If you add the same constant to the numerator and denominator, what is the relation between the new fraction and the original fraction? If I add a constant $\varepsilon < 1$ to the numerator and denominator of a fraction, is the new fraction always greater than the original?
That is, do I have 
$$
\frac{a}{b} \leq \frac{a+\varepsilon}{b+\varepsilon},\ \forall a,b\in\mathbb{R}
$$
 A: With calculus, define the function $$f(a,b)=\frac{a}{b}$$ Then the change resulting from the addition of a small $\epsilon$ will be approximately $$df\approx f_ada+f_bdb={1\over b}da-{a\over b^2}db={\epsilon\over b^2}(b-a)$$ This change will be positive if $\epsilon$ and $b-a$ have the same sign.
So the answer is no. For example, take $\epsilon=0.5$ and $a=2,b=1$. Then the original fraction is $2$ and the new one is $$\frac{2.5}{1.5}=\frac{5}{3}<2$$
A: To give you a feel for the geometry involved, consider $\displaystyle y={p\over q}$ and $\displaystyle z={p\over q}-{p+c\over q+c}={c(p-q)\over q(q+c)}$ where $0<c<1$. Without loss of generality, take $p=1$. In that case, the surface below on the left is $z(q,c)$ and on the right is the region in the $q$-$c$ plane where $z(q,c)>0$.

One can quickly see (graphically and analytically) the role that $0<q<1$ plays for any $c$ as well as the role of $c<-q$ for $q<0$. That is, assuming $c>0$, you need precisely two of these to hold:
$$
1-q>0, \quad q>0, \quad q+c>0.
$$
A quick analysis shows that the first two holding while the third fails is impossible. The first and third holding while the second fails leads to $-c<q<0$. The second and third holding while the first fails leads to $0<q<1$. 
Thus, $z(q,c)>0$ for $c>0$ when 
$$
\{-c<q<0\}\cup \{0<q<1\},
$$
just as the figure on the right shows.
PS I just realized you wanted to know when the new fraction was larger, not smaller. However, this can be easily deduced from the answer I gave (when the new fraction is smaller).
A: By high school long division (only one step is needed) we get $\frac{a}{b}$ with a remainder of $\varepsilon - \left(\frac{a}{b}\right)\varepsilon,$ which gives
$$\frac{a+\varepsilon}{b+\varepsilon} \;\; = \;\; \frac{a}{b} \; + \; \frac{\varepsilon \left(1 - \frac{a}{b}\right)}{b + \varepsilon}$$
Now it simply becomes a matter of whether $\frac{\varepsilon \left(1 - \frac{a}{b}\right)}{b + \varepsilon}$ is positive for the values you're interested in. Equivalently, look at the sign of $\frac{\varepsilon}{b}(b + \varepsilon)(b-a)$ or at the sign of $\varepsilon b(b + \varepsilon)(b-a).$
