State whether each of the following statements is true or false:

(i) Every whole number is a rational number.

(ii) Every integer is a rational number.

(iii) $\dfrac{2}{3} = \dfrac{2+4}{3+4}$

(iv) $\dfrac{-5}{6} \lt \dfrac{-6}{5}$

(v) $\dfrac{-3}{-4}$ is a negative rational number.

(vi) 0 is a whole number but it is not a rational number.

(i) Every whole number $p$ can be written as $\dfrac{p}{1}$, which is of the form $\dfrac{p}{q}$ where $p, q$ are integers and $q ≠ 0$. For example, $0 = \dfrac{0}{1} \text{ and } 7 = \dfrac{7}{1}$

Hence, the given statement is true.

(ii) Every integer $p$ can be written as $\dfrac{p}{1}$, which is of the form $\dfrac{p}{q}$ where $p, q$ are integers and $q ≠ 0$. For example, $-5 = \dfrac{-5}{1} \text{ and } 5 = \dfrac{5}{1}$.

Hence, the given statement is true.

(iii) An equivalent rational number is obtained by multiplying (or dividing) both the numerator and the denominator by the same non-zero number, not by adding the same number to both.

$\Rightarrow \dfrac{2 + 4}{3 + 4} = \dfrac{6}{7} \\[1em] \Rightarrow \dfrac{2}{3} ≠ \dfrac{6}{7} \quad [\text{as } 2 \times 7 ≠ 3 \times 6]$

Hence, the given statement is false.

(iv) Write both the rational numbers with the LCM of their denominators as the common denominator.

L.C.M. of 6 and 5 = 30

$\Rightarrow \dfrac{-5}{6} = \dfrac{(-5) \times 5}{6 \times 5} = \dfrac{-25}{30} \\[1em] \Rightarrow \dfrac{-6}{5} = \dfrac{(-6) \times 6}{5 \times 6} = \dfrac{-36}{30}$

As -36 < -25, so $\dfrac{-6}{5} \lt \dfrac{-5}{6}$.

Hence, the given statement is false.

(v) A rational number is positive if its numerator and denominator are either both positive or both negative.

$\dfrac{-3}{-4} = \dfrac{-3 \times (-1)}{-4 \times (-1)} = \dfrac{3}{4}$, which is a positive rational number.

Hence, the given statement is false.

(vi) 0 is a whole number and it can also be written as $\dfrac{0}{1}$, which is of the form $\dfrac{p}{q}$ where $q ≠ 0$, so 0 is a rational number as well.

Hence, the given statement is false.