Draw a number line and mark point O, representing zero, on it

Suppose point A represents 2 as shown in the figure Then OA = 2. Now, draw a right triangle OAB such that AB = 1. By Pythagoras theorem, we have

OB^{2} = OA^{2} + AB^{2}

=> OB^{2} = 2^{2 }+ 1^{2}

=> OB^{2} = 4 + 1 = 5 => OB = **√**5

Now, draw a circle with center O and radius OB. We find that the circle cuts the number line at A Clearly, OA_{1 }= OB = radius of circle = **√**5 Thus, A_{1} represents **√**5 on the number line. But, we have seen that **√**5 is not a rational number. Thus we find that there is a point on the number which is not a rational number. Now, draw a right triangle OA_{1}B_{1 }= AB = 1 ,Such that A_{1}B_{1 }= AB = 1 Again, by Pythagoras theorem, we have

Draw a circle with center O and radius OB_{1} = **√**6. This circle cuts the number line at A_{2} as shown in figure Clearly OA_{2} = OB_{1} = **√**6.Thus, A_{2} represents **√**6on the number line. Also, we know that **√**6 is not a rational number. Thus, A_{2} is a point on the number line not representing a rational number Continuing in this manner, we can represent **√**7 and **√**8 also on the number of lines as shown in the figure

Thus, OA_{3 }= OB_{2} = **√**7 and OA_{4} OB_{3} = **√**8