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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
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
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
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
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
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
(−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
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
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
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
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
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
