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Represent √6, √7, √8 on the number line.

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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 

 OB2 = OA2 + AB2

=> OB2 = 2+ 12

=> OB2 = 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, OA1 = OB = radius of circle = 5 Thus, A1 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  OA1B1 = AB = 1 ,Such that A1B1 = AB = 1  Again, by Pythagoras theorem, we have 

Draw a circle with center O and radius OB16. This circle cuts the number line at A2 as shown in figure Clearly  OA2 = OB1 = 6.Thus, A2 represents 6on the number line. Also, we know that 6 is not a rational number. Thus, A2 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= OB2  = 7 and  OA4 OB3 = 8

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