In this chapter we will learn method to convert irrational square root number into its decimal representation.

After completing the chapter, you will learn exact steps to find value of any square root irrational number.

Let us first start with the basics.

## Square root and irrational number

The square root of all positive integer (except perfect square ) is an irrational number as its **decimal value is non terminating and non repeating in nature**.

Here we will try to find the decimal value of given square root number.

As the decimal value is unending, it is not possible to calculate the infinite digits. Here **we will try to calculate the value up to two or three decimal places**.

## Calculating irrational number value

Let us consider number \mathtt{\sqrt{3}} for calculation.

To find the value, **follow the below steps;**

(a) **Introduce decimal point and write three pair of zeros**.

(b) **Select the first pair from the left.**

Selecting digit 3 as the first pair.

(c) **Find divisor which when multiplied by itself gives number less than or equal to 3**.

Using hit & trial method.**Let the divisor be 2.**

Multiply the number by itself.

⟹ 2 x 2

⟹ 4

Since 4 > 3, **this divisor is rejected**.

**Let divisor be 1.**

⟹ 1 x 1

⟹ 1

Since 1 < 3, **we select this divisor.**

**(d) Place 1 in both divisor and quotient. **

The number 1 x 1 = 1, will be placed in dividend for subtraction.

Now insert next pair of numbers for calculation.

Since the next pairs are after decimal point, introduce the decimal in quotient also.

**(e) Finding second divisor.**

This involves two steps;

(i)** Add previous divisor and quotient. Now put the number in new divisor area**.

Previous divisor = 1

Previous quotient = 1

Adding both numbers = 1 + 1 = 2**Putting number 2 in divisor area**.

(ii)** Find next digit of divisor.**

The digit must be such that multiplying the divisor with the digit should be less than or equal to 200.

Using hit and trial.**Let the digit be 8.**

The divisor becomes 28.

Multiply divisor with digit 8

28 x 8 = 224

Since 224 > 200, **digit 8 is rejected.**

**Let digit be 7.**

Divisor becomes 27.

Multiply 27 x 7 = 189

Since 189 < 200, **we accept this number.****Put number 7 in divisor and quotient**. Also put number 189 in dividend area for subtraction.

Now introduce next pair of zero for calculation.

(**f) Finding third divisor**

This involves two steps;

(i) **Add previous divisor and last quotient.**

Previous divisor = 27

Last quotient digit = 7

Adding both = 27 + 7 = 34**Put number 34 in divisor area.**

(ii)** Finding next digit of divisor**

The next digit should be such that the number formed if multiplied by same digit, it should be less than or equal to 1100.

The number is 3.

Number formed is 343.

Multiply 343 x 3 = 1029**Put number 3 in divisor and quotient** and put **1029 in dividend area for subtraction**.

Now put the next set of zeros on the dividend area.

**(g) Finding fourth divisor.**

This process again involves two steps;

(i) **Add previous divisor and quotient,**

Previous divisor = 343

Last quotient digit = 3

Adding both = 343 + 3 = 346

(ii) Finding next digit of divisor.

The next digit should be such that the number formed if multiplied by same digit should produce number less than 7100.

The digit is 2.

The **divisor formed 3462.**

Multiply 3462 x 2 = 6942

Place **digit 2 in divisor** **and quotient area**. Also **place 6942 in dividend area for subtraction**.

We will stop at this post as the process will go on and on.

Through the above process we found the approximate value of irrational number as 1.732.

Hence, \mathtt{\sqrt{3} =1.732} is the solution.

**Example 02**

Find the value of irrational number \mathtt{\sqrt{15} \ } up to two decimal places.

**Solution**

Follow the below steps;

(a) **Insert two pair of zeros after decimal point**. Form pair of two from left side.

**(b) Find first divisor.**

Find **largest number** which **when multiplied by itself produce number less than equal to 15**.

**Using number 3**.

Multiply number 3 by itself.

3 x 3 = 9

Since 9 < 15, we will **select number 3 for divisor**.**Put number 3 in divisor and quotient area**. Put number 9 in dividend area for subtraction.

Insert next pair of numbers.

Since the next pair are after decimal point, we will introduce decimal point in the quotient also.

**(c) Finding second divisor.**

This process involves two steps;**(i) Add previous divisor and quotient.**

Previous divisor = 3

Previous quotient = 3

Adding both = 3 + 3 = 6**Insert 6 in the divisor area.**

**(ii) Finding next digit of divisor**

Insert digit such that the number formed if multiplied by same digit will produce number less than 600.

**Let the last digit be 8.**

Number formed is 68.

Multiply 68 x 8 = 544

Since, 544 < 600, **we accept the digit 8.****Put digit 8 in divisor area and quotient**. Also Insert 544 in dividend area for subtraction.

Insert next pair of zeros for calculation.

**(d) Finding third divisor**

The process involves two steps;

(i) **Add previous divisor and quotient.**

Previous divisor = 68

Last quotient = 8

Adding both = 68 + 8 = 76

**Insert 76 in the divisor area.**

(b) **Now find next digit of divisor**

The digit should be such that the multiplication of number formed and the digit should be less than 5600.

The last digit is 7.

The number becomes 767.

Multiply 767 x 7 =5369.

Since 5369 < 5600, we accept the number.

Insert 7 in divisor and quotient. Also, subtract 5369 in dividend area.

We will stop the calculation at this point otherwise the process will go on and on.

Hence, 3.87 is the approximate value of \mathtt{\sqrt{15} \ }

**Next chapter** : **Comparing irrational numbers**