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Number Systems : Exercise 1.5 (Mathematics NCERT Class 9th)


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Q.1      Classify the following numbers as rational or irrational :
               (i) 2 - \sqrt 5                                      
            (ii) \left( {3 + \sqrt {23} } \right) - \sqrt {23}

               (iii) {{2\sqrt 7 } \over {7\sqrt 7 }}
            (iv) {1 \over {\sqrt 2 }}

               (v) 2\pi
Sol.

(i) 2 - \sqrt 5 is an irrational number being a difference between a rational and an irrational.
(ii) \left( {3 + \sqrt {23} } \right) - \sqrt {23} = 3 + \sqrt {23} - \sqrt {23} = 3, which is a rational number.
(iii) {{2\sqrt 7 } \over {7\sqrt 7 }} = {2 \over 7}, which is a rational number.
(iv)  - {1 \over {\sqrt 2 }} is an irrational number being the quotient of a rational and an irrational number .
(v) 2\pi is irrational being the product of rational and irrational number .


 Q.2      Simplify each of the following expressions :
            
(i) \left( {3 + \sqrt 3 } \right)\left( {2 + \sqrt 2 } \right)
            
(ii) \left( {3 + \sqrt 3 } \right)\left( {3 - \sqrt 3 } \right)
            
(iii) {\left( {\sqrt 5 + \sqrt 2 } \right)^2}
            
(iv) \left( {\sqrt 5 - \sqrt 2 } \right)\left( {\sqrt 5 + \sqrt 2 } \right)

Sol.

(i)   \left( {3 + \sqrt 3 } \right)\left( {2 + \sqrt 2 } \right)
         = 3 \times 2 + 3\sqrt 2 + 2\sqrt 3 + \sqrt 2 \times \sqrt 3
         = 6 + 3\sqrt 2 + 2\sqrt 3 + \sqrt 6
(ii)  \left( {3 + \sqrt 3 } \right)\left( {3 - \sqrt 3 } \right)
        = {\left( 3 \right)^2} - {\left( {\sqrt 3 } \right)^2}
         = 9 - 3 = 6
(iii) {\left( {\sqrt 5 + \sqrt 2 } \right)^2}
         {\left( {\sqrt 5 } \right)^2} + 2\sqrt 5 \sqrt 2 + {\left( {\sqrt 2 } \right)^2}
         5 + 2\sqrt {10} + 2 = 7 + 2\sqrt {10}
(iv)   \left( {\sqrt 5 - \sqrt 2 } \right)\left( {\sqrt 5 + \sqrt 2 } \right)
         = {\left( {\sqrt 5 } \right)^2} - {\left( {\sqrt 2 } \right)^2}
          = 5 - 2 = 3


Q.3      Recall , \pi is defined as the ratio of the circumference (say c) of a circle to its diameter (say d). That is \pi = {c \over d}. This seems to  contradict the fact that \pi is irrational. How will you resolve this contradiction?
Sol.         There is no contradiction as either c or d irrational and hence \pi is an irrational number.


Q.4      Represent \sqrt {9.3} on the number line.
Sol.

Mark the distance 9.3 units from a fixed point A on a given line to obtain a point B such that AB = 9.3 units. From B, mark a distance of 1 unit and mark the new point as C. Find the mid- point of AC and mark that point as O. Draw a semi-circle with centre O and radius OC. Draw a line perpendicular to AC passing through B and intersecting the semicircle at D.

3
Then BD = \sqrt {9.3} . To represent \sqrt {9.3} on the number line. Let us treat the line BC as the number line, with B as zero, C as 1 and so on. Draw an arc with centre B and radius BD, which intersects the number line in E. Then E represent \sqrt {9.3}


 

Q.5      Rationalise the denominators of the following :
            (i) {1 \over {\sqrt 7 }}
            (ii) {1 \over {\sqrt 7 - \sqrt 6 }}

               (iii) {{\rm{1}} \over {\sqrt 5 {\rm{ + }}\sqrt 2 }}
            (iv) {1 \over {\sqrt 7 - 2}}

Sol.

(i)    {1 \over {\sqrt 7 }} = {{1 \times \sqrt 7 } \over {\sqrt 7 \times \sqrt 7 }} = {{\sqrt 7 } \over 7}
(ii)   {1 \over {\sqrt 7 - \sqrt 6 }}  = {1 \over {\sqrt 7 - \sqrt 6 }} \times {{\sqrt 7 + \sqrt 6 } \over {\sqrt 7 + \sqrt 6 }}  = {{\sqrt 7 + \sqrt 6 } \over {{{\left( {\sqrt 7 } \right)}^2} - {{\left( {\sqrt 6 } \right)}^2}}}
          = {{\sqrt 7 + \sqrt 6 } \over {7 - 6}} = \sqrt 7 + \sqrt 6
(iii)  {1 \over {\sqrt 5 + \sqrt 2 }}
 = {{\rm{1}} \over {\sqrt 5 {\rm{ + }}\sqrt 2 }} \times {{\sqrt 5-\sqrt 2 } \over {\sqrt 5-\sqrt 2 }}  = {{\sqrt 5 - \sqrt 2 } \over {{{\left( {\sqrt 5 } \right)}^2} - {{\left( {\sqrt 2 } \right)}^2}}}
           = {{\sqrt 5 - \sqrt 2 } \over {5 - 2}} = {{\sqrt 5 - \sqrt 2 } \over 3}
(iv) {1 \over {\sqrt 7 - 2}}  = {1 \over {\sqrt 7 - 2}} \times {{\sqrt 7 + 2} \over {\sqrt 7 + 2}}
         = {{\sqrt 7 + 2} \over {{{\left( {\sqrt 7 } \right)}^2} - {{\left( 2 \right)}^2}}}
         = {{\sqrt 7 + 2} \over {7 - 4}}  = {{\sqrt 7 + 2} \over 3}



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