Looking for accurate SEBA Class 10 Maths Exercise 1.2 (Real Numbers) textual solutions for the 2026–27 academic year? Access complete step-by-step solutions covering proof of irrationality for numbers like √2, √3, √5, 3 + 2√5, and 1/√2 using the method of contradiction to score top marks in your HSLC board exams.
Q1. Prove that √5 is irrational.
Solution:
Given: The number √5.
To Prove: √5 is an irrational number.
Solution: We will prove this by the method of contradiction. Let us assume that √5 is a rational number.
Since it is a rational number, it can be written in the form of a/b, where:
a and b are integers.
b is not equal to 0.
a and b are co-prime (they have no common factors other than 1).
Therefore, we can write:
√5 = a / b
a = b√5
Squaring both sides of the equation:
a² = (b√5)²
a² = 5b² —- (Equation 1)
From Equation 1, we can see that 5 divides a². According to the theorem, if a prime number divides a², then it must also divide a.
Therefore, 5 divides a.
Since 5 divides a, we can write a as 5c for some integer c:
a = 5c
Now, substitute the value of a = 5c into Equation 1:
(5c)² = 5b²
25c² = 5b²
Dividing both sides by 5:
5c² = b²
b² = 5c² —- (Equation 2)
From Equation 2, we can see that 5 divides b². Following the same theorem, if a prime number divides b², it must also divide b.
Therefore, 5 divides b.
Conclusion: We found that 5 divides both a and b. This means a and b have at least 5 as a common factor.
However, this contradicts our initial assumption that a and b are co-prime (having no common factors other than 1). This contradiction has arisen because of our incorrect assumption that √5 is a rational number.
Answer: Therefore, by contradiction, √5 is an irrational number. Hence Proved.
Q2. Prove that 3 + 2√5 is irrational.
Solution:
Given: The expression 3 + 2√5.
We already know that √5 is an irrational number.
To Prove: 3 + 2√5 is an irrational number.
Solution: We will prove this by the method of contradiction. Let us assume that 3 + 2√5 is a rational number.
Since it is assumed to be rational, it can be written in the form of a/b, where:
a and b are integers.
b is not equal to 0.
a and b are co-prime.
Therefore, we can write:
3 + 2√5 = a / b
Rearranging the equation to isolate the term with the square root:
2√5 = (a / b) – 3
Taking the common denominator on the right side:
2√5 = (a – 3b) / b
Dividing both sides by 2:
√5 = (a – 3b) / 2b
Conclusion: Since a and b are integers, the expression (a – 3b) / 2b must be a rational number.
According to our equation, this means √5 must also be a rational number. However, this contradicts the well-established fact that √5 is an irrational number.
This contradiction has arisen because of our incorrect assumption that 3 + 2√5 is a rational number.
Answer: Therefore, by contradiction, 3 + 2√5 is an irrational number. Hence Proved.
Class 10 Maths Chapter 1 Exercise 1.2 Question Answer SEBA 2026-27
Q3. Prove that the following are irrationals:
i) 1 / √2
Soln:-
Prove that 1 / √2 is irrational.
Given: A real number 1 / √2.
To Prove: 1 / √2 is an irrational number.
Solution: Let us assume, to the contrary, that 1 / √2 is a rational number.
Therefore, we can find co-prime integers a and b (where b is not equal to 0) such that:
1 / √2 = a / b
By rearranging the terms (taking the reciprocal on both sides), we get:
√2 = b / a
Since a and b are integers, b / a is a rational number.
This implies that √2 must also be a rational number.
However, this contradicts the well-known fact that √2 is irrational.
This contradiction has arisen because of our incorrect assumption that 1 / √2 is rational.
Hence, we conclude that 1 / √2 is an irrational number. Hence Proved.
ii) 7√5
Soln:-
Prove that 7√5 is irrational.
Given: A real number 7√5.
To Prove: 7√5 is an irrational number.
Solution: Let us assume, to the contrary, that 7√5 is a rational number.
Therefore, we can find co-prime integers a and b (where b is not equal to 0) such that:
7√5 = a / b
By rearranging the terms (dividing both sides by 7), we get:
√5 = a / (7b)
Since a, 7, and b are integers, a / (7b) is a rational number.
This implies that √5 must also be a rational number.
However, this contradicts the well-known fact that √5 is irrational.
This contradiction has arisen because of our incorrect assumption that 7√5 is rational.
Hence, we conclude that 7√5 is an irrational number. Hence Proved.
iii) 6 + √2
Soln:-
Prove that 6 + √2 is irrational.
Given: A real number 6 + √2.
To Prove: 6 + √2 is an irrational number.
Solution: Let us assume, to the contrary, that 6 + √2 is a rational number.
Therefore, we can find co-prime integers a and b (where b is not equal to 0) such that:
6 + √2 = a / b
By rearranging the terms (subtracting 6 from both sides), we get:
√2 = (a / b) – 6 √2 = (a – 6b) / b
Since a and b are integers, (a – 6b) / b is a rational number.
