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

NCERT Solutions for Class 9 Maths Chapter 1 Number Systems Ex 1.5 are the part of NCERT Solutions for Class 9 Maths. Here you can find the NCERT Solutions for Class 9 Maths Chapter 1 Number System Ex 1.5.

Ex 1.5 Class 9 Maths Question 1.
Classify the following numbers as rational or irrational.

Solution:
(i) The difference between a rational number and an irrational number is always an irrational number.
∴ 2 – √5 is an irrational number.
(ii) (3 + √23) – √23 = 3 + √23 – √23 = 3
Here, 3 is a rational number. Thus, (3 + √23) – √23 is a rational number.
(iii) Since, 2√7/7√7 = 2/7, which is a rational number. Therefore, 2√7/7√7 is a rational number.
(iv) ∵ The numerator is a rational number and the denominator is an irrational number. The quotient of a rational divided by an irrational number is an irrational number.
∴ 1/√2 is an irrational number.
(v) ∵ 2π = 2 × π = Product of a rational number and an irrational number is an irrational number.
∴ 2π is an irrational number.

Ex 1.5 Class 9 Maths Question 2.
Simplify each of the following expressions.

Solution:
(i) (3 + √3)(2 + √2)
= 3(2 + √2) + √3(2 + √2)
= 6 + 3√2 + 2√3 + √6
Thus, (3 + √3)(2 + √2) = 6 + 3√2 + 2√3 + √6
(ii) (3 + √3)(3 – √3) = (3)² – (√3)²
= 9 – 3 = 6
Thus, (3 + √3)(3 – √3) = 6
(iii) (√5 + √2)² = (√5)² + (√2)² + 2(√5)(√2)
= 5 + 2 + 2√10 = 7 + 2√10
Thus, (√5 + √2)² = 7 + 2√10
(iv) (√5 – √2)(√5 + √2) = (√5)² – (√2)²

= 5 – 2 = 3
Thus, (√5 – √2) (√5 + √2) = 3

Ex 1.5 Class 9 Maths 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:
When we measure the length of a line with a scale or with any other device, we only get an approximate rational value, i.e., c and d both are approximate rational value.
∴ c/d is irrational and hence π is irrational.
Thus, there is no contradiction in saying that π is irrational.

Ex 1.5 Class 9 Maths Question 4.
Represent √9.3 on the number line.

Solution:
Draw a line segment AB = 9.3 units and extend it to C such that BC = 1 unit.
Find the mid-point of AC by bisecting the line segment AC and mark it as O.
Draw a semicircle taking O as centre and AO or OC as radius. Draw BD ⊥ AC.
Draw an arc taking B as centre and BD as radius meeting AC produced at E.

Now, BE = BD = √9.3 units