Friday, December 27, 2024
Google search engine
HomeData Modelling & AIWhy do we check up to the square root of a number...

Why do we check up to the square root of a number to determine if that number is Prime?

Why do we check up to the square root of a number to determine if that number is prime?

Prime numbers are natural numbers greater than 1 which are divisible only by 1 and itself. Examples include 2, 3, 5, 7, 11, 13, etc. (1 is neither prime nor composite).

Here is the code to check whether a number is prime or not.

As, the loop checks for each number until the sqrt(N), where N is the number to be checked, to be a factor of N. The reason for checking up to sqrt(n) only is explained as follows:-

Consider a natural number N (greater than 1) which exists as a product of two numbers a and b (assume a<=b) that is 

  • N = a * b  where, 1 < a ≤ b < n

By multiplying the relation a ≤ b by a, the relation becomes:

  • a2 ≤ ab     

By multiplying the relation a ≤ b by b, the relation becomes: 

  • ab ≤ b2

From the above-mentioned inequalities, the relation becomes: 

  • a2 ≤ ab ≤ b2

We know that N = a * b, so by replacing a * b with N, the following relation is obtained:

  • a2 ≤ N ≤ b2

Taking the square root of the equation, considering both a and b as positive numbers, we get:

  • a ≤ sqrt(N) ≤ b

So, according to the above relation, it can be concluded that one of the factors of a non-prime number will definitely be less than or equal to sqrt(N).

So, while checking from 2 to the sqrt(N), if we find a number that is a factor of the N, it would mean that the number has more than 2 factors, so that number would not be a prime number. 

Otherwise, if we do not find any factor of N, that implies that the N has only 2 factors, 1 and itself, thus it is a prime number.

Examples:

Example: N = 16

√16 = 4
We can factorize 16 in pairs as : (1, 16), (2, 8), (4, 4), (8, 2), (16, 1)
In every pair, it can be observed that one factor is less than the square root.

Example: N = 36

√36 = 6
We can factorize 36 in pairs as : (1, 36), (2, 18), (3, 12), (4, 9), (6, 6), (9, 4), (12, 3), (18, 2), (36, 1)
Again, it can be observed that one factor in every pair is less than the square root of n.

Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!

RELATED ARTICLES

Most Popular

Recent Comments