This implies that √2 must also be a rational number.
However, this contradicts the well-known fact that √2 is irrational.
This contradiction has arisen because of our incorrect assumption that 6 + √2 is rational.
Hence, we conclude that 6 + √2 is an irrational number. Hence Proved.
SEBA Class 10 Maths Exercise 1.2 Solutions
Q4. The product of a non-zero rational number and an irrational number is
(A) always irrational
(B) always rational
(C) always Integer
(D) rational or irrational
Answer
Given:
A non-zero rational number.
An irrational number.
To Find:
The nature of the product of these two numbers .
Solution:
Let us assume the opposite is true, and the product is a rational number.
If we divide this rational product by our original non-zero rational number, we should get our irrational number back.
However, dividing any rational number by another non-zero rational number must always result in a rational number.
This means our irrational number would have to be rational, which is impossible.
Because this creates an impossible contradiction, the product can never be rational. It must always be irrational.
Correct Option:
always irrational
Real Numbers Class 10 SEBA Ex 1.2 Solutions
Q5. √5 + √3 + 2 is
(A) a natural number
(B) an integer
(C) a rational number
(D) an irrational number
Answer
Given:
The expression: √5 + √3 + 2
To Find:
The nature of the given number (Options A, B, C, or D).
Solution:
We know that √5 and √3 are irrational numbers because they cannot be written as exact fractions.
The sum of two irrational numbers (√5 + √3) results in an irrational number.
Adding a rational number (2) to an irrational number still results in an irrational number.
Therefore, the entire expression √5 + √3 + 2 is an irrational number.
Correct Option:
(D) an irrational number
Q6. Assertion (A) √2 + √5 is an irrational number
Reason (R) : If p and q are prime positive integers, then √p + √q is an irrational number
(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation for (A)
(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation for (A)
(C) Assertion (A) is true but Reason (R) is false
(D) Assertion (A) is false but Reason (R) is true
Answer
Given:
Assertion (A): √2 + √5 is an irrational number
Reason (R): If p and q are prime positive integers, then √p + √q is an irrational number
To Find:
The correct option among A, B, C, and D.
Solution:
Assertion (A) is true because both 2 and 5 are prime numbers, and the sum of their square roots (√2 + √5) cannot be expressed as a simple fraction.
Reason (R) is also true as a general mathematical rule: the sum of the square roots of any two distinct prime numbers is always irrational.
Since √2 and √5 fit this exact rule perfectly, Reason (R) is the direct and correct explanation for Assertion (A).
Correct Option:
Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation for (A)
Prove Root 5 is Irrational Class 10 SEBA
Q7. Assertion (A) : √a is an irrational number, when a is a prime number.
Reason (R) : Square root of any prime number is an irrational number.
(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation for (A)
(B) Both Assertion (A) and Reason (R) are true and Reason (R) is not the correct explanation for A.
(C) Assertion (A) is true but Reason (R) is false.
(D) Assertion (A) is false but Reason (R) is true.
Answer
Given:
Assertion (A): √a is an irrational number, when a is a prime number.
Reason (R): Square root of any prime number is an irrational number.
To Find:
The correct option among A, B, C, and D.
Solution:
Assertion (A) is true because if a number ‘a’ is prime, its square root cannot be simplified into a whole number or a fraction, making it irrational.
Reason (R) is also true because it states a fundamental rule of mathematics: the square root of any prime number is always irrational.
Because Reason (R) is just a restatement of the exact rule that makes Assertion (A) true, it serves as the correct explanation.
Correct Option:
Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation for (A)
Q8. √2, √3, √5, √6, √7, √8, √10 are all irrationals
Which pair among them is like irrationals?
(A) √3, √6
(B) √8, √10
(C) √2, √8
(D) √7, √8
Answer:-
Given:
A list of irrational numbers: √2, √3, √5, √6, √7, √8, √10
Four options representing pairs of these numbers.
To Find:
The pair among them that represents “like irrationals” (or like radicals).
Solution:
Like irrationals are numbers that have the exact same irrational factor when they are simplified into their lowest terms.
Let us simplify the pairs to check their irrational parts:
For Option (C): √2 is already in its simplest form.
For √8, we can simplify it by breaking it into prime factors: √8 = √(4 * 2) = 2√2.
Since both √2 and 2√2 share the exact same irrational radical factor (√2), they are considered like irrationals.
Correct Option:
(C) √2, √8
Class 10 Maths Exercise 1.2 Irrational Numbers Proofs
📐 Updated Solutions Notice: This page features complete, step-by-step SEBA Class 10 Mathematics Chapter 1 (Real Numbers) Exercise 1.2 Textual Solutions updated for the 2026–27 academic session. All irrationality proofs, method of contradiction steps, and coprime assumption derivations strictly follow the latest revised Assam Board (SEBA/ASSEB) curriculum.
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