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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 $$ \frac{a}{b} \leq \frac{a+\varepsilon}{b+\varepsilon},\ \forall a,b\in\mathbb{R} $$

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Consider the case $a=b$. – Najib Idrissi Apr 28 '14 at 15:59
I edited the equation to be less than or equal – user118462 Apr 28 '14 at 16:02
Still not true: $a=2,b=1$. – Najib Idrissi Apr 28 '14 at 16:02
"This change will be positive if ϵ and b−a have the same sign." Is it the case then, that the above equation is true if a<b? – user118462 Apr 28 '14 at 16:04
so, would your answer be yes if I were to once again edit and say: for all a,b \in R such that a<b – user118462 Apr 28 '14 at 16:15

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 $$\frac{2.5}{1.5}=\frac{5}{3}<2$$

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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 $\varepsilon b(b + \varepsilon)(b-a).$

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This is interesting, since you come to a similar conclusion to the first answer using high school math: "This change will be positive if ϵ and b−a have the same sign." – user118462 Apr 28 '14 at 17:17
Can you just help me with my bottom line: I want to say that for \epsilon between 0 and 1, and a<b, (a+\epsilon)/(b+\epsilon) is strictly greater than a/b – user118462 Apr 28 '14 at 17:38

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 $z(q,c)>0$.

enter image description here

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).

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