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

The correct option is: always greater than the original fraction.

To understand why this is correct, let us represent the original proper fraction as:

$\frac{a}{b}$
where $a$ and $b$ are positive real numbers. Since the fraction is a proper fraction, the numerator must be strictly less than the denominator:

$a<b$
This inequality can also be written as:

$b-a>0$

Let $x$ be the positive quantity added to both the numerator and the denominator, where $x>0$. The resulting new fraction is:

$\frac{a+x}{b+x}$

To determine if the new fraction is greater than, equal to, or less than the original fraction, we can find the difference between the new fraction and the original fraction by subtracting the original fraction from the new fraction:

$\text{Difference}=\frac{a+x}{b+x}-\frac{a}{b}$

To subtract these fractions, we find a common denominator, which is $b(b+x)$:

$\text{Difference}=\frac{b(a+x)-a(b+x)}{b(b+x)}$

Now, let us expand the terms in the numerator:

$b(a+x)-a(b+x)=ab+bx-ab-ax$

Simplifying the expression by cancelling $ab$ and $-ab$ gives:

$bx-ax=x(b-a)$

Substituting this simplified numerator back into the difference equation, we get:

$\text{Difference}=\frac{x(b-a)}{b(b+x)}$

Next, let us analyze the sign of this difference expression step-by-step:
1. We are given that $x>0$ (positive).
2. Because the original fraction is proper, $a<b$, which means $(b-a)>0$ (positive).
3. The denominator $b(b+x)$ is a product of positive numbers, so it is also positive.

Since both the numerator $x(b-a)$ and the denominator $b(b+x)$ are positive, the entire difference is positive:

$\frac{x(b-a)}{b(b+x)}>0$

This tells us that:

$\frac{a+x}{b+x}-\frac{a}{b}>0$
Adding $\frac{a}{b}$ to both sides yields:

$\frac{a+x}{b+x}>\frac{a}{b}$

Therefore, the resulting fraction is always greater than the original fraction.