Question

Prove that following numbers are irrationals:
(i)2/√7 (ii) 3/2√5
(iii)4+√2 (iv) 5√2

Answer 1

Proven irrational

Step-by-step explanation:

(i) 2/√7

Let us assume 2/√7 is rational. Then it can be expressed as a/b where a and b are co-primes and b is not zero.

a / b = 2/√7

So a / 2b = √7

where a/2b is rational and √7 is irrational.

rational can't be equal to irrational.

So our assumption is incorrect.

So 2/√7 is irrational. Hence Proved.

(ii) 3/2√5

Let us assume 3/2√5 is rational. Then it can be expressed as a/b where a and b are co-primes and b is not zero.

a / b = 3/2√5

So 2a/3b = √5

where 2a/3b is rational and √5 is irrational.

rational can't be equal to irrational.

So our assumption is incorrect.

So 3/2√5 is irrational. Hence Proved.

(iii) 4+√2

Let us assume 4+√2 is rational. Then it can be expressed as a/b where a and b are co-primes and b is not zero.

a / b =  4+√2

So a/b - 4 = √2

where a/b - 4 is rational and √2 is irrational.

rational can't be equal to irrational.

So our assumption is incorrect.

So 4+√2 is irrational. Hence Proved.

(iv) 5√2

Let us assume 5√2 is rational. Then it can be expressed as a/b where a and b are co-primes and b is not zero.

a / b =  5√2

So a/5b = √2

where a/5b is rational and √2 is irrational.

rational can't be equal to irrational.

So our assumption is incorrect.

So 5√2 is irrational. Hence Proved.