On comparing fractions , fraction with smaller difference between numerator and denominator is greater than the other A text book proposed that "when comparing fractions ,if the  compared fractions's are such that numerator is smaller than denominator ,then fraction with more  difference(absolute) between numerator and denominator is the smallest among the fractions compared "   
And I found many text books support this. Even on looking at the video http://www.youtube.com/watch?v=rJz-f7uCBns#t=6m35s  , here he has used this idea , but for a case where numerator is greater than denominator .
But consider the case 
$2/3$   &  $ 20/30 $ , but as per the theory proposed 2/3 > 20/30.but actually they are the same.
Even taking a bit complex case 
if $2/7  = 0.285714  $ definitely we will be able to find another number with different difference but same answer as $ 3/0.2857514 = 10.5 $ so 
$ 3/ 10.5 = 0.285714 $
here difference is 7.5 but value of the ratio is still 0.285714 . 
So am I going wrong in understanding this concept preferred in many popular text books . If so please spot the error and help me by giving conditions when this fact holds good .
 A: I have never seen this claim in any textbook; in any case it's wrong. The claim seems to be that if you have two fractions $\frac{a}{b}$ and $\frac{c}{d}$ with $a < b$ and $c < d$, then $|a - b| < |c - d| \iff \frac{a}{b} < \frac{c}{d}$.
This is false. We can write $\frac{a}{b} = 1 - \frac{b-a}{b}$ and $\frac{c}{d} = 1 - \frac{d-c}{d}$, so you're comparing $\frac{b-a}{b}$ and $\frac{d-c}{d}$ (whichever is greater, the corresponding fraction is smaller). The first numerator may be smaller than the second, but the actual comparison of these fractions can of course go either way.
For example, 


*

*Here is one with $b - a < d - c$ but $\frac{a}{b} > \frac{c}{d}$: consider $\frac{2}{3} > \frac{3}{5}$.

*Here is one with $b - a < d - c$ but $\frac{a}{b} = \frac{c}{d}$: consider $\frac{2}{3} = \frac{4}{6}$.

*Here is one with $b - a < d - c$ but $\frac{a}{b} < \frac{c}{d}$: consider $\frac{2}{3} < \frac{5}{7}$.
So all results are possible; the test is nonsense.
Edit: Just for fun/completeness, here is a table showing pairs $(\frac{a}{b}, \frac{c}{d})$ with each possible combination of the two comparisions:
$$\begin{array}{c|c|c|c}
& \frac{a}{b}<\frac{c}{d} & \frac{a}{b}=\frac{c}{d} & \frac{a}{b}>\frac{c}{d}\\ 
\hline \\
b-a<d-c & \frac23,\frac57 & \frac23,\frac46 & \frac23,\frac35\\
\hline \\
b-a=d-c & \frac23,\frac34 & \frac23,\frac23 & \frac23,\frac12\\
\hline \\
b-a>d-c & \frac57,\frac23 & \frac46,\frac23 & \frac35,\frac23 \\
\end{array}$$
(If you want examples involving fractions greater than $1$, turn each of the fractions upside down. Each of the inequalities between the fractions will reverse direction, so you'll still have a complete set of examples.)

Edit: On looking at that segment of the video, it's possible (not very clear) that what he may have been saying is equivalent to the following claim, which is true: if you have two fractions $\frac{a}{b}$ and $\frac{c}{d}$ with $a > b$ and $c > d$, and if $a - b < c - d$ and $b > d$, then $\frac{a}{b} < \frac{c}{d}$. Proof:
$$\frac{a}{b} = 1 + \frac{a-b}{b} < 1 + \frac{c-d}{b} < 1 + \frac{c-d}{d} = \frac{c}{d}$$
With fractions less than $1$, the corresponding statement would be that if you have two fractions $\frac{a}{b}$ and $\frac{c}{d}$ with $a < b$ and $c < d$, and if $b - a < d - c$ and $b > d$, then $\frac{a}{b} > \frac{c}{d}$:
$$\frac{a}{b} = 1 - \frac{b-a}{b} > 1 - \frac{b-a}{d} > 1 - \frac{d-c}{d} = \frac{c}{d}$$
But these are so many conditions on the hypothesis that I wonder how often it will be useful.
A: The statement is incorrect, even without considering possibly special cases. Take for example $$\frac{901}{1000}<\frac{9011}{10000}.$$
A: Here is the comparison criterion for fractions in the unit interval. You omit a $\rm\,\color{#c00}{key\ hypothesis}.$
Lemma $\rm\ \ \color{#c00}{ A>a},\,\ \color{blue}{b\!-\!a > B\!-\!A}\ \Rightarrow\ A/B > a/b,\ \ $ for $\rm\ \ A/B,\,\color{#0a0}{a/b \,\in\, (0,1)},\ \ B,b> 0\:$
Proof $\rm\ \ \ \ \displaystyle \frac{A}B-\frac{a}b\, =\, \frac{A(b\!-\!a)-a(B\!-\!A)}{bB} \,=\, \frac{(\color{#c00}{A\!-\!a})(\color{#0a0}{b\!-\!a}) + a(\color{blue}{b\!-\!a\!-\!(B\!-\!A))}}{bB}\, >\, 0$
Remark $\ $ It is a special case of the continued fraction comparison algorithm. $ $Indeed
$$\rm\displaystyle \frac{A}B\, >\, \frac{a}b \iff \frac{b}a\, >\, \frac{B}A\ \stackrel{X\to\ X-1\!} {\iff}\ \frac{b-a}a\, >\, \frac{B-A}A \iff b\!-\!a\, >\, \frac{a}A\, (B-A)$$
A: The test is as follows:


*

*Given two Proper Fractions n/d & N/D where D>d, If (D-N) ≤ (d-n) then N/D>n/d 

*Given two Improper Fractions n/d and N/D where D>d, If (N-D) ≤ (n-d) then n/d>N/D
(Note: the conditions are sufficient but not necessary).
