In this chapter we will learn to compare two or more irrational numbers with solved examples.

After reading this chapter you will be able to arrange irrational numbers in ascending or descending orders.

## How to compare irrational numbers ?

Irrational numbers are generally given in the form of square roots, cube roots etc.

Finding the value of each irrational square root number will be complex and time consuming.

In order to **compare square root numbers**, **you have to square them to get integer value and then do the comparison**.

Similarly for **cube root number**, **take the cube of each number to get the integer value** and then do the comparison.

Given below is the thinking behind above methodology.

Let ” a ” and ” b” are the given numbers such that a > b.

If we square the above numbers, the relation between ” a” and ” b ” will remain the same.

\mathtt{\Longrightarrow \ a^{2} >b^{2}}

Hence, **on squaring / cubing all number numbers, the relation between numbers will not change**.

I hope you understood the above theory. Given below are some examples for your better understanding.

## Irrational number comparison – Solved examples

**Example 01**

Which number is greater \mathtt{\sqrt{5} \ or\ \sqrt{6}}

**Solution**

Both the irrational numbers are in square root form.

Square each of the numbers to get the integer values.

\mathtt{\left(\sqrt{5}\right)^{2} =\ 5}\\\ \\ \mathtt{\left(\sqrt{6}\right)^{2} =\ 6}

Now compare the integer values.

We know that 5 < 6.

Hence, \mathtt{\sqrt{5} \ < \ \sqrt{6}}

**Example 02**

Compare the irrational numbers \mathtt{\sqrt{45} \ \ and\ \sqrt{39}}

**Solution**

Both the irrational numbers are in square root form.

Take the square of the numbers to convert into integers.

\mathtt{\left(\sqrt{45}\right)^{2} =\ 45}\\\ \\ \mathtt{\left(\sqrt{39}\right)^{2} =\ 39}

We know that 45 > 39.

Hence, \mathtt{\sqrt{45} \ \ >\ \sqrt{39}}

**Example 03**

Compare the irrational numbers \mathtt{\sqrt[3]{51} \ \ \&\ \sqrt[3]{65}}

**Solution**

Both the numbers are in the form of cube roots.

Taking the cube of the numbers to get numbers in integer form .

\mathtt{\left(\sqrt[3]{51}\right)^{3} =\ 51}\\\ \\ \mathtt{\left(\sqrt[3]{65}\right)^{3} =\ 65}

We know that 51 < 65.

Hence, \mathtt{\sqrt[3]{51} \ \ < \ \sqrt[3]{65}}

**Example 04**

Which of the irrational number is greater, \mathtt{\sqrt[2]{59} \ \ or\ \sqrt[3]{76}}

**Solution**

One of the number is in square root form and other is in cube root form.

You can’t compare the numbers by squaring one number and cubing other numbers.

In this case you have to find the exact decimal value of irrational number.

\mathtt{\sqrt{59} =\ 7.681}\\\ \\ \mathtt{\sqrt[3]{65} =\ 4.02}

We know that 7.681 > 4.01.

Hence, \mathtt{\sqrt[2]{59} \ \ >\ \sqrt[3]{76}}

**Example 05**

Compare the irrational numbers \mathtt{5\sqrt{23} \ \ \&\ \ 6\sqrt{15}}

**Solution**

Both the numbers are in square root form.

Take the square of both the numbers to convert into integer form.

\mathtt{Squaring\ 5\sqrt{23}}\\\ \\ \mathtt{\Longrightarrow \ \left( 5\sqrt{23}\right){^{2}}}\\\ \\ \mathtt{\Longrightarrow \ 25\ \times \ 23}\\\ \\ \mathtt{\Longrightarrow \ 575}

\mathtt{Squaring\ 6\sqrt{15}}\\\ \\ \mathtt{\Longrightarrow \ \left( 6\sqrt{15}\right){^{2}}}\\\ \\ \mathtt{\Longrightarrow \ 36\ \times \ 15}\\\ \\ \mathtt{\Longrightarrow \ 540}

Now compare both the integers.

We know that 575 > 540.

Hence, \mathtt{5\sqrt{23} \ \ >\ \ 6\sqrt{15}}

**Example 06**

Compare the irrational numbers and arrange them in descending order.

\mathtt{4\sqrt{13} ,\ 8\sqrt{7} ,\ 3\sqrt{21} \ and\ \ 6\sqrt{15}}

**Solution**

All the numbers are present in square root form.

Square all the numbers to get the integer values.

\mathtt{Squaring\ 4\sqrt{13}}\\\ \\ \mathtt{\Longrightarrow \ \left( 4\sqrt{13}\right){^{2}}}\\\ \\ \mathtt{\Longrightarrow \ 16\ \times \ 13}\\\ \\ \mathtt{\Longrightarrow \ 208}

\mathtt{Squaring\ 8\sqrt{7}}\\\ \\ \mathtt{\Longrightarrow \ \left( 8\sqrt{7}\right){^{2}}}\\\ \\ \mathtt{\Longrightarrow \ 64\ \times \ 7}\\\ \\ \mathtt{\Longrightarrow \ 448}

\mathtt{Squaring\ 3\sqrt{21}}\\\ \\ \mathtt{\Longrightarrow \ \left( 3\sqrt{21}\right){^{2}}}\\\ \\ \mathtt{\Longrightarrow \ 3\ \times \ 21}\\\ \\ \mathtt{\Longrightarrow \ 63}

\mathtt{Squaring\ 6\sqrt{15}}\\\ \\ \mathtt{\Longrightarrow \ \left( 6\sqrt{15}\right){^{2}}}\\\ \\ \mathtt{\Longrightarrow \ 36\ \times \ 15}\\\ \\ \mathtt{\Longrightarrow \ 540}

Now comparing all the integers and arranging them in descending order.

540 > 448 > 208 > 63

Putting the corresponding irrational numbers.

\mathtt{6\sqrt{15} \ >\ 8\sqrt{7} >4\sqrt{13} >3\sqrt{21}}

**Example 07**

Compare the irrational numbers and arrange them in ascending order.

\mathtt{3\sqrt[3]{5} ,\ 2\sqrt[3]{7} \ and\ \ \sqrt[3]{9} \ \ \ }

**Solution**

Here all the irrational number are in the form of cube root.

Find cube of all the number to get integer value and then do the comparison.

\mathtt{Cubing\ \ 3\sqrt[3]{5}}\\\ \\ \mathtt{\Longrightarrow \ \left( 3\sqrt[3]{5}\right){^{3}}}\\\ \\ \mathtt{\Longrightarrow \ 27\ \times \ 5}\\\ \\ \mathtt{\Longrightarrow \ 135}

\mathtt{Cubing\ \ 2\sqrt[3]{7}}\\\ \\ \mathtt{\Longrightarrow \ \left( 2\sqrt[3]{7}\right){^{3}}}\\\ \\ \mathtt{\Longrightarrow \ 8\ \times \ 7}\\\ \\ \mathtt{\Longrightarrow \ 56}

\mathtt{Cubing\ \ \sqrt[3]{9}}\\\ \\ \mathtt{\Longrightarrow \ \left(\sqrt[3]{9}\right){^{3}}}\\\ \\ \mathtt{\Longrightarrow \ 9}

Comparing the integer values and arranging them in ascending order.

9 < 56 < 135

Hence, irrational number is arranged as;

\mathtt{\sqrt[3]{9} < \ 2\sqrt[3]{7} \ < 3\sqrt[3]{5}}

**Next chapter : Comparing rational & irrational number**