Question

Write a note for Indian mathematics Brahmaputra​

Answer 1

Answer:

Brahmagupta ( c. 598 – c. 668 CE) was an Indian mathematician and astronomer. Brahmagupta was the first to give rules to compute with zero. Brahmagupta was born in 598 CE according to his own statement.

Explanation:

brahmputra nhi brahmagupta

Answer 2

Explanation:

The Indian mathematician Brahmagupta (597-667 CE) appears to be the first to articulate the result that the product of two negative numbers is a positive number. He of course also gave the much easier result that the product of a positive number and a negative number is a negative number.

Brahmagupta was the greatest mathematician of his time and wrote on more advanced topics than arithmetic such as geometry, algebra and astronomy. He gave the solution to a general linear equation and to a general quadratic equation. But he also gave the rules for arithmetic operations involving zero. In fact, he was probably the first to treat zero as a number in its own right. However Brahmagupta made one mistake. He said zero divided by zero is always zero.

Brahmagupta was born and grew up in western India. In his time discovering and proving that the product of two negative numbers is a positive number was quite an intellectual achievement. Even today most people remember the product of negative being positive as a rule but do not have the least idea of how to prove it.

Consider a term of the form

(−1)·(−1) + ((−1)

It is true that [a·b + a] is equal to a·[b +1]. In the above term a=(−1) and b=(−1). This means that (b+1) is equal to zero and thus a·[b +1 is equal to zero. 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

Brahmagupta did not give a proof for the product of negative being positive.

The proper way to think of −a is as the additive inverse of a; i.e., the number whose sum with a gives the additive identity, zero. So the product of the additive inverses of a and b is the product of a and b. Note that the additive inverse of the additive inverse of a is a; i.e., −(−a) is a. Furthermore (−a) is equal to (−1)a. Thus −(−a) is equivalent to (−1)(−1)a. But −(−a) is equal to a and hence (−1)(−1)a=a and therefore (−1)(−1)=1.

A Generalization

A ring with a multiplicative identity is an algebraic structure possessing two associative operations analogous to addition and muliplication. Addition is commutative. Such a ring has a multiplicative identity as well an additive identity and each element has an additive inverse. Multiplication is distributive over addition; i.e., x(y+z)=xy+xz.

Proposition: The product of the additive inverses of two elements of a ring with a multiplicative identity is equal to the product of the elements: i.e., (-y)(-z)=yz.

Proof:

Let 0 be the additive identity. Note that y·0=0 because 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 the additive identity 0.s Likewise 0·y=0 because y+0·y=(1+0)y=1·y=y. Thus for any y, y+0·y=y so 0·y is the same as the additive identity 0.

Define x=(-y)(-z) + y(-z) + yz. The factoring the first two terms on the right gives

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

but [(-y) + y]=0

0(-z)=0

so x=yz

But the last two terms can also be factored

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

but [(-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)=1·1 but also that (-1)1=(-1) and 1(-1)=(-1).

All of this holds for fields, which are necessarily rings with a multiplicative identity.