In the number system, irrational numbers are numbers that can’t be written in fractional form. The different types of problems related to irrational numbers are comparison, representation on the number line, rationalizing the denominator and so on. All these types of irrational numbers questions are covered in the Worksheet on Irrational Numbers. So interested students have a look at them and start solving for scoring better marks in the exam.

The problems related to rationalizing the denominator helps to solve the functions of irrational numbers. Given some of the questions are involving calculations of irrational numbers.

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## Irrational Numbers Worksheet

Question 1:
Check if the below numbers are rational or irrational.
π, $$\frac { 1 }{ √2 }$$, $$\frac { √3 }{ 8 }$$, √5

Solution:

Since the irrational numbers non-repeating or non-terminating.
So, π, $$\frac { 1 }{ √2 }$$, $$\frac { √3 }{ 8 }$$, √5 are irrational numbers.

Question 2:
Determine whether the following numbers are rational or irrational.
$$\frac { 1 }{ 2 }$$, 8, $$\frac { 15 }{ 6 }$$, $$\frac { 76 }{ 5 }$$

Solution:

Since the decimal expansion of a rational number either repeats or terminates.
So, $$\frac { 1 }{ 2 }$$, 8, $$\frac { 15 }{ 6 }$$, $$\frac { 76 }{ 5 }$$ are rational numbers.

Question 3:
Compare √6 and √5

Solution:

Given two irrational numbers are √6 and √5
We know that if ‘p’ and ‘q’ are two numbers such that ‘q’ is greater than ‘p’, then q² will be greater than p². So, square the given numbers.
√6 = √6 x √6 = (√6)² = 6
√5 = √5 x √5 = (√5)² = 5
6 is greater than 5.
So, √6 is greater than √5.

Question 4:
Arrange the following irrational numbers in descending order.
√17, √3, √21, √10, √51

Solution:

Given irrational numbers are √17, √3, √21, √10, √51
We know that if ‘p’ and ‘q’ are two numbers such that ‘q’ is greater than ‘p’, then q² will be greater than p². So, square the given numbers.
√17 = √17 x √17 = (√17)² = 17
√3 = √3 x √3 = (√3)² = 3
√21 = √21 x √21 = (√21)² = 21
√10 = √10 x √10 = (√10)² = 10
√51 = √51 x √51 = (√51)² = 51
Arranging the descending order means placing the numbers from the greatest to the smallest.
So, 51 > 21 > 17 > 10 > 3
Hence, the descending order of numbers is √51 > √21 > √17 > √10 > √3.

Question 5:
Write the irrational numbers ∜6, √3 and ∛7 in ascending and descending orders.

Solution:

Given irrational numbers are ∜6, √3 and ∛7
Order of the irrational numbers are 4, 2, 3
The least common multiple of (4, 2, 3) = 12
So, we have to change the order of each number as 12
Change ∜5 as 12th root
∜6 = (4 x 3) √6³
= 12 √216
√3 = (2 x 6) √36
= 12 √729
∛7 = (3 x 4) √74
= 12 √2401
216 < 729 < 2401.
Therefore, ascending order is ∜6, √3, ∛7 and descending order is ∛7, √3 and ∜6.

Question 6:
Find an irrational number between √6 and 6.

Solution:

Given two real numbers are √6 and 6
A real number between √6 and 6 is $$\frac { √6 + 6 }{ 2 }$$ = ½√6 + 1
But 1 is a rational number and ½√6 is an irrational number. The sum of a rational number and an irrational number is irrational.
So, ½√6 + 1 is an irrational number that lies between √6 and 6.

Question 7:
Insert two irrational numbers between √13 and √19.

Solution:

Given two real numbers are √13 and √19
Consider the squares of √13 and √19
(√13)² = 13
(√19)² = 19
Since the numbers 14, 17 lie between 13 and 19 i.e between (√13)² and (√19)²
Therefore, √13 and √19 are √14 and √17.
Hence two irrational numbers between √13 and √19 are √14 and √17.

Question 8:
Rationalize $$\frac { 12 }{ 3√10 }$$

Solution:

Given fraction is $$\frac { 12 }{ 3√10 }$$
Since the given fraction has an irrational denominator, so we need to rationalize this and make it more simple. So, to rationalize this, we will multiply the numerator and denominator of the given fraction by root 10, i.e., √10. So,
$$\frac { 12 }{ 3√10 }$$ = $$\frac { 12 }{ 3√10 }$$ x $$\frac { √10 }{ √10 }$$
= $$\frac { 12√10 }{ 3(10) }$$
= $$\frac { 12√10 }{ 30 }$$
= $$\frac { 2√10 }{ 5 }$$
So, the required rationalized form is $$\frac { 2√10 }{ 5 }$$

Question 9:
Rationalize $$\frac { 25 }{ 16 + 2√31 }$$

Solution:

Given fraction is $$\frac { 25 }{ 16 + 2√31 }$$
Multiply numeration, denominator by (16 – 2√31)
$$\frac { 25 }{ 16 + 2√31 }$$ = $$\frac { 25 }{ 16 + 2√31 }$$ x $$\frac { 16 – 2√31 }{ 16 – 2√31 }$$  [(a + b)(a – b) = a² – b²]
= $$\frac { 25(16 – 2√31) }{ 16² – (2√31)² }$$
= $$\frac { 400 – 50√31 }{ 256 – 124 }$$
= $$\frac { 400 – 50√31 }{ 132 }$$
= $$\frac { 200 – 25√31 }{ 66 }$$
So, the required rationalized form is $$\frac { 200 – 25√31 }{ 66 }$$.

Question 10:
Rationalize $$\frac { 8 + √6 }{ √5 – 2√7 }$$

Solution:

Given fraction is $$\frac { 8 + √6 }{ √5 + 2√7 }$$
Since, the given problem has an irrational term in the denominator with addition format. So we need to rationalize using the method of multiplication by the conjugate. So,
$$\frac { 8 + √6 }{ √5 + 2√7 }$$ = $$\frac { 8 + √6 }{ √5 + 2√7 }$$ x $$\frac { √5 – 2√7 }{ √5 – 2√7 }$$
= $$\frac { (8 + √6)(√5 – 2√7) }{ 5 – 4(7) }$$
= $$\frac { 8√5 – 16√7 + √30 – 2√35 }{ 5 – 28 }$$
= $$\frac { 8√5 – 16√7 + √30 – 2√35 }{ -23 }$$
So, the required rationalized form is $$\frac { -8√5 + 16√7 – √30 + 2√35 }{ 23 }$$.