Irrational Numbers and Irrationality Class 10th

Introduction

Irrational Numbers – The numbers that cannot be written in the form of p/q, are called Irrational Numbers. Where p and q are integers and q ≠ 0. Irrational numbers are denoted by T. Examples – √2, √3, √5, π, etc.

In this section, we shall study the proof of the irrationality of an irrational number with the help of The Fundamental Theorem of Arithmetic. The proof of irrationality is based on a technique which is called proof by contradiction. This is the technique, in which we prove the irrationality by the wrong assumption.

Before we prove irrationality, there is a theorem to be studied, and whose proof is based on the fundamental theorem of arithmetic.

Theorems Based on Irrationality

Theorem 1) If p divides a², then p also divides a. where p is a prime number and a is a positive integer.

Proof – Since a is a positive integer. Let the prime factorization of a be as follows

a = p₁×p₂×p₃…………….p_n

Where p₁, p₂, p₃, …… p_n are prime numbers (may be different or same)

Therefore, a² = (p₁×p₂×p₃…………….p_n)(p₁×p₂×p₃…………….p_n) = p₁²×p₂²×p₃²…………….p_n²

Now, it is given that p divides a². So, by the fundamental theorem of arithmetic, p will be one of the prime factors of a².

But prime factors of a² are p₁, p₂, p₃, …… p_n so p is one of the prime factors p₁, p₂, p₃, …… p_n.

Since a = p₁×p₂×p₃…………….p_n  

Therefore, p divides a.         Hence Proved.

Theorem 2) Prove that √2 is an irrational number.

Proof – Using the contradiction, let √2 be a rational number.

So, for integers a and b, we can write     

√2 = a/b    where b ≠ 0 and a, b = coprime numbers [not any common factor other than 1]

√2b = a

By squaring both sides,

(√2b)² = a²

2b² = a²   

Or  b² = a²/2,  it means 2 divides a² therefore, 2 also divides a [by theorem 1]   ………………..(1)

So, we can write for any integer c, a = 2c   …………………….(2)

Putting the value of a from equation (2) to (1)

We get,    b² = (2c)²/2

b² = 4c²/2

b² = 2c²

Or  b²/2 = c²  it means 2 divides b² therefore, 2 also divides b [by theorem 1]   ………………(3)

From equations (1) and (3), 2 divides a, and 2 divides b this means a and b have at least 2 as a common factor.

But this contradicts (opposes) the fact that a and b have no common factor other than 1.

It shows that our assumption that √2 is a rational number is wrong.

So, we conclude that √2 is an irrational number. Hence Proved.

Let’s have some examples based on this theorem.

Some Examples

Example 1) Prove that √3 is an irrational number.

Solution – Let √3 be a rational number.

So, for integers a and b, we can write,     

√3 = a/b    where b ≠ 0 and a, b = coprime numbers [not any common factor other than 1]

√3b = a

By squaring both sides

(√3b)² = a²

3b² = a²   

Or   b² = a²/3  it means 3 divides a² therefore, 3 also divides a [by theorem 1]   ………………..(1)

So, we can write for any integer c, a =3c    …………………….(2)

Putting the value of a from equation (2) to (1)

We get,     b² = (3c)²/3

b² = 9c²/3

b² = 3c²

Or  b²/3 = c²  it means 3 divides b² therefore, 3 also divides b [by theorem 1]   ………………(3)

From equations (1) and (3), 3 divides a, and 3 divides b this means a and b have at least 3 as a common factor.

But this contradicts (opposes) the fact that a and b have no common factor other than 1.

It shows that our assumption that √3 is a rational number is wrong.

So, we conclude that √3 is an irrational number. Hence Proved.

Example 2) Prove that √5 is an irrational number.

Solution – Let √5 be a rational number.

So, for integers a and b, we can write,     

√5 = a/b    where b ≠ 0 and a, b = coprime numbers [not any common factor other than 1]

√5b = a

By squaring both sides

(√5b)² = a²

5b² = a²   

Or    b² = a²/5  it means 5 divides a² therefore, 5 also divides a [by theorem 1]   ………………..(1)

So, we can write for any integer c, a = 5c    …………………….(2)

Putting the value of a from equation (2) to (1)

We get,     b² = (5c)²/5

b² = 25c²/5

b² = 5c²

Or   b²/5 = c²  it means 5 divides b² therefore, 5 also divides b [by theorem 1]   ………………(3)

From equations (1) and (3), 5 divides a, and 5 divides b this means a and b have at least 5 as a common factor.

But this contradicts (opposes) the fact that a and b have no common factor other than 1.

It shows that our assumption that √5 is a rational number is wrong.

So, we conclude that √5 is an irrational number. Hence Proved.

Note – We know that the addition and subtraction of rational numbers and irrational numbers are also irrational numbers and multiplication and Division of non-zero rational numbers and irrational numbers are also irrational numbers.

Example 3) Prove that 5 + √3 is irrational.

Solution – Let 5 + √3 be a rational number.

So, for integers a and b,      

5 + √3 = a/b where b ≠ 0 and a, b = coprime numbers [not any common factor other than 1]

√3 = a/b – 5

√3 = (a – 5b)/b …………………….(1)

Since a, b, and 5 are integers, so (a – 5b)/b is a rational number.

So, from equation (1), √3 will be a rational number.

But this contradicts the fact that √3 is an irrational number.

It shows that our assumption that 5 + √3 is a rational number is wrong.

So, we conclude that 5 + √3 is irrational. Hence Proved.

Example 4) Prove that 3√5 is irrational.

Solution – Let 3√5 be a rational number.

So, for integers a and b,      

3√5 = a/b where b ≠ 0 and a, b = coprime numbers

√5 = a/3b …………………….(1)

Since a, b, and 3 are integers, so a/3b is a rational number.

So, from equation (1), √5 will be a rational number.

But this contradicts the fact that √5 is an irrational number.

It shows that our assumption that 3√5 is a rational number is wrong.

So, we conclude that 3√5 is irrational. Hence Proved.

Example 5) Show that √3 + √5 is irrational.

Solution – Let √3 + √5 be rational.

So, √3 + √5 = a/b where b ≠ 0 and a, b = coprime numbers

√5 = a/b – √3

Squaring both sides, (√5)² = (a/b – √3)²

5 = a²/b² – 2√3a/b + 3                       [∵ (a – b)² = a² – 2ab + b²]

2√3a/b = a²/b² + 3 – 5

2√3a/b = a²/b² – 2

2√3a/b = (a² – 2b²)/b²

√3 = (a² – 2b²)/b² × b/2a

√3 = (a² – 2b²)/2ab     …………………….(1)

Since a, b, and 2 are integers, so (a² – 2b²)/2ab will be rational.

So, from equation (1), √3 will be a rational number.

But this contradicts the fact that √3 is an irrational number.

It shows that our assumption that √3 + √5 is rational is wrong.

Therefore, √3 + √5 is irrational. Hence Proved.

Irrational Numbers and Irrationality Class 10^th in Hindi

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