RBSE Solutions for Class 9 Maths Chapter 1 Number Systems Ex 1.5

Rajasthan Board RBSE Solutions for Class 9 Maths Chapter 1 Number Systems Ex 1.5 Textbook Exercise Questions and Answers.

RBSE Class 9 Maths Solutions Chapter 1 Number Systems Exercise 1.5

Question 1.
Classify the following numbers as rational or irrational:
(i) 2 – 4√5
Answer:
2 – √5 in an irrational number being a difference between a rational and an irrational.

(ii) (3 + √23) – √23
Answer:
(3 + √23) – √23 = 3 + √23 – √23 = 3, which is a rational number.

(iii) \(\frac{2 \sqrt{7}}{7 \sqrt{7}}\)
Answer:
\(\frac{2 \sqrt{7}}{7 \sqrt{7}}\) , which is a rational number.

(iv) \(\frac{1}{\sqrt{2}}\)
Answer:
\(\frac{1}{\sqrt{2}}\) is irrational being the quotient of a rational and an irrational.

(v) 2π
Answer:
2π is irrational being the product of a rational and an irrational.

Question 2.
Simplify each of the following expressions :
(i) (3 + √3)(2 + √2)
Answer:
(3 + √3)(2 + √2) = 3 × 2 + 3√2 + 2√3 + √2 × √3 = 6 + 3√2 + 2√3 + √6

(ii) (3 + √3)(3 – √3)
Answer:
(3 + √3 )(3 – √3) = (3)² – (√3)² = 9 – 3 = 6

(iii) (√5 + √2)²
Answer:
(√5 + √2)² = (√5)² + 2 √5 √2+ (√2 )² = 5 + 2√10 + 2 = 7 + 2√10

(iv) (√5 – √2)(√5 + √2)
Answer:
(√5 – √2)(√5 + √2) = (√5)² – (√5)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 = \(\frac{c}{d}\). This seems to contradict the fact that π is irrational. How will you resolve this contradiction?
Answer:
There is no contradiction since when we measure a length with a scale or any other device, we only get an approximate value. So, we may not realise that either c or d irrational So, either c or d is irrational and hence, \(\frac{c}{d}\) is irrational, i.e. π is irrational.

Question 4.
Represent √9.3 on the number line.
Answer:
RBSE Solutions for Class 9 Maths Chapter 1 Number Systems Ex 1.5 1
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 semi-circle at D. Then BD = √9.3. To represent √9.3 on the number line, let us treat the line BC as the number line, with B as 0 and C as 1. Draw an arc with centre B and radius BD, which intersects the number line at E. Then, E represents √9.3.