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- Class : 10th
- Chapter : 1
- Subject : Mathematics
- Topic : Real Number Exercise 1.2
- Resource : Notes, Important Question & Practice Paper
Chapter 1 Real Number Exercise 1.2
Q.1 Prove that √5 is irrational.
Solution :- Let us assume, to the contrary, that √5 is rational.
That, we can find integrate a and b (≠ 0) such that √5 = a/b
Suppose a and b have a common factor other than 1, then we can divide by the common factor, and assume that a and b are coprime.
So, b√5 = a.
Squaring on both sides and rearranging we get 5b² = a².
Therefore a² is divisible by 5, and by theorem 1.3 follows that a is also divisible by 5.
So, we can write a = 5c for some integer c.
Substituting for a, we get 5b² = 25c², that
b² = 5c².
This means that b² is divisible by 3, and so b also divisible by 5 (using Theorem 1.3 with p = 5).
Therefore a and b have at least 5 as a common factor.
But this contradiction has arisen because of our incorrect assumption that √5 is rational.
So, we conclude that √5 is irrational.
Q.2 Prove that 3+ 2√5 is irrational.
Solution :- We have to prove 3 + √5 is irrational
Let us assume the opposite,
i.e., 3 + √5 can be written in the form a/b
Where a and b (b ≠ 0) are coprime (no common factor other than 1)
Hence,
3 + 2√5 = a/b
2√5 = a/b – 3
2√5 = a – 3b/ b
2√5 = 1/2 × a – 3b/b
√5 = a- 3b/2b
Here, a – 3b is rational number But √5 is irrational
Since,
Rational ≠ Number
This is contradiction
Our assumption is incorrect
Therefore, 3 + 2√5 is irrational
Hence Proved,
Q.3 Prove that the following are irrationals :
(i). 1/√2
(ii). 7√5
(iii). 6+√2
(i). 1/√2
Solution :-
We have to prove 1/√2 is irrational
Let us assume the opposite,
i.e, 1/√2 is rational
Hence, 1/√2 can be written in the form a/b
Where a and b (b ≠ 0) are co-prime (no common factor other than 1)
Hence,
1/√2 = a/b
b/a = √2
Here, b/a is rational number
But √2 is irrational
Since,
Rational ≠ irrational
This is a contradiction
Our assumption is incorrect
Therefore, 1√2 is irrational
Hence Proved
(ii). 7√5
Solution :-
We have to prove 7√5 is irrational
Let us assume the opposite,
i.e., 7√5 is rational
Hence, 7√5 can be written in the form a/b
Where a and b (b ≠ 0) are co-prime (no common factor other than 1)
Hence,
7√5 = a/b
√5 = 1/7 × a/b
Here, a/7b is a rational number
But √5 is irrational
Since,
Rational ≠ irrational
This is a contradiction
Our assumption is incorrect
Therefore, 7√5 is irrational
Hence Proved
(iii). 6+√2
Solution :-
We have to prove 6+√2 is irrational
Let us assume the opposite,
i.e., 6+√2 is rational
Hence, 6+√2 can be written in the form a/b
Where a and b (b ≠ 0) are co-prime (no common factor other than 1)
Hence,
6+√2 = a/b
√2 = a/b – 6
Here, a – 6b/b is a rational number
But √5 is irrational
Since,
Rational ≠ irrational
This is a contradiction
Our assumption is incorrect
Therefore, 6+√2 is irrational
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