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Exercise 1.5

1.  Classify the following numbers as rational or irrational :
(i) 2-  $\sqrt 5$

Ans. – 2-  $\sqrt 5$  is an irrational number, as difference of a rational and an irrational number is irrational.

(ii) (3 + $\sqrt {23}$ ) – $\sqrt {23}$

Ans. – $(3 + \sqrt {23} ) - \sqrt {23} = 3 + \sqrt {23} - \sqrt {23} = 3$, is a rational number.

(iii) ${{2\sqrt 7 } \over {7\sqrt 7 }}$

Ans. – ${{2\sqrt 7 } \over {7\sqrt 7 }} = {2 \over 7}$ , is a rational number.

(iv) ${1 \over {\sqrt 2 }}$

Ans. –  ${1 \over {\sqrt 2 }} = {{\sqrt 2 } \over 2}$ is an irrational number, as divisors of an irrational number by a non-zero rational number is irrational.

(v) 2π

Ans. – 2π  irrational number, as it is an irrational number and multiplication of a rational and an irrational number is irrational.

2. Simplify each of the following expressions :
(i) (3 + √5 ) (2 + √2)

Ans. –  $(3 + \sqrt 3 )(2 + \sqrt 2 ) = 6 + 3\sqrt 2 + 2\sqrt 3 + \sqrt 6$

(ii) (3 + √3) (3 – √3)

Ans. – $(3 + \sqrt 3 )(3 - \sqrt 3 ) = {(3)^2} - {(\sqrt 3 )^2} = 9 - 3 = 6$

(iii) ( √5 + √2)²

Ans. –  ${(\sqrt 5 + \sqrt 2 )^2} = {(\sqrt 5 )^2} + 2.\sqrt 5 .\sqrt 2 + {(\sqrt 2 )^2}$
$= 5 + 2\sqrt {10} + 2 = 7 + 2\sqrt {10}$

(iv) (√5 – √2) (√5 + √2)

Ans. – $(\sqrt 5 - \sqrt 2 )(\sqrt 5 + \sqrt 2 ) = {(\sqrt 5 )^2} - {(\sqrt 2 )^2} = 5 - 2 = 3$

3. Recall, π it is defined as the ratio of the circumference (say c) of a circle to its diameter (say d). That is, π =  ${c \over d}$
This seems to contradict the fact that r is irrational. How will you resolve this contradiction?

Sol. On measuring c with any device, we get only approximate measurement. Therefore, π is an irrational.

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

Sol.
$\sqrt {9.3} = \sqrt {9.3 \times 1}$   Let OA = 9.3 and OB = 1.
With AB as diameter draw a semicircle. Draw OP perpendicular to AB, meeting the semicircle at P. Then OP = $\sqrt {9.3}$ . With O as centre and OP as radius draw an arc to meet the number line at Q on the positive side. Then, OQ = $\sqrt {9.3}$and the point Q thus obtained represents
$\sqrt {9.3}$.

5. Rationalise the denominators of the following.
(i) ${1 \over {\sqrt 7 }}$

Ans. –  ${1 \over {\sqrt 7 }} = {1 \over {\sqrt 7 }} \times {{\sqrt 7 } \over {\sqrt 7 }} = {{\sqrt 7 } \over 7}$

(ii) ${1 \over {\sqrt 7 - \sqrt 6 }}$

Ans. –  ${1 \over {\sqrt 7 - \sqrt 6 }} = {{\sqrt 7 + \sqrt 6 } \over {(\sqrt 7 - \sqrt 6 )(\sqrt 7 + \sqrt 6 )}} = {{\sqrt 7 + \sqrt 6 } \over {7 - 6}} = \sqrt 7 + \sqrt 6 .$

(iii) ${1 \over {\sqrt 5 + \sqrt 2 }}$

Ans. –  ${1 \over {\sqrt 5 + \sqrt 2 }} = {{\sqrt 5 - \sqrt 2 } \over {(\sqrt 5 + \sqrt 2 )(\sqrt 5 - \sqrt 2 )}} = {{\sqrt 5 - \sqrt 2 } \over {5 - 2}} = {{\sqrt 5 - \sqrt 2 } \over 3}.$
(iv) ${1 \over {\sqrt 7 + 2}}$
Ans. – ${1 \over {\sqrt 7 - 2}} = {{\sqrt 7 + 2} \over {(\sqrt 7 - 2)(\sqrt 7 + 2)}} = {{\sqrt 7 + 2} \over {7 - 4}} = {{\sqrt 7 + 2} \over 3}.$