Effect of adding a constant to both Numerator and Denominator

Question

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?

Thanks in advance!

Answer 1

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.

Answer 2

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.

Answer 3

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.

Answer 4

$$ \lim_{x\to 1} f(x)= \frac{a+x}{b+x} $$ where $a b \ne 0$.

As neither $a$ nor $b$ are zero the function $f(x)$ is continuous and differentiable for all $x$ belongs to $\mathbb{R}^{+}$ (positive real numbers).

As $x\to \infty$, $f(x) \to 1$.

As in our case $ x>0$ and $x\to \infty$ we observe that, if $\frac{a}{b}<1$ , $f(x)$ tends to $+1$ thereby $\frac{a}{b} < f(x)$ and if $\frac{a}{b} > 1$ , $f(x)$ tends to $+1$ thereby $\frac{a}{b} < f(x)$. Hope this helps!

Answer 5

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.

Answer 6

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.

Answer 7

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

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

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

Answer 8

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.

Answer 9

This may be an old post, but nearly every answer thusfar has been only partially correct. This is primarily because the the question itself provided an incomplete basis for everyone to build off of. The user Bill Dubuque was probably the closest to getting the correct answer, as the Wikipedia page for the mediant provides the full list of conditions for what was provided to be true. There is infact only a tiny interval in which what was stated holds true for $\frac{a-x}{b-x}$. To illustrate why, we go about the process as everyone else had and stop at $a(b-x) < b(a-x)$. The following cannot be claimed to be true for all x: $$ a(b-x) < b(a-x) \implies \frac{a}{b}<\frac{a-x}{b-x} $$ The reason is that we cannot divide $(b-x)$ and maintain the direction of the inequality without knowing that what we are dividing by is positive. Therefore we need the condition that $b>x$.