$p$ and $q$ are two non-zero unequal rational numbers. Which of the following can't be equal to 1?

(i) $\dfrac{p - q}{q - p}$

(ii) $\dfrac{p + q}{q + p}$

(iii) $\dfrac{pq}{qp}$

(iv) $\dfrac{p}{q} \div \dfrac{q}{p}$

1. only (i)

2. only (iv)

3. (i) and (ii)

4. (i) and (iv)

Examine each expression separately, keeping in mind that $p$ and $q$ are non-zero and $p ≠ q$.

(i) As $p ≠ q$, the numerator $p - q$ is non-zero and $q - p = -(p - q)$

$\Rightarrow \dfrac{p - q}{q - p} = \dfrac{p - q}{-(p - q)} = -1$

So, this expression is always equal to -1 and can never be equal to 1.

(ii) Here the numerator and the denominator are the same, as $p + q = q + p$ by the commutative law of addition.

$\Rightarrow \dfrac{p + q}{q + p} = 1$

So, this expression is equal to 1.

(iii) Here the numerator and the denominator are the same, as $pq = qp$ by the commutative law of multiplication.

$\Rightarrow \dfrac{pq}{qp} = 1$

So, this expression is equal to 1.

(iv) Dividing,

$\Rightarrow \dfrac{p}{q} \div \dfrac{q}{p} \\[1em] \Rightarrow \dfrac{p}{q} \times \dfrac{p}{q} \\[1em] \Rightarrow \dfrac{p^2}{q^2}$

This is equal to 1 whenever $p^2 = q^2$, which happens for $p = -q$. For example, taking $p = 2$ and $q = -2$, we get $\dfrac{p^2}{q^2} = \dfrac{4}{4} = 1$

So, this expression can be equal to 1.

Thus, only expression (i) can never be equal to 1.

Hence, Option 1 is the correct option.