Rational and Irrational Numbers both real numbers and both of them have different properties. A rational number is a real number that can be expressed in the form of simple fraction p/q, whereas an irrational number is also a real number but it cannot be expressed in the form p/q, where p and q both are integers, and q not equal to ‘0’.

The numbers having non-terminating and non-repeating decimal representations are also known as irrational numbers. Irrational numbers include numbers such as √2, ????, e (Euler’s number). In this article we would be delving into the practice Questionss related to rational numbers.

Properties of Irrational Numbers

Some of the common properties of irrational numbers are:

• Addition: Addition of irrational numbers would be same as that of normal addition. Example: √4 + √6 = 2 + √6 (as √4 = 2).
• Subtraction: Subtraction option would be same as that of normal arithmetic subtraction. Example: √4 – √7 = 2 – √7 (as √4 = 2).
• Multiplication: For the multiplication of irrational numbers one needs to multiply the radicals. Example: √a + √b = √ab
• Division: Division of irrational numbers involves rationalizing the denominator. Example: √4/√6 = (√4/√6) * (√6/√6) = √24/6 = (2√6)/6.

Solved Questions Irrational numbers

Question 1: Prove that √2 is irrational.

Solution:

Initially assuming that √2 is rational. Then √2 = p/q, where p and q both are integers having no common factors.

Now, squaring on both the sides,

[Tex]\therefore 2 = \frac{p^2}{q^2}[/Tex]

∴ 2q² = p²

This shows that p² is even which means that p is also even. So, let p = 2k

∴ 2q² = (2k)² = 4k²

∴ q² = 2k²

This also signifies that q is also even. This both the statements contradicts the definition that p and q have no common factors. So, we conclude that √2 is irrational.

Question 2: Prove that √3 is irrational.

Solution:

Initially assuming that √3 is rational. Then √3 = p/q, where p and q both are integers having no common factors.

Now, squaring on both the sides,

[Tex]\therefore 3 = \frac{p^2}{q^2}[/Tex]

∴ 3q² = p²

This shows that p2 is divisible by 3 which also means that p is also divisible by 3. So, let p = 3k

∴ 3q² = (3k)² = 9k²

∴ q² = 3k²

This also signifies that q is also divisible by 3. This both the statements contradicts the definition that p and q have no common factors. So, we conclude that √3 is irrational.

Question 3: Simplify [Tex]\sqrt{7+4\sqrt{3}}[/Tex].

Solution:

Let us consider [Tex]\sqrt{7+4\sqrt{3}} = \sqrt{a} + \sqrt{b}[/Tex].

Now, squaring on both the sides,

[Tex]\therefore 7 + 4\sqrt{3} = a+b+2\sqrt{ab}[/Tex]

On equating expressions on both the sides, we get

a + b = 7 and [Tex]2\sqrt{ab} = 4\sqrt{3} [/Tex]

So, [Tex] \sqrt{ab} = 2\sqrt{3}[/Tex]

∴ ab = 12

Solving for a and b

a = 3, and b = 4 or a = 4, and b = 3

Now, substituting the value of a and b

[Tex]\therefore \sqrt{7+4\sqrt{3}} = \sqrt{a} + \sqrt{b} = \sqrt{3} + 2[/Tex]

Question 4: Simplify [Tex]\sqrt{9 + 2\sqrt{20}} – \sqrt{9 – 2\sqrt{20}}[/Tex]

Solution:

Let us consider [Tex]\sqrt{9 + 2\sqrt{20}} = \sqrt{a} + \sqrt{b}[/Tex] and [Tex]\sqrt{9 – 2\sqrt{20}} = \sqrt{a} – \sqrt{b}[/Tex]

Now, squaring on both the sides for both the equations

[Tex]\therefore 9 + 2\sqrt{20} = a + b + 2\sqrt{ab}[/Tex]

Also, [Tex] 9 – 2\sqrt{20} = a + b – 2\sqrt{ab}[/Tex]

So, on comparing both the equations we get:

[Tex]a + b = 9 and 2\sqrt{ab} = 2\sqrt{20}[/Tex]

So, [Tex]\sqrt{ab} = \sqrt{20}[/Tex]

∴ ab = 20

Now, solving for a and b we get:

a = 5, and b = 4 or a = 4, and b = 5

[Tex]\therefore \sqrt{9 + 2\sqrt{20}} – \sqrt{9 – 2\sqrt{20}}= \sqrt{a} + \sqrt{b} – (\sqrt{a} – \sqrt{b})[/Tex]

[Tex]\therefore \sqrt{9 + 2\sqrt{20}} – \sqrt{9 – 2\sqrt{20}}= 2\sqrt{b}[/Tex]

Substituting the value of b,

[Tex]\therefore \sqrt{9 + 2\sqrt{20}} – \sqrt{9 – 2\sqrt{20}}= 2\sqrt{4} = 4[/Tex]

or [Tex]\therefore \sqrt{9 + 2\sqrt{20}} – \sqrt{9 – 2\sqrt{20}}= 2\sqrt{5}[/Tex]

Question 5: Simplify [Tex]\sqrt{6+4\sqrt{2}}[/Tex].

Solution:

Let us consider [Tex]\sqrt{6+4\sqrt{2}} = \sqrt{a} + \sqrt{b}[/Tex].

