Question

10th class 1 chapter real numbers exercise 1.3 solution for cbse

Answer 1

Real Number

NextPreviouschapter List

Exercise 1.3 (NCERT)

Question 1: Prove that  is irrational.

Answer: Let us assume the contrary, i.e.  is irrational.

Thus, there can be two integers a and b (b≠0) and a and b are coprime so that;



Squaring on both sides, we get;



This means that a2 is divisible by 5 and hence a is also divisible by 5.

This contradicts our earlier assumption that a and b are coprime, because we have found 5 as at least one common factor of a and b.

This also contradicts our earlier assumption that  is irrational.


Question 2: Prove that  is irrational.

Answer: Let us assume to the contrary, i.e.  is irrational.

Thus, there can be two integers a and b (b≠0) and a and b are coprime so that;




Since a and b are rational, so  is rational and hence,  is rational.

But this contradicts the fact that  is irrational.

This happened because of our faulty assumption.



Question 3: Prove that following are irrationals:



Answer: Let us assume to the contrary, i.e.  is rational.

Thus, there can be two integers a and b (b≠0) and a and b are coprime so that;



Squaring on both sides, we get;

\

This means that b2 is divisible by 2 and hence a is also divisible by 2.

This contradicts our earlier assumption that a and b are co-prime, because 2 is at least one common factor of a and b.

This also contradicts our earlier assumption that  is rational.

Hence,  is irrational proved.



Answer: Let us assume to the contrary, i.e.  is rational.

There can be two integers a and b (b≠0) and a and b are coprime, so that;



Squaring on both sides, we get;



This means that a2 is divisible by 245; which means that a is also divisible by 245.

This contradicts our earlier assumption that a and b are coprime, because 245 is at least one common factor of a and b.

This happened because of our faulty assumption and hence,  is irrational proved.



Answer: Let us assume to the contrary, i.e.  is rational.

Thus, there can be two integers a and b (b≠0) and a and b are coprime so that;



Since a and b are rational, so  is rational and hence,  is rational.

But this contradicts the fact that  is irrational.

This happened because of our faulty assumption.

Hence,  is irrational proved.




Hope it will help you

Answer 2

Answer:

prove that √5 is an irrational number.

Assume √5 is a rational number

Let √5 = p/q where p and q is co-prime

Then √5 =p =→ 5q2 =p2

Since, 5 divides p2, so it will divide p also

let p =5r

Then p2-25r (squaring on both sides)

→ 5q2 = 25r2

→5q2 =25r2. (from I)

→q2 = 5r

Since 5 divides q2 , so it will divide q also. Thus , 5 is a common factor of both p and q .

This contradicts our assumption that √5 is an irrational number

hence √5 is irrational .Hence proved