KnowledgeBoat Logo

Mathematics

Prove that the following numbers are irrational:

(i) 8\sqrt{8}

(ii) 14\sqrt{14}

(iii) 23\sqrt[3]{2}

Rational Irrational Nos

8 Likes

Answer

(i) 8\sqrt{8} can be written as 222\sqrt{2}, now we are going to show that 2\sqrt{2} is an irrational number.

Let 2\sqrt{2} be a rational number, then

2=pq,\sqrt{2} = \dfrac{p}{q},

where p, q are integers, q ≠ 0 and p, q have no common factors (except 1)

2=p2q2p2=2q2….(i)\Rightarrow 2 = \dfrac{p^2}{q^2} \\[1.5em] \Rightarrow p^2 = 2q^2 \qquad \text{….(i)}

As 2 divides 2q2, so 2 divides p2 but 2 is prime

2 divides p(Theorem 1)\Rightarrow 2 \text{ divides } p \qquad \text{(Theorem 1)}

Let p = 2m, where m is an integer.

Substituting this value of p in (i), we get

(2m)2=2q24m2=2q22m2=q2(2m)^2 = 2q^2 \\[1.5em] \Rightarrow 4m^2 = 2q^2 \\[1.5em] \Rightarrow 2m^2 = q^2 \\[1.5em]

As 2 divides 2m2, so 2 divides q2 but 2 is prime

2 divides q(Theorem 1)\Rightarrow 2 \text{ divides } q \qquad \text{(Theorem 1)}

Thus, p and q have a common factor 2. This contradicts that p and q have no common factors (except 1).

Hence, 2\sqrt{2} is not a rational number. So, we conclude that 2\sqrt{2} is an irrational number.

Since, product of non-zero rational number and an irrational number is an irrational number.

And 2\sqrt{2} is an irrational number this implies that 222\sqrt{2} = 8\bold{\sqrt{8}} is an irrational number.

(ii) Suppose that 14\sqrt{14} is a rational number, then

14=pq,\sqrt{14} = \dfrac{p}{q},

where p, q are integers, q ≠ 0 and p, q have no common factors (except 1)

14=p2q2p2=14q2….(i)\Rightarrow 14 = \dfrac{p^2}{q^2} \\[1.5em] \Rightarrow p^2 = 14q^2 \qquad \text{….(i)}

As 2 divides 14q2, so 2 divides p2 but 2 is prime

2 divides p(Theorem 1)\Rightarrow 2 \text{ divides } p \qquad \text{(Theorem 1)}

Let p = 2k, where k is some integer.

Substituting this value of p in (i), we get

(2k)2=14q24k2=14q22k2=7q2(2k)^2 = 14q^2 \\[1.5em] \Rightarrow 4k^2 = 14q^2 \\[1.5em] \Rightarrow 2k^2 = 7q^2 \\[1.5em]

As 2 divides 2k2, so 2 divides 7q2

\Rightarrow 2 divides 7 or 2 divides q2

But 2 does not divide 7, therefore, 2 divides q2

\Rightarrow 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, 14\sqrt{14} is not a rational number. So, we conclude that 14\bold{\sqrt{14}} is an irrational number.

(iii) Suppose that 23\sqrt[3]{2} = pq\dfrac{p}{q}, where p, q are integers , q ≠ 0 , p and q have no common factors (except 1)

2=(pq)3p3=2q3….(i)\Rightarrow 2 = \Big(\dfrac{p}{q}\Big)^3 \\[1.5em] \Rightarrow p^3 = 2q^3 \qquad \text{….(i)}

As 2 divides 2q3 \Rightarrow 2 divides p3

\Rightarrow 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

\phantom{\Rightarrow}(2k)3 = 2q3
\Rightarrow 8k3 = 2q3
\Rightarrow 4k3 = q3

As 2 divides 4k3 \Rightarrow 2 divides q3

\Rightarrow 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 23\sqrt[3]{2} cannot be expressed as pq\dfrac{p}{q}, where p, q are integers, q > 0, p and q have no common factors (except 1).

23\bold{\sqrt[3]{2}} is an irrational number.

Answered By

3 Likes


Related Questions