Ancient Indian Mathematics is different from Western Mathematics

2x2 in western Maths is adding 2 two times

2x2 in Ancient Maths is adding 2 two times to Zero

Now many will say what's the difference but it is

1/n

Western copied maths from Indians but could never understand the original concept

Most significant is Zero and it's uses

Western took zero as only place holder but in Indian Mathematics it was a number

Brahmgupta first to give concepts for uses of Zero and Negative Numbers
2/n

He defined Zero and Negative Numbers

He was first to say that product of two negative numbers is positive

In western Maths
-5<-1

In Indian Maths
-5>-1

West understood negative numbers as less than zero and lost the originality of concept

3/n

But Brahmgupta defined them As example if a protrusion of 1 meter is positive than depth of 1 meter is negative

How a 5 meter depth is less than 1 meter depth

He defined negative numbers as opposite entity of positive number not as numbers less than zero

4/n

Now you know why product of two negative numbers is a positive

And

Original Concept of calculations with negative numbers

Consider term

(−1)·(−1) + (−1)

It is true that

[a·b + a] = a·[b +1]

In above term
a=(−1) and b=(−1)

(b+1)= 0

and

a·[b +1]=0

5/n

But if

(−1)·(−1) + (−1) = 0
then
(−1)·(−1) = +1

Then in general for a and b positive

(−a)·(−b) = (−1)·(−1)a·b = a·b

6/n

The proper way to think of '−a' is as the additive inverse of 'a'

whose sum with 'a' gives additive identity zero

So product of additive inverses of a and b is product of a and b

Additive inverse of additive inverse of 'a' is 'a'

−(−a) is 'a'

7/n

Furthermore (−a) = (−1)a

Thus −(−a) = (−1)(−1)a

But −(−a) = a

hence (−1)(−1)a=a

and therefore (−1)(−1)=1

8/n

Multiplicative identity as well additive identity and each element has an additive inverse

Multiplication is distributive over addition
x(y+z)=xy+xz

Product of additive inverses of two elements of with a multiplicative identity is equal to product of elements
(-y)(-z)=yz

9/n

Let 0 be additive identity

y·0=0

y+y·0=y·1 + y·0 = y(1 + 0) = y·1 = y
Thus for any y
y+y·0=y

which means that y·0 is the same as additive identity 0

Define x=(-y)(-z) + y(-z) + yz

factoring first two terms on right

x = [(-y) + y](-z) + yz
[(-y) + y]=0
0(-z)=0
so x=yz

10/n

But the last two terms can be factored

x=(-y)(-z) + y(-z) + yz
x=(-y)(-z) + y[(-z) + z]
[(-z) + z]=0
and y0=0
so
x = (-y)(-z)

Since x=yz and x=(-y)(-z)

(-y)(-z) = yz

Note that this not only covers

(-1)(-1)=1·

But also that

(-1)1=(-1)

and

1(-1)=(-1)

11/n

@Dharma_Yoddhaa

@rightwingchora @VedicWisdom1 @vedicvishal @TIinExile @monicanitin @ShefVaidya @jcrabtree @mathematicsprof @Aabhas24