Correct Option (2)

Let the original proper fraction be represented as $\frac{a}{b}$ , where $a$ and $b$ are positive integers such that $a < b$ . This condition signifies that the fraction's value is less than 1.

Let the positive quantity by which both the numerator and denominator are increased be $k$ , where $k > 0$ . The resulting fraction is $\frac{a + k}{b + k}$ .

To compare the original fraction $\frac{a}{b}$ with the new fraction $\frac{a + k}{b + k}$ , we can examine their difference:

$\frac{a + k}{b + k} - \frac{a}{b} = \frac{b (a + k) - a (b + k)}{b (b + k)}$

$= \frac{a b + b k - a b - a k}{b (b + k)}$

$= \frac{b k - a k}{b (b + k)}$

$= \frac{k (b - a)}{b (b + k)}$

Given that $\frac{a}{b}$ is a proper fraction, $a < b$ , which implies that $(b - a)$ is a positive quantity. Additionally, $k > 0$ , $b > 0$ , and $(b+k) > 0$ . Therefore, the entire expression $\frac{k (b - a)}{b (b + k)}$ is always positive.

Since the difference $\frac{a+k}{b+k} - \frac{a}{b} > 0$ , it follows that $\frac{a+k}{b+k} > \frac{a}{b}$ . Thus, the resulting fraction is always greater than the original proper fraction.

Incorrect Options:

Option 1 (always less than the original fraction): This statement is incorrect. As demonstrated by the algebraic analysis, adding the same positive quantity to both the numerator and denominator of a proper fraction consistently increases its value, moving it closer to 1.
Option 3 (always equal to the original fraction): This statement is incorrect. For the fractions to be equal, the difference $\frac{k (b - a)}{b (b + k)}$ would need to be zero. This is not possible because $k > 0$ and $(b - a) > 0$ for a proper fraction.
Option 4 (such that nothing can be claimed definitely): This statement is incorrect. A definite conclusion can be drawn based on the mathematical properties of proper fractions and the specified operation. The resulting fraction is consistently greater than the original.