Irrational numbers are defined as the numbers in the number system which can’t be represented in the form of a fraction. The general questions related to irrational numbers are comparison, rationalizing and so on.

To compare two square or cube root numbers ‘a’ and ‘b’ such that ‘a’ is greater than ‘b’, then a² is greater than b². The same concept is applied for the comparison between rational and irrational numbers. Have a look at the following sections and solve those problems for the understanding of the concept.

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## Examples on Irrational Numbers

Example 1:
Compare 5 and √11

Solution:
The given numbers are 5, √11
Among the numbers given, one of them is rational while another one is irrational. So, to make a comparison between them, we will raise both of them to the same power such that the irrational one becomes rational. So,
5² = 5 x 5 = 25
(√11)² = √11 x √11 = 11
25 > 11
So, 5 > √11.

Example 2:
Compare √17 and √13

Solution:
Given numbers are √17, √13
Since the given numbers are not the perfect square roots of any number, so they are irrational numbers. To compare them, we will first compare them into rational numbers and then perform the comparison. So,
(√17)² = √17 x √17 = 17
(√13)² = √13 x √13 = 13
17 > 13
So, √17 > √13.

Example 3:
Arrange the following numbers in the ascending order
∛11, √5, √3, 3, √7, 2√2

Solution:
Given irrational numbers are ∛11, √5, √3, 3, √7, 2√2
The order of given irrational numbers are 3, 2, 2, 1, 2, 2
The least common multiple of (3, 2, 2, 1, 2, 2) = 6
Multiply every number to the power of 6. So,
(∛11)⁶ = ∛11 x ∛11 x ∛11 x ∛11 x ∛11 x ∛11 = 121
(√5)⁶ = √5 x √5 x √5 x √5 x √5 x √5 = 125
(√3)⁶ = √3 x √3 x √3 x √3 x √3 x √3 = 27
(3)⁶ = 3 x 3 x 3 x 3 x 3 x 3 = 279
(√7)⁶ = √7 x √7 x √7 x √7 x √7 x √7 = 343
(2√2)⁶ = 2√2 x 2√2 x 2√2 x 2√2 x 2√2 x 2√2 = 512
Arranging the ascending order means placing the numbers from least to greatest.
27 < 121 < 125 < 279 < 234 < 512
Hence, the ascending order of number is √3 < ∛11 < √5 < 3 < √7 < 2√2.

Example 4:
Arrange the following numbers in the descending order
√11, √2, √17, √13, √5, √3

Solution:
Given irrational numbers are √11, √2, √17, √13, √5, √3
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.
(√11)² = √11 x √11 = 11
(√2)² = √2 x √2 = 2
(√17)² = √17 x √17 = 17
(√13)² = √13 x √13 = 13
(√5)² = √5 x √5 = 5
(√3)² = √3 x √3 = 3
Arranging the descending order means placing the numbers from the greatest to the smallest.
So, 17 > 13 > 11 > 5 > 3 > 2
Hence, the descending order of numbers is √17 > √13 > √11 > √5 > √3 > √2.

Example 5:
Compare ∛15 and ∛17

Solution:
Since the given numbers are not the perfect cube roots. So, to make a comparison between them first need to convert them into rational numbers and then perform the comparison. So,
(∛15)³ = ∛15 x ∛15 x ∛15 = 15
(∛17)³ = ∛17 x ∛17 x ∛17 = 17

Example 6:
Rationalize $$\frac { 1 }{ √5 }$$.

Solution:
Given fraction is $$\frac { 1 }{ √5 }$$
Multiply both numerator and denominator by √5 to rationalize the fraction.
$$\frac { 1 }{ √5 }$$ = $$\frac { 1 }{ √5 }$$ x $$\frac { √5 }{ √5 }$$
= $$\frac { √5 }{ 5 }$$
So, the rationalized fraction is $$\frac { √5 }{ 5 }$$

Example 7:
Rationalize $$\frac { 10 }{ √3 + 4 }$$

Solution:
Given fraction is $$\frac { 10 }{ √3 + 4 }$$
Since the given fraction contains an irrational denominator, so we need to convert it into a rational denominator so that calculations may become easier and simplified. To do so we will multiply both numerator and denominator by the conjugate of the denominator. So,
$$\frac { 10 }{ √3 + 4 }$$ = $$\frac { 10 }{ √3 + 4 }$$ x $$\frac { √3 – 4 }{ √3 – 4 }$$
= $$\frac { 10(√3 – 4) }{ 3 – 4 }$$ = $$\frac { 10(√3 – 4) }{ -1 }$$
= -10(√3 – 4)
So, the rationalized fraction is -10(√3 – 4).

Example 8:
Rationalize $$\frac { 2 }{ 14 – √3 }$$

Solution:
Given fraction is $$\frac { 2 }{ 14 – √3 }$$
Since the given fraction contains an irrational denominator, so we need to convert it into a rational denominator so that calculations may become easier and simplified. To do so we will multiply both numerator and denominator by the conjugate of the denominator. So,
$$\frac { 2 }{ 14 – √3 }$$ = $$\frac { 2 }{ 14 – √3 }$$ x $$\frac { 14 + √3 }{ 14 + √3 }$$
= $$\frac { 2(14 + √3) }{ 14² – 3 }$$
= $$\frac { 2(14 + √3) }{ 193 }$$
So, the rationalized fraction is $$\frac { 2(14 + √3) }{ 193 }$$