Prove that the following number is irrational 8

Suppose 8 is a rational number let 8 = p q , where p and q are integers , q > 0 and p , q have no common factor Squaring both sides we get , 8 = p 2 q 2 ⇒ p 2 = 8 q 2 therefore 8 q 2 is divisible by 8 p 2 is also divisible by 8 ⇒ p is divisible by 8 let p = 8 k where k is an integer, squaring both sides , p 2 = 64 k 2 , 64 k 2 = 8 q 2 therefore 64 k 2 is divisible by 8 p 2 is divisible by 8 means p is divisible by 8 hence p , q are both divisible by 8 so our supposition is wrong and 8 is an irrational number

Prove that the following number is irrational 14

suppose 14 is a rational number let 14 = p q , where p and q are integers , q > 0 and p , q have no common factor squaring both sides we get , 14 = p 2 q 2 ⇒ 14 q 2 = p 2 . . . . . . . . . . . . . . . . . . . . . 1 therefore p 2 is divisible by 2 ⇒ p is divisible let p = 2 m substituting the value of p in 1 4 m 2 = 14 q 2 ⇒ 2 m 2 = 7 q 2 Squaring both sides, therefore q 2 is divisible by 2 q is divisible by 2 ⇒ p , q have 2 as the common factor which is not possible and 14 is an irrational number.

Prove that 3 is an Irrational number . Hence show that 5 - 3 is an Irrational number ?

Let 3 be a rational number, then 3 = p q , where p, q are integers, q ≠ 0 and p , q have no common factors. ⇒ 3 = p 2 q 2 ⇒ p 2 = 3 q 2 . . . . i As 3 divides 3 q 2 , so 3 divides p 2 but 3 is prime. ⇒ 3 divides p Let p = 3 m, where m is an integer. Substituting this value of p in i , we get 3 m 2 = 3 q 2 ⇒ 9 m 2 = 3 q 2 ⇒ 3 m 2 = q 2 As 3 divides 3 m 2 , so 3 divides q 2 but 3 is prime. ⇒ 3 divides q Thus, p and q have a common factor 3 . This contradicts that p a n d q have no common factors (except 1 ) Hence, 3 is not a rational number. So, we conclude that 3 is an irrational number. Now in 5 - 3 , 5 is rational and - 3 is an irrational number. Also we know the sum or difference of a rational and irrational number is always irrational. Hence, 5 - 3 is an irrational number.

Show that 1 3 - 5 is an irrational number.

Given, 1 3 - 5 Rationalise the denominator and apply the identity a - b a + b = a 2 - b 2 1 3 - 5 = 1 3 - 5 × 3 + 5 3 + 5 = 3 + 5 9 - 5 = 3 + 5 4 Let us assume that 1 3 - 5 is a rational number, then 3 + 5 = r (say), where r is a rational number. ⇒ 3 + 5 = r ⇒ 5 = r - 3 As r is a rational number and r ≠ 0 , so r - 3 is also rational number. ⇒ 5 is a rational number. But this contradicts that 5 is an irrational number. Hence, our assumption is wrong. Therefore, 1 3 - 5 is an irrational number.

Rationalise the denominator of the following : 10 2 2 + 3

Given: 10 2 2 + 3 Multiply and divide 2 2 - 3 = 10 2 2 + 3 × 2 2 - 3 2 2 - 3 = 10 × 2 2 - 10 3 2 2 2 - 3 2 = 20 2 - 10 3 2 2 × 2 - 3 = 20 2 - 10 3 4 × 2 - 3 = 20 2 - 10 3 5 = 4 2 - 2 3 ⇒ 2 2 2 - 3