**Question:**

State whether the following statements are true or false? Justify your answer.

(i) $\frac{\sqrt{2}}{3}$ is a rational number.

(ii) There are infinitely many integers between any two integers.

(iii) Number of rational numbers between 15 and 18 is finite.

(iv) There are numbers which cannot be written in the form $\frac{p}{q}$, $q \neq 0, p$ and $q$ both are integers.

(v) The Square of an irrational number is always rational.

(vi) $\frac{\sqrt{12}}{\sqrt{3}}$ is not a rational number as $\sqrt{12}$ and $\sqrt{3}$ are not integers.

(vii) $\frac{\sqrt{15}}{\sqrt{3}}$ is written in the form $\frac{p}{q}, q \neq 0$ and so it is a rational number.

**Solution:**

(i) False, here $\sqrt{2}$ is an irrational number and 3 is a rational number, we know that when we divide irrational number by non-zero rational number it will always give an irrational number.

(ii) False, because between two consecutive integers (likel and 2), there does not exist any other integer.

(iii) False, because between any two rational numbers there exist infinitely many rational numbers.

(iv) True, because there are infinitely many numbers which cannot be written in the form $p / q, q \neq 0 . p, q$ both are integers and these numbers are called irrational numbers.

(v) False, e.g.. Let an irrational number be $\sqrt{2}$ and $\sqrt[4]{2}$

(a) $(\sqrt{2})^{2}=2$, which is a rational number.

(b) $(\sqrt[4]{2})^{2}=\sqrt{2}$, which is not a rational number.

Hence, square of an irrational number is not always a rational number.

(vi) False, $\frac{\sqrt{12}}{\sqrt{3}}=\frac{\sqrt{4 \times 3}}{\sqrt{3}}=\frac{\sqrt{4} \times \sqrt{3}}{\sqrt{3}}=2 \times 1=2$, which is a rational number.

(vii) False, $\frac{\sqrt{15}}{\sqrt{3}}=\frac{\sqrt{5 \times 3}}{\sqrt{3}}=\frac{\sqrt{5} \times \sqrt{3}}{\sqrt{3}}=\sqrt{5}$, which is an irrational number.

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