In this chapter we will learn to represent irrational number on number line with solved examples.

We know that in number line, number are expressed in the form of integers and decimals.

So if we are provided with irrational number in the form of square root, it gets difficult to represent in the number line.

Given below are problems where we have represented the irrational number (in form of square root) step by step.

**Question 01**

Represent \mathtt{\sqrt{5}} in number line.

**Solution**

(a) Express irrational number in form of difference of square.

Let;

\mathtt{\sqrt{5} =\sqrt{a^{2} -b^{2}} \ }

Where;

\mathtt{a\ =\ \frac{5+1}{2} \ =\ 3}\\\ \\ \mathtt{b\ =\ \frac{5-1}{2} =\ 2}

Putting the values of ” a ” & ” b ” in above expression;

\mathtt{\sqrt{5} =\sqrt{3^{2} -2^{2}} \ }

(b) The expression \mathtt{\sqrt{5} =\sqrt{3^{2} -2^{2}} \ } is in form of Pythagoras equation.

We will use it to locate the irrational number on number line.

(i) From point O cut an arc of 2 cm and name it point A

(ii) From point A, cut an arc of 3 cm on line OM and name the point AB.

AB = 3 cm

(iii) Now join the point OB.

As per Pythagoras theorem, the length of OB will be \mathtt{\sqrt{5}} units.

Now you can measure arc OB using divider and mark it on number line.

**Question 02**

Plot \mathtt{\sqrt{8}} on number line.

**Solution**

Follow the below steps;

(a) Express the irrational number as difference of square.

\mathtt{\sqrt{8} =\sqrt{a^{2} -b^{2}}}

Where;

\mathtt{a\ =\ \frac{8+1}{2} \ =\ 4.5}\\\ \\ \mathtt{b\ =\ \frac{8-1}{2} =\ 3.5}

Putting the values of “a” and “b” in above expression.

\mathtt{\sqrt{8} =\sqrt{( 4.5)^{2} -( 3.5)^{2}}}

(b) The given expression is in form of Pythagoras theorem.

Follow the below steps to locate the point graphically.

(i) From point O, cut arc A of length 3.5 cm

(ii) from point A, cut an arc of length 4.5 cm on line OM and name it point B.

Here the length of line OB = \mathtt{\sqrt{8}} units.

**Question 03**

Represent \mathtt{\sqrt{12.5} \ } on number line.

**Solution**

Follow the below steps:

(a) Express the irrational number in form of difference of square.

\mathtt{\sqrt{12.5} \ \ \ =\sqrt{a^{2} -b^{2}} \ }

Where;

\mathtt{a\ =\ \frac{12.5+1}{2} \ =\ 6.75}\\\ \\ \mathtt{b\ =\ \frac{12.5-1}{2} =\ 5.75}

Putting the values of “a” & “b” in above expression.

\mathtt{\sqrt{12.5} =\sqrt{( 6.75)^{2} -( 5.75)^{2}} \ }

(b) The above expression is in form of Pythagoras theorem.

Using the expression to draw the irrational number graphically.

(i) From point O cut an arc of 5.75 cm and name it point A.

(ii) From point A, cut an arc of 6.75 cm on line OM and name it point B.

The length of line OB will be equal to \mathtt{\sqrt{12.5} \ } units.

**Question 04**

Represent \mathtt{\sqrt{3.5} \ \ } on number line.

**Solution**

Follow the below steps;

(a) Expressing the irrational number in form of difference of square;

\mathtt{\sqrt{3.5} \ \ \ =\sqrt{a^{2} -b^{2}} \ }

Where;

\mathtt{a\ =\ \frac{3.5+1}{2} \ =\ 2.25}\\\ \\ \mathtt{b\ =\ \frac{3.5-1}{2} =\ 1.25}

Putting the values in expression;

\mathtt{\sqrt{3.5} =\sqrt{( 2.25)^{2} -( 1.25)^{2}} \ }

(b) The equation is in form of Pythagoras expression;

Plotting the expression graphically;

(i) From point O, cut an arc of 1.25 cm and name it point A.

(ii) From point A, cut an arc of 2.25 cm at line OM.

Here the line OB measures \mathtt{\sqrt{3.5}} units