# Can adding 1 to the numerator and denominator on both sides of an inequation change the outcome?

I'm computing a fraction from a database when both numerator and denominator can be zero. To solve this problem I thought of adding 1 to each.

I know I can add 1 only to the denominator, but this is for optimization of resources and adding 1 to the denominator favors tasks which have a low denominator.

Because 3/3 == 4/4, but 3/4 > 4/5 and thus the task with 4 will get the resources because the program will think it has more to complete.

This brings me to my question:

If I know that

$$\frac{a}{b} > \frac {c}{d}$$

Can $$\frac{a+1}{b+1} < \frac {c+1}{d+1}$$ happen, even once?

The above formula translates to

$$a+d > c+b+(bc-ad)$$

and this is where I'm stuck.

• This won't be true for suppose $\frac{-2}{5} \gt \frac{-1}{4}$ – gemspark Aug 24 at 13:20

$$\frac{7}{10}\gt\frac{2}{3}$$ but $$\frac{8}{11}\lt \frac{3}{4}$$

• could you explain how you found this? – Hachiloni Aug 24 at 13:40
• @Hachiloni I restricted myself to positive numbers, just in case those were more useful to you. Then I recalled that adding $1$ to both numerator and denominator in those circumstances brings the fraction closer to $1$ (from below or above - fairly easy to prove rigorously), but, following the intuition as in Yves Daoust's graphical answer, I expected that it will "affect" more the fractions with the smaller numerator/denominator. – Stinking Bishop Aug 24 at 13:44
• Thus, I had to find a fraction $\gt 1$ with "small" numerator/denominator that is only slightly bigger than another one with "large" numerator/denominator - and then check what happens when I add $1$. Pretty much the first pair of fractions I tried worked. – Stinking Bishop Aug 24 at 13:44
• (My claim from above: let $a,b>0, \frac{a}{b}>1$, i.e. $a>b$. Then $\frac{a+1}{b+1}<\frac{a}{b}$ because $(a+1)b=ab+b<ab+a=a(b+1)$. – Stinking Bishop Aug 24 at 13:47
• I guess I forgot to limit the question to fractions smaller than 1. I hoped you'd have a formula so I could look at it – Hachiloni Aug 24 at 13:50

++++++++++

• Are not "The ticks show one unit" 5 words? :) – user Aug 24 at 13:39
• could you explain the graph? What are the colors? what are the axis? two answers state two different sets, how do I see it here? – Hachiloni Aug 24 at 13:39
• @Hachiloni: I want to leave this answer "without words". – Yves Daoust Aug 24 at 13:42
• @user: the editor did not let me write only the title. But I will modify. – Yves Daoust Aug 24 at 13:43
• @Hachiloni The small blue triangle has the corners at $(0,0)$, $(b,a)$ and $(b+1, a+1)$. The slope of the sides coming out of $(0,0)$ is, respectively, $\frac{a}{b}, \frac{a+1}{b+1}$, the slope of the third side is $1$. You can see geometrically how the slope is "pushed up" (towards $1$, as it originally was $<1$) when we switch from $(b,a)$ to $(b+1, a+1)$. The same for the green triangle, which is all about $c,d$. – Stinking Bishop Aug 24 at 13:52

Assume $$a,b,c,d>0$$ then we have that

$$\frac{a}{b} > \frac {c}{d} \iff ad-bc>0$$

and

$$\frac{a+1}{b+1} > \frac {c+1}{d+1} \iff ad-cb+a+d-c-b >0$$

which fails when

$$c+b-a-d>ad-cb \iff c+b+bc>a+d+ad$$

that is for example

$$\frac{4}{7}\gt\frac{1}{2}, \quad \frac{5}{8}\lt\frac{2}{3}$$

indeed in this case

$$c+b+bc=1+7+7=15>a+d+ad=4+2+8=14$$