bL, aR.bR+aL.bL) .. Now both terms belong to W

= ((aR.bR+aL.bL) – (aL.bR+aR.bL))

Let’s check.

              8 x 3 = (2, 10) x (3, 6)

= (2x6+10x3, 10x6+3x2)

= (12+30, 60+6)

= (42, 66)

= 24

Another example:

1 x -1 = (5, 4) x (4, 5)

= (5x5+4x4, 4x5+5x4)

= (41, 40)

= -1

Another:

A x 0 = (aL, aR) x (a, a)

= (aL.a+aR.a, aR.a+aL.a) è Both terms of the tuple are identical – hence:

= 0.

A x 1 = (aL, aR) (a, a+1)

= (aL.(a+1)+aR.a, aL.a+aR.(a+1))

= (aL.a+aL+aR.a, aL.a+aR.a+aR)

Removing common factor from both terms of the tuple:

= (aL, aR)  è Same as A we started with

And finally:

-1 x -1 = (4, 5) x (3, 2)

 = (4x2+5x3, 5x2+4x3)

= (21, 22)

= +1

And generically:

-1 x -1 = (a+1, a) x (b+1, b)

     = ((a+1)b+a(b+1), ab+(a+1)(b+1))

     = (ab+b+ab+a, ab+ab+a+b+1)

     = (ab+ab+a+b+1)-(ab+b+ab+a)

= 1

Thus without making assumption ever that -1 x -1 = +1 we get the desired result.

 This is NOT a rigorous mathematical proof – just a practical one!

Abhijit Tongaonkar/ 6 Aug 2026