NUMBER SYSTEMS – Exercise 1.5 – (MATHEMATICS) – 9th Class

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Exercise 1.5

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

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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}

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

(iii) {{2\sqrt 7 } \over {7\sqrt 7 }}

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

(iv) {1 \over {\sqrt 2 }}

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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

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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)

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

(ii) (3 + √3) (3 – √3)

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

(iii) ( √5 + √2)²

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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)

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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?

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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.

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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 }}

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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 }}

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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 }}

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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}}

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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}.

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