Prove: 23 + 528 = 2 3 + 11

Given, 23 + 528 = 2 3 + 11 LHS, 23 + 528 = 23 + 4 2 × 3 × 11 = 23 + 4 33 = 12 + 11 + 4 33 = 12 2 + 2 × 12 × 11 + 11 2 = 12 + 11 2 = 12 + 11 = 2 3 + 11 = RHS Hence proved

The number N = 5 3 + 50 5 - 24 75 - 5 2 when simplified reduces to :

Rational which is not an integer

Prove that the number 2 - 3 5 is irrational.

Assuming 2 - 3 5 to be rational. Hence 2 and 2 - 3 5 are rational. ∴ 2 - ( 2 - 3 5 ) = 2 - 2 + 3 5 = 3 5 , which is supposed to be rational number [ ∵ Difference of two rationals is rational] 1 3 × 3 5 = 5 , which is supposed to be a rational number [ ∵ Product of two rational is rational] This contradicts the fact that 5 is irrational. The contradiction arises by assuming 2 - 3 5 is rational. Hence, 2 - 3 5 is irrational.

Prove that 2 3 - 1 is an irrational number.

Assuming x = 2 3 − 1 be a rational number. x = 2 3 − 1 x 2 = ( 2 3 − 1 ) 2 x 2 = ( 2 3 ) 2 + ( 1 ) 2 − 2 ( 2 3 ) ( 1 ) x 2 = 12 + 1 − 4 3 x 2 − 13 = − 4 3 13 − x 2 4 = 3 Since x is rational number, x 2 is also a rational number. 13 − x 2 is a rational number 13 − x 2 4 is a rational number 3 is a rational number But 3 is an irrational number, which is a contradiction. Hence, our assumption is wrong. Thus, ( 2 3 − 1 ) is an irrational number.

Prove that the number 6 is irrational.

Assuming 6 to be rational. We know that, 6 = 3 × 2 Hence, 3 , 2 are both rational. This contradicts the fact that 3 , 2 are irrational. The contradiction arises by assuming 6 is rational. Hence, 6 is irrational.

Show that 2 3 is irrational.

Assuming 2 3 to be a rational number. ∴ 2 3 = p q where p and q are some integers and q ≠ 0 , also HCF ( p , q ) = 1 . . . ( 1 ) ⇒ 2 q = 3 p ⇒ ( 2 q ) 2 = ( 3 p ) 2 ⇒ 2 q 2 = 9 p 2 p 2 is divisible by 2 p is divisible by 2 . . . ( 2 ) Let p = 2 m , where m is some integer. ⇒∴ 2 q = 3 p ⇒ 2 q = 3 ( 2 m ) ⇒ ( 2 q ) 2 = 3 ( 2 m ) 2 ⇒ 2 q 2 = 4 9 p 2 ⇒ q 2 = 2 9 p 2 ⇒ q 2 is divisible by 2 q is divisible by 2 . . . ( 3 ) From 2 and 3 , 2 is a common factor of both p and q , which contradicts 1 . Hence, our assumption is wrong. Thus, 2 3 is irrational.

Prove that the number 3 5 is irrational:

Assuming 3 5 to be rational. ∴ 1 3 × 3 5 = 1 5 which is supposed to be a rational number [ ∵ Product of two rationals is rational] This contradicts the fact that 1 5 is irrational. ∴ 1 × 5 5 × 5 = 1 5 5 So, if 1 5 is rational, then 1 5 5 is rational ∴ 5 × 1 5 5 = 5 which is supposed to be a rational number [ ∴ Product of two rationals is rational ] But, 5 is irrational. Hence, 1 5 is irrational The contradiction arises by assuming 3 5 is rational. Hence, 3 5 is irrational.

Prove that 2 7 is irrational.

Given, 2 7 = 2 7 × 7 7 = 2 7 7 Let 2 7 7 be a rational number. ∴ 2 7 7 = p q , where p and q are some integers and HCF ( p , q ) = 1 . . . . ( 1 ) 2 7 q = 7 p ( 2 7 q ) 2 = ( 7 p ) 2 7 4 q 2 = 49 p 2 4 q 2 = 7 p 2 q 2 is divisible by 7 q is divisible by 7 . . . . ( 2 ) Let q = 7 m , where m is some integer. ∴ 2 7 q = 7 p [ 2 7 ( 7 m ) ] 2 = ( 7 p ) 2 343 4 m 2 = 49 p 2 7 4 m 2 = p 2 p 2 is divisible by 7 p is divisible by 7 . . . . ( 3 ) From 2 and 3 , 7 is a common factor of both p and q , which contradicts 1 . Hence, our assumption is wrong. Thus, 2 7 is irrational.

Prove that the number 3 7 is irrational.

Assuming 3 7 to be rational. Then, 1 3 × 3 7 = 7 which is supposed to be a rational number [∵Product of two rational is rational] This contradicts the fact that 7 is irrational. The contradiction arises by assuming 3 7 is rational. Hence, 3 7 is irrational.

Prove that 4 - 5 2 is an irrational number.

Assuming 4 - 5 2 to be a rational number. x = 4 - 5 2 x 2 = ( 4 - 5 2 ) 2 x 2 = ( 4 2 + ( 5 2 ) 2 - 2 ( 4 ) ( 5 2 ) ) ⇒ x 2 = 16 + 50 - 40 2 x 2 - 66 = - 40 2 ⇒ 66 - x 2 40 = 2 Since x is a rational number, x 2 is also a rational number 66 - x 2 is a rational number 66 - x 2 40 is a rational number But 2 is an irrational number, which is a contradiction. Hence, our assumption is wrong. Thus, 4 - 5 2 is an irrational number.