Who found zero?

Education Journalist | Study Abroad Lead

The concept of zero as a number was independently discovered by ancient civilizations in several parts of the world, including the Babylonians, Mayans, and Indians.

The Indian mathematician and astronomer Brahmagupta is often credited with being one of the first to use zero as a placeholder and to define arithmetic rules for its use in the 7th century.
Brahmagupta was a mathematician and astronomer who lived in India during the Golden Age of Indian mathematics.
Aryabhatta is credited with the concept of using zero in the decimal system.
Brahmagupta is credited for mathematical operations associated with zero such as addition or subtraction.

aryabhata and brahmagupta

The invention of zero revolutionized mathematics, allowing for the development of more advanced mathematical concepts and techniques. Today, zero is considered one of the most important numbers in mathematics and science and is used in numerous fields, including engineering, physics, and computer science.

As Aryabhatta predated Brahmagupta, it is often said that Aryabhatta found zero.

Read More:

Important Concepts
Relation Between HCF and LCM	Decimal expansion of rational numbers	Operations on Rational Numbers
Trigonometric Functions	Number System Conversion	Number System Important Questions
Addition And Subtraction Of Integers	Multiplication And Division Of Integers	Integers As Exponents

CBSE CLASS XII Related Questions

1.
For two vectors \(\vec{a}\) and \(\vec{b}\):

Assertion (A): \[ |\vec{a}\times\vec{b}|^2+(\vec{a}\cdot\vec{b})^2 = |\vec{a}|^2|\vec{b}|^2 \] Reason (R): \[ |\vec{a}\times\vec{b}| = (\vec{a}\cdot\vec{b})\tan\theta, \quad \theta\neq\frac{\pi}{2}. \]

Both Assertion (A) and Reason (R) are true and the Reason (R) is the correct explanation of the Assertion (A).
Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
Assertion (A) is true, but Reason (R) is false.
Assertion (A) is false, but Reason (R) is true.

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2.
In a school, the probability of holding a debate competition is \( \frac{1}{3} \) and that of a quiz competition is \( \frac{2}{3} \). In the two participating teams, A has 4 girls and 6 boys and B has 7 girls and 3 boys. If a debate competition is held, the students are selected from team A and for the quiz competition they are selected from team B. If only two students are to be chosen from the teams, then find the probability that one will be a girl and the other a boy.

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3.
The least value of \[ f(x)=e^{-x} \] in the interval \[ [0,3] \] is:

\( e^{-3} \)
\( -1 \)
\( 1 \)
\( -e^3 \)

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4.
Check whether \[ f:\mathbb{R}-\{3\}\rightarrow\mathbb{R} \] defined as \[ f(x)=\frac{x-2}{x-3} \] is onto or not.

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5.
Find: \[ \int \frac{x^2}{(x^2+9)(x^2+16)}\,dx \]

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6.
For \[ f(x)=x+\frac{1}{x}, \quad x\neq 0. \]

local maximum value is 2
local minimum value is \( -2 \)
local maximum value is \( -2 \)
local minimum value \( < \) local maximum value

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Who found zero?

CBSE CLASS XII Related Questions

1.
For two vectors \(\vec{a}\) and \(\vec{b}\):

Assertion (A): \[ |\vec{a}\times\vec{b}|^2+(\vec{a}\cdot\vec{b})^2 = |\vec{a}|^2|\vec{b}|^2 \] Reason (R): \[ |\vec{a}\times\vec{b}| = (\vec{a}\cdot\vec{b})\tan\theta, \quad \theta\neq\frac{\pi}{2}. \]

3.
The least value of \[ f(x)=e^{-x} \] in the interval \[ [0,3] \] is:

4.
Check whether \[ f:\mathbb{R}-\{3\}\rightarrow\mathbb{R} \] defined as \[ f(x)=\frac{x-2}{x-3} \] is onto or not.

5.
Find: \[ \int \frac{x^2}{(x^2+9)(x^2+16)}\,dx \]

6.
For \[ f(x)=x+\frac{1}{x}, \quad x\neq 0. \]

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Who found zero?

CBSE CLASS XII Related Questions

1.For two vectors \(\vec{a}\) and \(\vec{b}\): Assertion (A): \[ |\vec{a}\times\vec{b}|^2+(\vec{a}\cdot\vec{b})^2 = |\vec{a}|^2|\vec{b}|^2 \] Reason (R): \[ |\vec{a}\times\vec{b}| = (\vec{a}\cdot\vec{b})\tan\theta, \quad \theta\neq\frac{\pi}{2}. \]

3. The least value of \[ f(x)=e^{-x} \] in the interval \[ [0,3] \] is:

4.Check whether \[ f:\mathbb{R}-\{3\}\rightarrow\mathbb{R} \] defined as \[ f(x)=\frac{x-2}{x-3} \] is onto or not.

5. Find: \[ \int \frac{x^2}{(x^2+9)(x^2+16)}\,dx \]

6. For \[ f(x)=x+\frac{1}{x}, \quad x\neq 0. \]

Comments

SUBSCRIBE TO OUR NEWS LETTER

1.
For two vectors \(\vec{a}\) and \(\vec{b}\):

Assertion (A): \[ |\vec{a}\times\vec{b}|^2+(\vec{a}\cdot\vec{b})^2 = |\vec{a}|^2|\vec{b}|^2 \] Reason (R): \[ |\vec{a}\times\vec{b}| = (\vec{a}\cdot\vec{b})\tan\theta, \quad \theta\neq\frac{\pi}{2}. \]

3.
The least value of \[ f(x)=e^{-x} \] in the interval \[ [0,3] \] is:

4.
Check whether \[ f:\mathbb{R}-\{3\}\rightarrow\mathbb{R} \] defined as \[ f(x)=\frac{x-2}{x-3} \] is onto or not.

5.
Find: \[ \int \frac{x^2}{(x^2+9)(x^2+16)}\,dx \]

6.
For \[ f(x)=x+\frac{1}{x}, \quad x\neq 0. \]