Prove that 6 is an irrational number.

Let 6 is an rational number, then 6 = p q , where p , q are integers, q ≠ 0 and p , q have no common factors (Except 1 ) ⇒ 6 = p 2 q 2 p 2 = 6 q 2 . . . . . . . ( i ) As 2 divides 6 q 2 , So 2 divides p 2 and 2 is prime ⇒ 2 divides p . Let p = 2 k where k is some integer. Substitute this value of p in E q ( i ) we have: 2 k 2 = 6 q 2 4 k 2 = 6 q 2 3 q 2 = 2 k 2 As 2 divides 2 k 2 , so 2 divides 3 q 2 ⇒ 2 divides 3 or q 2 , but 2 does not divide 3 . ⇒ 2 divides q 2 That is 2 divides q . Thus, p and q have a common factor 2 . This contradicts that p and q have no common fact (Except 1 ) . Therefore, 6 is not a rational number. Hence, 6 is an irrational number.

Prove that 6 is an irrational number.

Let 6 is an rational number, then 6 = p q , where p , q are integers, q ≠ 0 and p , q have no common factors (Except 1 ) ⇒ 6 = p 2 q 2 p 2 = 6 q 2 . . . . . . . ( i ) As 2 divides 6 q 2 , So 2 divides p 2 and 2 is prime ⇒ 2 divides p . Let p = 2 k where k is some integer. Substitute this value of p in E q ( i ) we have: 2 k 2 = 6 q 2 4 k 2 = 6 q 2 3 q 2 = 2 k 2 As 2 divides 2 k 2 , so 2 divides 3 q 2 ⇒ 2 divides 3 or q 2 , but 2 does not divide 3 . ⇒ 2 divides q 2 That is 2 divides q . Thus, p and q have a common factor 2 . This contradicts that p and q have no common fact (Except 1 ) . Therefore, 6 is not a rational number. Hence, 6 is an irrational number.

Prove that 1 11 is an irrational number.

Let 1 11 is an rational number, then 1 11 = p q , where p , q are integers, q ≠ 0 and p , q have no common factors (Except 1 ) ⇒ 1 11 = p 2 q 2 11 p 2 = q 2 . . . . . . . ( i ) As 11 divides 11 p 2 , So 11 divides p 2 and 11 is prime ⇒ 11 divides p . Let q = 11 k where k is some integer. Substitute this value of q in E q ( i ) we have: 11 k 2 = 11 p 2 121 k 2 = 11 p 2 p 2 = 11 k 2 As 11 divides 11 k 2 , so 11 divides q 2 ⇒ 11 is a prime. ⇒ 11 divides q Thus, p and q have a common factor 11 . This contradicts that p and q have no common fact (Except 1 ) . Therefore, 1 11 is not a rational number. Hence, it is proved that 1 11 is an irrational number.

Prove that 2 is an irrational number. Hence, show that 3 - 2 is an irrational number.

Let 2 is an rational number, then 2 = p q , where p , q are integers, q ≠ 0 and p , q have no common factors (Except 1 ) ⇒ 2 = p 2 q 2 2 q 2 = p 2 . . . . . . . ( i ) As 2 divides 2 q 2 , So 2 divides q 2 and 2 is prime ⇒ 2 divides q . Let p = 2 k where k is some integer. Substitute this value of p in E q ( i ) we have: 2 k 2 = 2 q 2 4 k 2 = 2 q 2 q 2 = 2 k 2 As 2 divides 2 k 2 , so 2 divides p 2 ⇒ 2 is a prime. ⇒ 2 divides p Thus, p and q have a common factor 2 . This contradicts that p and q have no common fact (Except 1 ) . Therefore, 2 is not a rational number. It is irrational. Let 3 - 2 is rational number equal to r . ⇒ 3 - 2 = r ⇒ 3 - r = 2 If 3 , r are rational numbers, then 2 is also rational number. This contradicts the fact that 2 is irrational. Thus, we get 3 - 2 is irrational. Hence, it is proved that 2 and 3 - 2 are irrational.

Prove that 3 is an irrational number. Hence, show that 2 5 3 is an irrational number.

Let 3 is an rational number, then 3 = p q , where p , q are integers, q ≠ 0 and p , q have no common factors (Except 1 ) ⇒ 3 = p 2 q 2 3 q 2 = p 2 . . . . . . . ( i ) As 3 divides 3 q 2 , So 3 divides q 2 and 3 is prime ⇒ 3 divides q . Let p = 3 k where k is some integer. Substitute this value of p in E q ( i ) we have: 3 k 2 = 3 q 2 9 k 2 = 3 q 2 q 2 = 3 k 2 As 3 divides 3 k 2 , so 3 divides p 2 ⇒ 3 is a prime. ⇒ 3 divides p Thus, p and q have a common factor 3 . This contradicts that p and q have no common fact (Except 1 ) . Therefore, 3 is not a rational number. It is irrational. Let 2 5 3 is rational number equal to r . ⇒ 2 5 3 = r ⇒ 5 2 r = 3 If 5 2 , r are rational numbers, then 3 is also rational number. This contradicts the fact that 3 is irrational. Thus, we get 2 5 3 is irrational. Hence, it is proved that 3 and 2 5 3 are irrational.

Prove that 5 is an irrational number. Hence, show that - 3 + 2 5 is an irrational number.

Let 5 is an rational number, then 5 = p q , where p , q are integers, q ≠ 0 and p , q have no common factors (Except 1 ) ⇒ 5 = p 2 q 2 5 q 2 = p 2 . . . . . . . ( i ) As 5 divides 5 q 2 , So 5 divides q 2 and 5 is prime ⇒ 5 divides q . Let p = 5 k where k is some integer. Substitute this value of p in E q ( i ) we have: 5 k 2 = 5 q 2 25 k 2 = 5 q 2 q 2 = 5 k 2 As 5 divides 5 k 2 , so 5 divides p 2 ⇒ 5 is a prime. ⇒ 5 divides p Thus, p and q have a common factor 5 . This contradicts that p and q have no common fact (Except 1 ) . Therefore, 5 is not a rational number. It is irrational. Let - 3 + 2 5 is rational number equal to r . ⇒ - 3 + 2 5 = r ⇒ r + 3 2 = 5 If 3 , 2 , r are rational numbers, then 5 is also rational number. This contradicts the fact that 5 is irrational. Thus, we get - 3 + 2 5 is irrational. Hence, it is proved that 5 and - 3 + 2 5 are irrational.

Prove that the following number is irrational: 2 3 - 7

Let 2 3 - 7 is rational number equal to r . ⇒ 2 3 - 7 = r ⇒ r + 7 2 = 3 If 2 , 7 , r are rational numbers, then 3 is also rational number. This contradicts the fact that 3 is irrational. Thus, we get 2 3 - 7 is irrational. Hence, it is proved that 2 3 - 7 are irrational.