NumberSystems Exercise 1.5 and 1.6|NCERT Solutions for Class 9th Maths

Exercise 1.5

Question 1 
Classify the following numbers as rational or irrational:
(i) 2 - √5
(ii) (3 + √23) - √23
(iii) 2√7/7√7
(iv) 1/√2
(v) 2π
Solution

Question 2
Simplify each of the following expressions:
(i) (3 + √3) (2 + √2)
(ii) (3 + √3) (3 - √3)
(iii) (√5 + √2)²
(iv) (√5 - √2) (√5 + √2)
Solution
(i) (3 + √3) (2 + √2)
= 3 × 2 + 2 + √3 + 3√2+ √3 ×√2
= 6 + 2√3 +3√2 + √6
(ii) (3 + √3) (3 - √3)
Now as (a + b) (a - b) = a² - b²
= 3² - (√3)²
= 9 - 3= 6
(iii) (√5 + √2)² 
Now as (a + b)² = a² + b² + 2ab
= (√5)² + (√2)² + 2 ×√5 × √2
= 5 + 2 + 2 × √5× 2 
= 7 +2√10

(iv) (√5 - √2) (√5 + √2)
Now as (a + b) (a - b) = a² - b²
= (√5)² - (√2)² 
= 5 – 2= 3
Question 3
 Recall, π is defined as the ratio of the circumference (say c) of a circle to its diameter (say d). That is, π = c/d. This seems to contradict the fact that π is irrational. How will you resolve this contradiction?
Solution
There is no contradiction. When we measure a value with a scale, we only obtain an approximate value. We never obtain an exact value. Therefore, we may not realize that either c or d is irrational. The value of π is almost equal to 22/7 or 3.142857...
Question 4.
Represent √9.3 on the number line.
Question 5
 Rationalize the denominators of the following:
(i) 1/√7
(ii) 1/ (√7-√6)
(iii) 1/ (√5+√2)
(iv) 1/ (√7-2)
Solution

 

Exercise 1.6

Question 1.
Find:
(i) 64^1/2
(ii) 32^1/5
(iii) 125^1/3
Solution
i) 64^1/2 =(2⁶)^1/2  =2^{6 X1/2}  = 2³ =8
ii) 32^1/5  =(2⁵)^1/5 = 2
iii) 125^1/3 = (5³)^1/3 =5
Question 2
Find:
(i) 9^3/2
(ii) 32^2/5
(iii) 16^3/4
(iv) 125^-1/3
Solution
i) 9^3/2 = (3²)^3/2 = 27
ii) 32^2/5 = (2⁵)^2/5 =4
iii) 16^3/4 = (2⁴)^3/4 =8
iv) 125^-1/3 = 1/125^1/3 = 1/(5³)^1/3 =1/5
Question 3.
Simplify:
(i) 2^2/3.2^1/5
(ii) (1/3³)⁷
(iii) 11^1/2/11^1/4
(iv) 7^1/2.8^1/2

Solution
i) 2^2/3.2^1/5 =2 ^{2/3+ 1/5} = 2^10+3/15 =2^13/15
ii) (1/3³)⁷
=1/3^3X7 = 1/3²¹ =3 ^-21
 iii) 11^1/2/11^1/4
= 11^{1/2 -1/4}  = 11^1/4
 iv) 7^1/2.8^1/2
=(7X8)^1/2 = 56^1/2
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