Write the natural numbers from $102$ to$113$. What fraction of them are prime numbers?

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Hint:Write the set of natural numbers from $102$ to $113$ and we have to choose a set of prime numbers from $102$ to $113$ and calculate what fraction of them are prime numbers.

Natural numbers start with $1$ and $2,{\text{ }}3,{\text{ }}4$…etc., are called natural numbers
Here we have to write from $102$ to $113$ and prime numbers from $102$ to$113$.
The number is said to be a prime number, it has only two divisors one is $1$ and another is itself.
E.g. $2,3,5,7,11,..$ etc., these are the prime number, it has only two divisors.
The number which is not prime is called composite numbers (except $1,{\text{ }}1$ is neither prime nor composite).

First the set of natural numbers from $102$ to $113$ is
$\{ 102,103,104,105,106,107,108,109,110,111,112,113\}$
Then we split into two sets, they are composite and prime numbers.
Composite numbers$= \{ 102,104,105,106,108,110,111,112\}$
Prime number$= \{ 103,107,109,113\}$
Consider the prime numbers set, it has $4$ elements (numbers)
Also, the set of natural number from $102$ to $113$ has $12$ elements (numbers)
Then, we get there are$\;4$ numbers are prime numbers out of $12$ numbers.
Fraction of prime numbers is equal to the total number of prime numbers from $102$ to $113$ divided by the total numbers of natural numbers.
Number of prime numbers is $4$; total number of numbers is $12$
Then, we get fraction of prime number $= \dfrac{4}{{12}}$
On Simplification,
We get fraction of prime number $= \dfrac{1}{3}$
Hence, $\dfrac{1}{3}$ fraction of them is prime numbers.

Note:Sometimes, we may think every odd number is a prime number, and every number ending with $1$ or $7$ are prime numbers but it is not true. Every odd number is not a prime number but every prime number is odd except the number $2$. Also, $2$ is only a prime number which is an even number, all other even numbers are divisible by $2$.