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

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. 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) ${1 \over {\sqrt 7 - 2}}$ $= {1 \over {\sqrt 7 - 2}} \times {{\sqrt 7 + 2} \over {\sqrt 7 + 2}}$
$= {{\sqrt 7 + 2} \over {{{\left( {\sqrt 7 } \right)}^2} - {{\left( 2 \right)}^2}}}$
$= {{\sqrt 7 + 2} \over {7 - 4}}$ $= {{\sqrt 7 + 2} \over 3}$