Question

P.T √5+√7 is an irrational number​

Answer 1

Answer:

Lets prove first√5 as an irrational number

Let √5 be an rational number

√5=p/q where p and q are integers and q not equals to zero and p and q have no common factors(except 1)

Squaring both the sides

5=p^2/q^2

p^2=5q^2 (1)

As 5 divides 5q^2,so 5 divides q^2 but 5 is prime

5 divides p

Let p=5k where k is an integer

Substituting the values of p in(1)

(5k)^2=5q^2

25k^2=5q^2

q^2=5k^2

As 5 divides 5k^2 so 5 divides k^2 but 5 is prime

5 divides q

Thus p and q have common factor 5

But,this contradicts the fact that p and q have no common factors(except 1)

Hence our supposition is wrong,√5 is an irrational number.

Lets prove√7 as irrational number

Let √7 be an rational number

√7=p/q where p and q are integers and q not equals to zero and p and q have no common factors(except 1)

Squaring both the sides

7=p^2/q^2

p^2=7q^2 (1)

As 7 divides 7q^2,so 7 divides q^2 but 7 is prime

7 divides p

Let p=7k where k is an integer

Substituting the values of p in(1)

(7k)^2=7q^2

49k^2=7q^2

q^2=7k^2

As 7 divides 7k^2 so 7 divides k^2 but 7 is prime

5 divides q

Thus p and q have common factor 7

But,this contradicts the fact that p and q have no common factors(except 1)

Hence our supposition is wrong,√7 is an irrational number.

Since,the sum of irrational number is always irrational

Hence,√5+√7 is an irrational number.