Prime Numbers: Overview, Questions, Preparation

A prime number is a natural number greater than 1, having only two factors, i.e., 1 and the integer itself.

For example, 7 ( as 7/1 =7, 7/7= 1 and it doesn’t have any other factor)

How to check for Prime number?

The simplest way to check whether a number is prime or not is the trial method, i.e., check whether n is a multiple of any integer between 2 and √n. If yes, then n is not a prime number. Otherwise, it’s a prime number.

Prime Factorisation

To write a number as a product of prime numbers is called prime factorisation.

Example: 594 = 2×3×3×3×11

                           = 2×3^3×11

The Fundamental Theorem of Arithmetic

The theorem states that every composite number can be expressed (factorised) as a prime product. This factorisation is unique, apart from the order in which the prime factors occur.

Example: 32760 = 2×2×2×3×3×5×7×13

                                = 2^3×3^2×5×7×13

The chapter Real numbers include a detailed explanation of the Fundamental Theorem of Arithmetic. The chapter covers it’s statement, examples, applications, and related questions. 

It also includes some other theorems related to prime numbers.

For the year 2021, the Prime number topic in Class 10 would hold the weightage of 4 marks.

1. Find the HCF and LCM of 12, 15, and 21 by prime factorisation method.

Solution.

We have , 

                12 = 2^2 × 3

                15 = 3 × 5

                21 = 3 × 7

3^1 is the smallest power of common factor 3

So, 

    HCF (12,15,21) = 3

2^2,3,5,7are the greatest powers of prime factors 2,3,5,7 respectively involved in 12, 15 ,21

So, 

   LCM (12,15,21) = 2^2×3×5×7

                              = 420

2. Prove that √5 is irrational.

Solution.

Let us assume that √5 is rational.

So, we can find integers r and s (not equal to 0) such that √5 = r/s

Suppose r and s have a common factor other than 1.

Then we divide by common factor to get 

                    √5 = a/b, where a and b are coprime

So,

       b√5 = a

Squaring on both sides,

       (b√5)^2 = a^2

        5b^2 = a^2              -------(1)

This means,

              5 divides a^2

5 divides a         .........By theorem ( Let p be prime, p divides a^2 => p divides a)

So,

       a = 5c          for some integer c

Squaring on both sides,

a^2 = (5c)^2

a^2 = 25 c^2
b^2 = 5 c^2
5 divides b^2
5 divides b              By theorem explained above

Therefore, a and b holds at least 5 as a common factor.

But this opposes the fact that a and b have no common factors other than 1.

So, our assumption is wrong

√5 is irrational.

3. Express each number as a product of its prime factors:

Solution.

Finding the LCM of 140, we will get the product of its prime factor.

Therefore, 140 = 2 × 2 × 5 × 7 × 1 = 22×5×7.

Q: What are the Prime numbers between 1 to 100?

Q: What if a Natural number is not Prime?

Q: What are the smallest even and odd Prime Numbers?

Q: Who gives the proof of the Fundamental Theorem of Arithmetic?

Q: What is the application of the Fundamental theorem of Arithmetic?