Now, squaring on both the sides,

[Tex]\therefore 6 + 4\sqrt{2} = a+b+2\sqrt{ab}[/Tex]

On equating expressions on both the sides, we get

a + b = 6 and [Tex]2\sqrt{ab} = 4\sqrt{2} [/Tex]

So, [Tex]\sqrt{ab} = 2\sqrt{2}[/Tex]

∴ ab = 8

Solving for a and b

a = 4, and b = 2 or a = 2, and b = 4

Now, substituting the value of a and b

[Tex]\therefore \sqrt{6+4\sqrt{2}} = \sqrt{a} + \sqrt{b} = \sqrt{2} + 2[/Tex]

Question 6: Simplify [Tex](\sqrt{5}+\sqrt{2})\cdot(\sqrt{5}-\sqrt{2})[/Tex]

Solution:

Multiplying the terms

[Tex]\therefore \sqrt{5}\cdot(\sqrt{5}+\sqrt{2})- \sqrt{2}\cdot(\sqrt{5}+\sqrt{2})[/Tex]

[Tex]\therefore (\sqrt{5})^2 + \sqrt{10} – (\sqrt{10}+(\sqrt{2})^2)[/Tex]

[Tex]\therefore (\sqrt{5})^2 – (\sqrt{2})^2[/Tex]

∴ 5 – 2 = 3

Question 7: Prove that √5 is irrational.

Solution:

Initially assuming that √5 is rational. Then √5 = p/q, where p and q both are integers having no common factors.

Now, squaring on both the sides,

[Tex]\therefore 5 = \frac{p^2}{q^2}[/Tex]

∴ 5q² = p²

This shows that p² is divisible by 5 which also means that p is also divisible by 5. So, let p = 5k

∴ 5q² = (5k)² = 25k²

∴ q² = 5k²

This also signifies that q is also divisible by 5. This both the statements contradicts the definition that p and q have no common factors. So, we conclude that √5 is irrational.

Question 8: Prove that √19 is irrational.

Solution:

Initially assuming that √19 is rational. Then √19 = p/q, where p and q both are integers having no common factors.

Now, squaring on both the sides,

[Tex]\therefore 19 = \frac{p^2}{q^2}[/Tex]

∴ 19q² = p²

This shows that p2 is divisible by 3 which also means that p is also divisible by 3. So, let p = 19k

∴ 19q² = (19k)² = 361.k²

∴ q² = 19k²

This also signifies that q is also divisible by 19. This both the statements contradicts the definition that p and q have no common factors. So, we conclude that √19 is irrational.

Question 9: Simplify [Tex](\sqrt{5}+\sqrt{2})\cdot(\sqrt{7}-\sqrt{2})[/Tex]

Solution:

Multiplying the terms

[Tex]\therefore \sqrt{7}\cdot(\sqrt{5}+\sqrt{2})- \sqrt{2}\cdot(\sqrt{5}+\sqrt{2})[/Tex]

[Tex]\therefore \sqrt{35}+\sqrt{14}-(\sqrt{10}+\sqrt{4})[/Tex]

[Tex]\therefore \sqrt{35}+\sqrt{14}-\sqrt{10}-2[/Tex]

Question 10: Simplify [Tex]\sqrt{20-2\sqrt{19}}.[/Tex]

Solution:

Let us consider [Tex]\sqrt{20-2\sqrt{19}} = \sqrt{a} – \sqrt{b}[/Tex].

Now, squaring on both the sides,

[Tex]\therefore 20 – 2\sqrt{19} = a+b-2\sqrt{ab}[/Tex]

On equating expressions on both the sides, we get

a + b = 20 and [Tex]2\sqrt{ab} = 2\sqrt{19} [/Tex]

So, [Tex]\sqrt{ab} = \sqrt{19}[/Tex]

∴ ab = 19

Solving for a and b

a = 19, and b = 1

(As the resulting expression could not equate to negative so a ≠ 1 and b ≠ 19)

Now, substituting the value of a and b

[Tex]\therefore \sqrt{20-2\sqrt{19}} = \sqrt{a} – \sqrt{b} = \sqrt{19} – 1[/Tex]

Read More,

• Rational and Irrational Numbers
• Rational Numbers
• Irrational Numbers

Irrational Numbers Practice Questions

Question 1: Show that √12 is irrational.

Question 2: Show that √13 is irrational

Question 3: Prove that √17 is irrational

Question 4: Determine if [Tex]\frac{3}{5}[/Tex] is rational or irrational ?

Question 5: Determine if 0.12345678910……(concatenation of natural numbers) are rational or irrational ?

Question 6: Is 0.8888888…..(recurring decimal) rational or irrational ?

Question 7: Is 0.1467914679……(repeating decimal) rational or irrational ?

Question 8: Find [Tex]\sqrt{6+2\sqrt{5}} -\sqrt{6-2\sqrt{5}}[/Tex].

Question 9: Compute [Tex]\sqrt{11+2\sqrt{10}} +\sqrt{11-2\sqrt{10}}[/Tex].

Question 10: Evaluate [Tex]\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}[/Tex].

Practice Questions on Irrational Numbers – FAQs

What are irrational numbers?

Irrational Number are number which cannot be expressed in the form p/q, where p and q both are integers, and q not equal to ‘0’.

What are some examples of Irrational numbers?

√2, √3, √5,  π  are some example of irrational numbers.

Is the product of two irrational also irrational?

The product of two irrational numbers is not necessarily irrational; it can be either rational or irrational.

Example:[Tex]\sqrt{2} \times \sqrt{2} = 4 [/Tex]which is not irrational

Example:[Tex]\sqrt{2} \times \sqrt{3} = \sqrt6[/Tex] which is not irrational

Is the sum of two irrational also irrational?

The sum of two irrational numbers can be either rational or irrational.

Example: [Tex]\sqrt{2} + (\sqrt{2} -2) = 2[/Tex] which is not irrational

Example: [Tex]2\sqrt{2} + 3\sqrt{2} = 5\sqrt{2}[/Tex] which is not irrational