The Smallest Prime Number Is

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The smallest prime number is 2. Prime numbers are positive integers greater than 1 that have exactly two positive divisors, 1 and the number itself. In other words, a prime number is a number that is only divisible by 1 and itself.

Examples of other prime numbers are 3, 5, 7, 11, and 13. Whereas, 4, 6, 8, and 9 are not prime because they have other divisors in addition to 1 and themselves. Prime numbers play an essential role in number theory and have many applications in cryptography and other areas of mathematics.

Also check: Prime Numbers from 1 to 1000

How to Determine A Prime Number?

To determine if a number is prime, you can use several methods, including trial division, the Sieve of Eratosthenes, and probabilistic algorithms.

  • Trial Division involves dividing the number in question by each integer greater than 1 and less than or equal to the square root of the number. If the number is prime, then it will not have any divisors other than 1 and itself.
  • Sieve of Eratosthenes is an algorithm for finding all prime numbers up to a given limit. It works by marking all multiples of each prime number as composite (not prime), and then repeatedly marking the next smallest unmarked number as prime.
  • Probabilistic algorithms use statistical methods to determine if a number is likely to be prime. These methods can be faster than trial division or the Sieve of Eratosthenes, but they are not guaranteed to always give the correct answer.

In summary, finding prime numbers is a question of determining the factors of a given number. There are several methods for doing this, each with its own trade-offs in terms of accuracy, speed, and complexity.

Also check:

CBSE CLASS XII Related Questions

  • 1.
    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

    • 2.
      A function \[ f:\mathbb{R}-\left\{\frac{3}{5}\right\} \to \mathbb{R}-\left\{\frac{3}{5}\right\} \] is defined as \[ f(x)=\frac{3x+2}{5x-3}. \] Show that \(f\) is one-one and onto.


        • 3.

          If \[ B(\operatorname{adj} B)= \begin{bmatrix} \frac{1}{3} & 0 & 0\\ 0 & \frac{1}{3} & 0\\ 0 & 0 & \frac{1}{3} \end{bmatrix}, \] then the value of \[ \det(B^{-1}) \] is: 

            • \(\frac{1}{3}\)
            • \(\frac{1}{9}\)
            • \(3\)
            • \(9\)

          • 4.

            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.

            • 5.
              Assertion (A) : In an experiment of throwing an unbiased die, the probability of getting a prime number given that number appearing on the die being odd is \( \frac{2}{3} \).
              Reason (R) : For any two events \( A \) and \( B \), \( P(A|B) = \frac{P(A \cup B)}{P(B)} \).

                • 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 and Reason (R) is false.
                • Assertion (A) is false and Reason (R) is true.

              • 6.

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

                  CBSE CLASS XII Previous Year Papers

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