Prove that the following numbers are irrational: (i) √8 (ii) √14 (iii) ∛2

Question

KnowledgeBoat · Accepted Answer

(i) can be written as , now we are going to show that is an irrational number. Let be a rational number, then where p, q are integers, q ≠ 0 and p, q have no common factors (except 1) As 2 divides 2q², so 2 divides p² but 2 is prime Let p = 2m, where m is an integer. Substituting this value of p in (i), we get As 2 divides 2m², so 2 divides q² but 2 is prime Thus, p and q have a common factor 2. This contradicts that p and q have no common factors (except 1). Hence, is not a rational number. So, we conclude that is an irrational number. Since, product of non-zero rational number and an irrational number is an irrational number. And is an irrational number this implies that = is an irrational number. (ii) Suppose that is a rational number, then where p, q are integers, q ≠ 0 and p, q have no common factors (except 1) As 2 divides 14q², so 2 divides p² but 2 is prime Let p = 2k, where k is some integer. Substituting this value of p in (i), we get As 2 divides 2k², so 2 divides 7q² 2 divides 7 or 2 divides q² But 2 does not divide 7, therefore, 2 divides q² 2 divides q (Theorem 1) Thus, p and q have a common factor 2. This contradicts that p and q have no common factors (except 1). Hence, our supposition is wrong. Therefore, is not a rational number. So, we conclude that is an irrational number. (iii) Suppose that = , where p, q are integers , q ≠ 0 , p and q have no common factors (except 1) As 2 divides 2q³ 2 divides p³ 2 divides p (using generalisation of theorem 1) Let p = 2k , where k is an integer. Substituting this value of p in (i), we get (2k)³ = 2q³ 8k³ = 2q³ 4k³ = q³ As 2 divides 4k³ 2 divides q³ 2 divides q (using generalisation of theorem 1) Thus, p and q have a common factor 2. This contradicts that p and q have no common factors (except 1). Hence, our supposition is wrong. It follows that cannot be expressed as , where p, q are integers, q > 0, p and q have no common factors (except 1). ∴ is an irrational number.

Mathematics

Prove that the following numbers are irrational: (i) √8 (ii) √14 (iii) ∛2

Rational Irrational Nos

Related Questions

Prove that the reciprocal of an irrational number is irrational.