Real Numbers Exercise 1.2

                                                                        Exercise 1.2 

Q-1 Prove that √5 is irrational.

Solution:  To prove √5 is irrational, let us assume that √5 is a rational number. Therefore √5 will in form of p/q i.e.

√5 = p/q , -------------------(1)

where q ≠ 0 and p, q are some co-prime integers.

Note: if two numbers are co-prime to each other, its mean that they have only one common factor which is "1".

Square in both sides of equation (1) 

(√5)² = (p/q)²

=> 5=p²/q²

 => q²=p²/5--------------(2)

since q is an integer, it means p² will be divisible by 5 and hence p also will be divisible by 5, let p=5r for some integer " r ",

put p=5r in equation (2)

q²=( 5r )² / 5

=>q²=5r²

=>r²=q²/5 

(since, r is an integer and so r² will also be an integer, therefore q² is a multiple of 5 and therefor q also be a multiple of 5 ). 

Hence, we found that 5 is a common factor of p and q, which contradict our assumption that p/q are co-prime. Therefore, our assumption is wrong and √5 is not a rational number.

Hence it is proved that √5 is an irrational number.

Q(2) Prove that  3 +  2√5 is irrational.

Solution : to prove 3 +  2√5 is irrational let's assume that 3 +  2√5 is a rational number, 

i.e. let

( 3 +  2√5 ) = p/q -----------------(1) where q # 0 and p, q are both co prime numbers.

=>3-p/q = 2√5

=>(3/2) - (p/2q) = √5

since p and q are integers , therefore

(3/2) - (p/2q) is a rational number and therefore √5 is a rational number, which contradict the fact that √5 is an irrational number, so our assumption is wrong, therefore 3 + 2√5 is irrational number.

Q(3)  Prove that following are irrationals:

(i) 1/√2        (II)  7√5        (iii) 6+√2

Solution:

(i) 1/√2

let's assume that 1/√2 is a rational number , so

1/√2 = p/q , q # 0, for some co-prime integer p and q.

√2 = q/p  

√2 is irrational number which we can prove like in question 1, therefore 1/√2 is an irrational number.

(ii) 7√5

solution:  let 7√5 is a rational number so let's assume that:

7√5 = p/q , q # 0 , for some coprime integer p and q.

=> √5 = p / 7q

since p, q and 7 are integers, so p / 7q is a rational number and therefore √5 is a rational number, which contradict the fact that √5 is irrational, hence our assumption is wrong, therefore 7√5 is an irrational number.

(iii)    6+√2 

Solution: let 6+√2 is a rational no in the form of p/q , q # 0 for some co-prime integer p and q. i.e.

 6+√2 = p/q

=> √2 = p/q -6 

since p, q and 6 are integers, so (p/q - 6) is a rational number and so √2 is a rational number, which contradict the fact that √2 is irrational, so our assumption is wrong, therefore 6+√2 is irrational.

Please share your valuable feedback in comment