If the numerator and denominator of a proper fraction are increased by the same positive quantity which is greater than zero, the resulting fraction is

Correct Option (2)

Let the original proper fraction be represented as $\frac{a}{b}$, where $a$ and $b$ are positive integers such that . 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 . 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:

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

$$=\frac{ab+bk-ab-ak}{b(b+k)}$$

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

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

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

Since the difference , it follows that . 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 and 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.