Ncert Math chapter 1 solutions class 9th exercises 1.5 –
1.1 Operations on Real Numbers:
You have learnt, in earlier classes, that rational numbers satisfy the commutative, associative and distributive laws for addition and multiplication. Moreover, if we add, subtract, multiply or divide (except by zero) two rational numbers, we still get a rational number (that is, rational numbers are “closed” with respect to addition, subtraction, multiplication and division). It turns out that irrational numbers also satisfy the commutative, associative and distributive laws for addition and multiplication. However, the sum, difference, quotients and products of irrational numbers are not always irrational.
For example, (√6) + (-√6) , (√2)-(√2), (√3)·(√3) and √17/√17 are rationals.
Let us look at what happens when we add and multiply a rational number with an irrational number. For example, √3 is irrational. What about 2+ √3 and 2√3? Since √3 has a non-terminating non-recurring decimal expansion, the same is true for 2+√3 and 2√3. Therefore, both 2+ √3 and 2√3 are also irrational numbers.
EXERCISE 1.5
Questions: 1
Classify the following numbers as rational or irrational;
- (1) 2-√5
- (2) (3+√23) -√23
- (3) 2√7/7√7
- (4) 1/√2 (5)
Solution:
Questions: 2
Simplify each of the following expressions
- (1) (3+√3)(2+√2)
- (2) (3+√3) (3-√3)
- (3) (√5+ √2)²
- (4) (√5-√2)(√5+√2)
Solution:
Questions:3
Recall, it is defined as the ratio of the circumference (say c) of a to its diameter
(say d). That is = This seems to contradict the fact th sirrational. How will bpubched
you resolve this contradiction?
Solution:
Question 4:
Represent √9.3 on the number line.
Solution:
Question 5:
Rationalise the denominators of the following:
(1) 1/√7 (2) 1/√7-√6 (3) 1/√5+√2 (4) 1/√7-2
Solution:
Class 9th Ncert book chapter 1 exercises 1.5 all example:-
Example 1:
Check whether 7√5.√2+21, -2 are irrational numbers or not.
Solution:
√52.236… √2 = 1.4142…, π = 3.1415…
Example 2:
Add 2√2+5√3 and √2-3√3.
Solution:
(2√2+5√3)+(√2 −3√3) = (2√2 + √2) + (5√3-3√3)
= (2+1)√2+(5-3)√3- 2+2√3
Example 3:
Multiply 6√5 by 2√5.
Solution:
6√5 x 2√3 =6×2× √5 × √5 = 12x 5 = 60
Example 4:
Divide 8√15 by 2√3.
Solution:
8√15÷2√3- 8√3× √5 2√3 4√5
