Looking for accurate SEBA Class 9 Maths Exercise 1.4 (Number Systems) textual solutions for the 2026–27 academic year? Access complete step-by-step solutions covering operations on real numbers, simplifying radical expressions, rationalizing the denominator, applying identities for square roots, and solving laws of exponents for real numbers to score top marks in your annual exams.
EXERCISE 1.4
1. Classify the following numbers as rational or irrational:
(i) 2 – √5
= 2 – 2.236067…
= -0.236067… (Non-terminating non-repeating)
Therefore It is an irrational number.
(ii) (3 + √23) – √23
= 3 + √23 – √23
= 3
= 3/1
(iii) 2√7 / 7√7
= 2/7
Therefore It is a rational number.
(iv) 1/√2
= 1 × √2 / √2 × √2
= √2 / 2
Therefore It is an irrational number.
(v) 2π
Solution: Since π is irrational,
2 × irrational = irrational
Therefore It is an irrational number.
SEBA Class 9 Maths Exercise 1.4 Solutions
2. Simplify each of the following expressions:
(i) (3 + √3)(2 + √2)
= 3(2 + √2) + √3(2 + √2)
= 6 + 3√2 + 2√3 + √6
(ii) (3 + √3)(3 – √3)
= (3)² – (√3)²
= 9 – 3
= 6
(iii) (√5 + √2)²
= (√5)² + 2(√5)(√2) + (√2)²
= 5 + 2√10 + 2
= 7 + 2√10
(iv) (√5 – √2)(√5 + √2)
= (√5)² – (√2)²
= 5 – 2
= 3
(v) (3√5 – 4√3)²
= (3√5)² – 2(3√5)(4√3) + (4√3)²
= (9 × 5) – 24√15 + (16 × 3)
= 45 – 24√15 + 48
= 93 – 24√15
(vi) (√7 – 6)(√3 – √7)
= √7(√3 – √7) – 6(√3 – √7)
= √21 – 7 – 6√3 + 6√7
= √21 – 6√3 + 6√7 – 7
(vii) (2 + √6)(4 + √6)
= 2(4 + √6) + √6(4 + √6)
= 8 + 2√6 + 4√6 + 6
= 14 + 6√6
Class 9 Maths Chapter 1 Exercise 1.4 Question Answer SEBA 2026-27
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:
Let c be the circumference and d be the diameter of a circle.
π = c/d
When we measure length with a scale or any other device, we only get an approximate rational value.
Therefore, either c or d or both are irrational numbers, which makes c/d irrational.
Hence, there is no contradiction.
Rationalize the Denominator Class 9 SEBA Ex 1.4 Solutions
4. Represent √9.3 on the number line.
SEBA Class 9 Maths Chapter 1 Exercise 1.4 Textbook Answers
5. Rationalise the denominators of the following:
(i) 1/√7
(ii) 1/(√7 – √6)
(iii) 1/(√5 + √2)
(iv) 1/(√7 – 2)
(v) 13/√5
(vi) (6 + √5)/(6 – √5)
(vii) (5 + √3)/(3 + √6)
(viii) (2√3 – 3√2)/(3√2 + 2√3)
(ix) (√(a+x) – √(a-x))/(√(a+x) + √(a-x))
6. (i) If x = 1 + √2, then show that (x – 1/x)³ = 8
(ii) Simplify: 1/(1 + √2) + 1/(√2 + √3) + 1/(√3 + √4)
(iii) Add the rational numbers 1.27 with 7 recurring and 0.13 with 3 recurring.
7. 4 + √5 and 4 – √5 are two irrational numbers.
(P) (4 + √5) + (4 – √5) is rational number
(Q) (4 + √5) × (4 – √5) is rational number
Which is correct?
(A) P is true, Q is false
(B) P is false, Q is true
(C) Both P and Q are true
(D) Both P and Q are false
Solution: Given irrational numbers: 4 + √5 and 4 – √5
(P) (4 + √5) + (4 – √5)
= 4 + 4 + √5 – √5
= 8 (which is a Rational number)
Therefore P is true.
(Q) (4 + √5) × (4 – √5)
= (4)² – (√5)²
= 16 – 5
= 11 (which is a Rational number)
Therefore Q is true.
Correct Option: (C) Both P and Q are true
Class 9 Maths Laws of Exponents Real Numbers Solutions
📐 Updated Solutions Notice: This page features complete, step-by-step SEBA Class 9 Mathematics Chapter 1 (Number Systems) Exercise 1.4 Textual Solutions updated for the 2026–27 academic session. All operations on real numbers, radical identity simplifications, rationalization of denominators, and exponent rules(a^p * a^q = a^(p+q), (a^p)^q = a^(pq)) strictly follow the latest revised Assam Board (SEBA) curriculum.
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