Real Numbers, Class 10, Mathematics - Solutions of NCERT (CBSE) Math Exercise 1.3

 


Class 10, NCERT (CBSE) solution of Mathematics  

Chapter 1, Real Numbers

NCERT solution for Math Textbook Exercise 1.3

(Page 14)
Q 1: Prove that √5 is irrational.
Solution:
Let us assume √5 is a rational number.
Then, we can find two positive integers a, b (b ≠ 0) such that



Suppose, a and b have a common factor other than 1. Then we can divide them by the common factor, and assume that a and b are co-prime.
or, a = √5b
or, a2 = 5b2
Therefore, a2 is divisible by 5 and so, it can be said that a is divisible by 5.
Now suppose,
a = 5k, where k is an integer.




From the above expression, b2 is divisible by 5 and hence, b is divisible by 5.
This implies that a and b have at least 5 as a common factor.
And this contradicts the fact that a and b are co-prime.
Hence, √5 is an irrational number.
Q2: Prove that 3 + 2√5 is irrational.
Solution:
Let us assume on the contrary that 3 + 2√5 is rational.
Therefore, there exist co-prime positive integers say a, b (b ≠ 0) such that




 






This contradicts the fact that √5 is irrational. So, our assumption that 3 + 2√5 is rational is false.
Therefore, 3 + 2√5 is an irrational number.
Q3: Prove that the following are irrationals:




Solution:
(i)
 



Therefore, we can have two integers a, b (b ≠ 0) such that









(ii) Let us assume on the contrary that 7√5 is rational.
Then, we can find two integers a, b (b ≠ 0) such that for some integers a and b







Therefore, √5 must be rational.
This contradicts to the fact that √5 is irrational.
Or, our assumption that 7√5 is rational is false. Hence, 7√5 is irrational.
(iii)
Let 6 + √2 be rational.
Now, let there be two integers a, b (b ≠ 0) such that

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