# Adding a different constant to numerator and denominator

Suppose that $a$ is less than $b$ , $c$ is less than $d$.

What is the relation between $\dfrac{a}{b}$ and $\dfrac{a+c}{b+d}$? Is $\dfrac{a}{b}$ less than, greater than or equal to $\dfrac{a+c}{b+d}$?

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Note that if $b$ and $d$ have the same sign, then
$$\frac{a}{b}-\frac{a+c}{b+d}=\frac{ad-bc}{b(b+d)}$$
and
$$\frac{a+c}{b+d}-\frac{c}{d}=\frac{ad-bc}{d(b+d)}$$
also have the same sign.

Therefore, if $b$ and $d$ have the same sign, then $\dfrac{a+c}{b+d}$ is between $\dfrac{a}{b}$ and $\dfrac{c}{d}$.

Comment: As Srivatsan points out, if $b$ and $d$ are both positive,
$$\frac{a}{b}\lesseqqgtr\frac{a+c}{b+d}\text{ if }\frac{a}{b}\lesseqqgtr\frac{c}{d}$$

One nice thing to notice is that
$$\frac{a}{b}=\frac{c}{d} \Leftrightarrow \frac{a}{b}=\frac{a+c}{b+d}$$
no matter the values of $a$, $b$, $c$ and $d$. The $(\Rightarrow)$ is because $c=xa, d=xb$ for some $x$, so $\frac{a+c}{b+d}=\frac{a+xa}{b+xb}=\frac{a(1+x)}{b(1+x)}=\frac{a}{b}$. The other direction is similar.

The above is pretty easy to remember, and with that intuition in mind it is not hard to imagine that
$$\frac{a}{b}<\frac{c}{d} \Leftrightarrow \frac{a}{b}<\frac{a+c}{b+d}<\frac{c}{d}$$
and similar results.