Prove that 2 + 5 is an irrational number.

Let us assume 2 + 5 is a rational number 2 + 5 = p q , q ≠ 0 5 = p q - 2 5 = p - 2 q q p - 2 q 2 is a rational number but, 5 is a irrational number. ∴ Our assumption was wrong 2 + 5 is an irrational number.

Prove that 2 + 5 is an irrational number.

Let us assume 2 + 5 is a rational number 2 + 5 = p q , q ≠ 0 5 = p q - 2 5 = p - 2 q q p - 2 q 2 is a rational number but, 5 is a irrational number. ∴ Our assumption was wrong 2 + 5 is an irrational number.

Rationalise the denominator and simplify : 5 + 3 5 - 3

Given that, 5 + 3 5 - 3 Divide and multiply both numerator and denominator by 5 + 3 we get, 5 + 3 5 - 3 × 5 + 3 5 + 3 = 5 + 3 2 5 + 3 5 - 3 = 5 + 3 + 2 5 3 5 2 - 3 2 [Using a + b 2 = a 2 + b 2 + 2 a b and a + b a - b = a 2 - b 2 ] = 8 + 2 15 5 - 3 = 2 4 + 15 2 = 4 + 15

Prove that 3 + 5 is an irrational number.

Let us assume that 3 + √ 5 is a rational number ⇒ 3 + √ 5 = a b where, b ≠ 0 and a , b are integers. ⇒ √ 5 = a b - 3 ⇒ √ 5 = a - 3 b b Since a , b are integers. Hence, a – 3 b is also integer. ⇒ a - 3 b b is rational which cannot be possible. ∵ a - 3 b b = 5 which is an irrational number. I t is contradicting our assumption. So, the assumption was wrong. Hence, 3 + √ 5 is an irrational number.

Prove that 3 is an irrational number.

Let 3 be a rational number ∴ 3 = p q , where p , q are integers and q ≠ 0 and p and q are coprimes. Coprime number is a set of numbers which have no common divisor other than 1 p = 3 q Taking square on both sides, we get p 2 = 3 q 2 p 2 = 3 q 2 q 2 = p 2 3 we get p is the divisor of 3 ∴ p 3 = a ⇒ p = 3 a Taking square on both sides, we get p 2 = 9 a 2 Here, p 2 = 3 q 2 ⇒ 3 q 2 = 9 a 2 ⇒ q 2 = 9 a 2 3 ⇒ a 2 = q 2 3 ⇒ q is the factor of 3 . This contradicts out assumption. Hence, 3 is an irrational number.

Rationalise the denominator and simplify : 5 + 3 5 - 3

Given that, 5 + 3 5 - 3 Divide and multiply both numerator and denominator by 5 + 3 we get, 5 + 3 5 - 3 × 5 + 3 5 + 3 = 5 + 3 2 5 + 3 5 - 3 = 5 + 3 + 2 5 3 5 2 - 3 2 [Using a + b 2 = a 2 + b 2 + 2 a b and a + b a - b = a 2 - b 2 ] = 8 + 2 15 5 - 3 = 2 4 + 15 2 = 4 + 15

Prove that 3 is an irrational number.

Let 3 be a rational number ∴ 3 = p q , where p , q are integers and q ≠ 0 and p and q are coprimes. Coprime number is a set of numbers which have no common divisor other than 1 p = 3 q Taking square on both sides, we get p 2 = 3 q 2 p 2 = 3 q 2 q 2 = p 2 3 we get p is the divisor of 3 ∴ p 3 = a ⇒ p = 3 a Taking square on both sides, we get p 2 = 9 a 2 Here, p 2 = 3 q 2 ⇒ 3 q 2 = 9 a 2 ⇒ q 2 = 9 a 2 3 ⇒ a 2 = q 2 3 ⇒ q is the factor of 3 . This contradicts out assumption. Hence, 3 is an irrational number.

If U = 0 , 1 , 2 , 3 , 4 and A = 1 , 4 , B = 1 , 3 show that A ∪ B ' = A ' ∩ B '

Given that, If U = 0 , 1 , 2 , 3 , 4 and A = 1 , 4 , B = 1 , 3 Then, A ∪ B = 1 , 3 , 4 A ∪ B ' = 0 , 2 ..................(i) Now, A ' = 0 , 2 , 3 and B ' = 0 , 2 , 4 A ' ∩ B ' = 0 , 2 .......................(ii) From (i) and (ii) we get, A ∪ B ' = A ' ∩ B ' Hence, proved.