Question

Prove that the following are irrational :
i ) 1/√2
ii ) 7√5
iii) 6+√2​

Answer 1

CONCEPT :

The rational number is a number that is expressed as the ratio of two integers, where the denominator should not be equal to zero, whereas an irrational number cannot be expressed in the form of fractions. Rational numbers are terminating decimals but irrational numbers are non-terminating.

SOLUTION :

$\implies$Let us assume 1/√2 is a rational number

$\implies$Let us assume 1/√2 = r where r is a rational number

$\implies$On further calculation we get

1/r = √2

$\implies$Since r is a rational number, 1/r = √2 is also a rational number

$\implies$But we know that √2 is an irrational number

So our supposition is wrong.

$\mapsto$ Hence, 1/√2 is an irrational number.

(ii) 7√5

Solution:

$\implies$Let’s assume on the contrary that 7√5 is a rational number. Then, there exist positive integers a and b such that

$\implies$7√5 = a/b where, a and b, are co-primes

$\implies$√5 = a/7b

$\implies$ √5 is rational [∵ 7, a and b are integers ∴ a/7b is a rational number]

$\implies$This contradicts the fact that √5 is irrational. So, our assumption is incorrect.

$\mapsto$ Hence, 7√5 is an irrational number.

(iii) 6 + √2

Solution:

$\implies$Let’s assume on the contrary that 6+√2 is a rational number. Then, there exist co prime positive integers a and b such that

$\implies$6 + √2 = a/b

$\implies$ √2 = a/b – 6

$\implies$ √2 = (a – 6b)/b

$\implies$ √2 is rational [∵ a and b are integers ∴ (a-6b)/b is a rational number]

$\implies$This contradicts the fact that √2 is irrational. So, our assumption is incorrect.

$\mapsto$ Hence, 6 + √2 is an irrational number.

Answer 2

Step-by-step explanation:

°____________________°

• The rational number is a number that is expressed as the ratio of two integers, where the denominator should not be equal to zero, whereas an irrational number cannot be expressed in the form of fractions. Rational numbers are terminating decimals but irrational numbers are non-terminating.

SOLUTION :

\implies⟹ Let us assume 1/√2 is a rational number

\implies⟹ Let us assume 1/√2 = r where r is a rational number

\implies⟹ On further calculation we get

1/r = √2

\implies⟹ Since r is a rational number, 1/r = √2 is also a rational number

\implies⟹ But we know that √2 is an irrational number

So our supposition is wrong.

\mapsto↦ Hence, 1/√2 is an irrational number.

(ii) 7√5

Solution:

\implies⟹ Let’s assume on the contrary that 7√5 is a rational number. Then, there exist positive integers a and b such that

• \implies⟹ 7√5 = a/b where, a and b, are co-primes

• \implies⟹ √5 = a/7b

• \implies⟹ √5 is rational [∵ 7, a and b are integers ∴ a/7b is a rational number]

\implies⟹ This contradicts the fact that √5 is irrational. So, our assumption is incorrect.

\mapsto↦ Hence, 7√5 is an irrational number.

(iii) 6 + √2

Solution:

\implies⟹ Let’s assume on the contrary that 6+√2 is a rational number. Then, there exist co prime positive integers a and b such that

\implies⟹ 6 + √2 = a/b

\implies⟹ √2 = a/b – 6

\implies⟹ √2 = (a – 6b)/b

• \implies⟹ √2 is rational [∵ a and b are integers ∴ (a-6b)/b is a rational number]

• \implies⟹ This contradicts the fact that √2 is irrational. So, our assumption is incorrect.

• \mapsto↦ Hence, 6 + √2 is an irrational number

hope this helps you ✌️