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Mathematics

Prove that 3\sqrt{3} is an irrational number. Hence show that 5 - 3\sqrt{3} is an irrational number.

Rational Irrational Nos

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Answer

Let 3\sqrt{3} be a rational number, then

3=pq,\sqrt{3} = \dfrac{p}{q},

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

3=p2q2p2=3q2….(i)\Rightarrow 3 = \dfrac{p^2}{q^2} \\[1.5em] \Rightarrow p^2 = 3q^2 \qquad \text{….(i)}

As 3 divides 3q2, so 3 divides p2 but 3 is prime

3 divides p(Theorem 1)\Rightarrow 3 \text{ divides } p \qquad \text{(Theorem 1)}

Let p = 3m, where m is an integer.

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

(3m)2=3q29m2=3q23m2=q2(3m)^2 = 3q^2 \\[1.5em] \Rightarrow 9m^2 = 3q^2 \\[1.5em] \Rightarrow 3m^2 = q^2 \\[1.5em]

As 3 divides 3m2, so 3 divides q2 but 3 is prime

3 divides q(Theorem 1)\Rightarrow 3 \text{ divides } q \qquad \text{(Theorem 1)}

Thus, p and q have a common factor 3. This contradicts that p and q have no common factors (except 1).

Hence, 3\sqrt{3} is not a rational number. So, we conclude that 3\sqrt{3} is an irrational number.

Suppose that 535 - \sqrt{3} is a rational number, say r.

Then, 535 - \sqrt{3} = r (note that r ≠ 0)

3=r53=5r\Rightarrow - \sqrt{3} = r - 5 \\[1.5em] \Rightarrow \sqrt{3} = 5 - r \\[1.5em]

As r is rational and r ≠ 0, so 5 - r is rational

3\Rightarrow \sqrt{3} is rational

But this contradicts that 3\sqrt{3} is irrational. Hence, our supposition is wrong.

53\bold{5 - \sqrt{3}} is an irrational number.

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