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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!

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    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.2013-06-16

8 Answers 8

8

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.

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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.

1

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.

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    (I just edited/tweaked you formatti$n$g 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.)2011-06-18
1

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

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

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

$a

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

Then $-x<0$. We thus have

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

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

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    @amWhy: Thanks! corrected.2011-06-18
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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.

0

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.

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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.

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$ \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!

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    Welcome to Mathematics StackExchange. We use MathJax to represent data and equations. An excellent guide is at https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference.2017-07-30