# Effect of adding a constant to both Numerator and Denominator

I was reading a text book and came across the following:

If a ratio $a/b$ is given such that $a \gt b$, and given $x$ is a positive integer, then $$\frac{a+x}{b+x} \lt\frac{a}{b}\quad\text{and}\quad \frac{a-x}{b-x}\gt \frac{a}{b}.$$

If a ratio $a/b$ is given such that $a \lt b$, $x$ a positive integer, then $$\frac{a+x}{b+x}\gt \frac{a}{b}\quad\text{and}\quad \frac{a-x}{b-x}\lt \frac{a}{b}.$$

I am looking for more of a logical deduction on why the above statements are true (than a mathematical "proof"). I also understand that I can always check the authenticity by assigning some values to a and b variables.

Can someone please provide a logical explanation for the above?

-
I think you used the "logic" tag again in a previous question that was unrelated to logic so I'm posting a comment, the logic tag is used for questions that are related to mathematical logic (for more information see the tag description). From the way you are posing the question it looks to me that you could use the "intuition" tag (though I'm not sure). – Apostolos Jun 18 '11 at 20:45
Just as a note, assigning some values to $a$ and $b$ would not be a way to check the 'authenticity' (by which I assume you mean 'truth') of those statements; if you found $a$ and $b$ such that the statement failed to hold you would have shown the general statement false, but no amount of testing with constants can prove it to be true. Of course it can be a helpful way to build intuition about the question, though. – dbmag9 Jun 16 '13 at 9:10

Let $a>b>0$ and $x>0$. Because $a>b$ and $x$ is positive, we have that $ax>bx$. Therefore $ab+ax>ab+bx$. Note that $ab+ax=a(b+x)$ and $ab+bx=b(a+x)$, so our inequality says that $$a(b+x)>b(a+x).$$ Dividing, we have that $$\frac{a}{b}>\frac{a+x}{b+x}.$$ The other inequalities have a similar explanation.

-

HINT $\$ View it as a mediant; geometrically, the diagonal of the parallelogram with sides being the vectors $\rm\:(a,b),\ (x,x)\:,\:$ noting that the slope of the diagonal lies between the slopes of the sides.

-

When $a>b,\;\; \text{with}\;\; x \in \mathbb Z, x > 0$ $$f(x)=\frac{a+x}{b+x}=1+\frac{a-b}{b+x}$$ is decreasing w.r.t. $x>-b$. It is an intuitive explanation, but I am not sure whether this is your logical explanation.

-
(I just edited/tweaked you formatting to make your function larger (in the sense of being easier to view)! Feel free to "roll back" to what you had, if you prefer.) – amWhy Jun 18 '11 at 20:40

For the first question. If $x>0$, we have:

$$a>b\Rightarrow ax>bx\Rightarrow ab+ax>ab+bx\Rightarrow a(b+x)>b(a+x)$$

$$\Rightarrow \frac{a}{b}>\frac{a+x}{b+x}\Leftrightarrow \frac{a+x}{b+x}<% \frac{a}{b}.$$

Then $-x<0$. We thus have

$$a>b\Rightarrow -ax<-bx\Rightarrow ab-ax<ab-bx\Rightarrow a(b-x)<b(a-x)$$

$$\Rightarrow \frac{a}{b}<\frac{a-x}{b-x}\Leftrightarrow \frac{a-x}{b-x}>% \frac{a}{b}.$$

For the second question. If $x>0$, we have:

$$a<b\Rightarrow ax<bx\Rightarrow ab+ax<ab+bx\Rightarrow a(b+x)<b(a+x)$$

$$\Rightarrow \frac{a}{b}<\frac{a+x}{b+x}\Leftrightarrow \frac{a+x}{b+x}>% \frac{a}{b}.$$

Then $-x<0$. We thus have

$$a<b\Rightarrow -ax>-bx\Rightarrow ab-ax>ab-bx\Rightarrow a(b-x)>b(a-x)$$

$$\Rightarrow \frac{a}{b}>\frac{a-x}{b-x}\Leftrightarrow \frac{a-x}{b-x}<% \frac{a}{b}.$$

-
Did you mean in your first implication, to write $a > b \rightarrow ax > bx$? – amWhy Jun 18 '11 at 20:57
@amWhy: Thanks! corrected. – Américo Tavares Jun 18 '11 at 21:07

Ok. Here is my intuition. Since the people have already added the formal proofs, i'll only give the intuition. Consider two guys a and b. a is a rich man and b is a poor man. Now you give both equal amount of money x. How is the relative monetary status of both changed? for a it doesn't add as much as it improves the state of b. Therefore the relative superiority of a over b has decreased.

-

Let $a,b,x>0$ and $a>b$. Then

$$\frac{a+x}{b+x}=\frac{a+\frac{a}{b}x}{b+x}+\frac{(1-\frac{a}{b})x}{b+x}=\frac{a}{b}+(1-\frac{a}{b})\frac{x}{b+x}\leq\frac{a}{b}.$$

The other assertions can be shown similarly.

-

How about something along these lines: Think of a pot of money divided among the people in a room. In the beginning, there are a dollars and b persons. Initially, everyone gets a/b>1 dollars since a>b. But new people are allowed into the room at a fee of 1 dollar person. The admission fees are put into the pot. The average will at always be greater than 1 but since each new person is not charged what he (or she) is getting back, the average will have to drop and so [ \frac{a+x}{b+x}<\frac ab.] Similar reasoning applies to the other inequalities.

-