Worksheet on Irrational Numbers | Practice Problems on Irrational Numbers

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

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.

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.

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.

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.

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.

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.

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 \(\frac { 2√10 }{ 5 } \)

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 \(\frac { 200 – 25√31 }{ 66 } \).

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 \(\frac { -8√5 + 16√7 – √30 + 2√35 }{ 23 } \).