Find the HCF and LCM of 8 9 , 10 27 and 16 81

Given fractions are 8 9 , 10 27 , 16 81 HCF ( 8 9 , 10 27 , 16 81 ) = H C F o f t h e n u m e r a t o r s L C M o f t h e d e n o min a t o r s = H C F ( 8 , 10 , 16 ) L C M ( 9 , 27 , 81 ) Performing prime factorisation 8 = 2 3 10 = 2 × 5 16 = 2 4 So, the HCF ( 8 , 10 , 16 ) = 2 and LCM ( 9 , 27 , 81 ) = 3 4 = 81 Thus, HCF 8 9 , 10 27 , 16 81 = 2 81 And, LCM 8 9 , 10 27 , 16 81 = L C M o f n u m e r a t o r s H C F o f d e n o m i n a t o r s = L C M 8 , 10 , 16 H C F 9 , 27 , 81 Performing prime factorisation we get, 9 = 3 2 27 = 3 3 81 = 3 4 So, the HCF ( 9 , 27 , 81 ) = 3 2 = 9 and LCM ( 8 , 10 , 16 ) = 80 Thus, L C M 8 9 , 10 27 , 16 81 = 80 9 ⇒ H C F ( 8 9 , 10 27 , 16 81 ) = 2 81 and L C M ( 8 9 , 10 27 , 16 81 ) = 80 9

Prove that 3 is an irrational number.

Consider that 3 be a rational and let its simplest form be a b . Then, a and b are integers having no common factor other than 1 , and b ≠ 0 . Now, 3 = a b ⇒ 3 = a 2 b 2 . .... (On squaring both sides) ⇒ 3 b 2 = a 2 ⇒ 3 divides a 2 . . . . [ ∵ d i v i d e s 3 b 2 ] ⇒ 3 divides a .... [ ∵ 3 is prime and 3 divides a 2 ⇒ 3 divides a ] Let a = 3 c for some integer c. Putting a = 3 c in (i) we get 3 b 2 = 9 c 2 ⇒ b 2 = 3 c 2 ⇒ 3 divides b 2 . . . . [ ∵ 3 divides 3 c 2 ] ⇒ 3 divides b .... [ ∵ 3 is prime and 3 divides a 2 ⇒ 3 divides a] Thus, 3 is a common factor of a and b . But, this contradicts the fact that a and b have no common factor other than 1 . The contradiction arises by assuming that 3 is rational. Hence, 3 is an irrational.

Show that 4 + 3 2 is irrational.

Consider that 4 + 3 2 be a rational. Then 4 and 3 2 are rational. ⇒ 4 + 3 2 − 4 is rational .... ( ∵ difference of two rationals is rational) ⇒ 3 2 is rational ⇒ 2 is rational ..... ( ∵ 3 is rational) This contradicts the fact that 2 is irrational. The contradiction arises by assuming that 4 + 3 2 is rational. Hence, 4 + 3 2 is irrational.