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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 .

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As for the second example, I think the fractions are implicitly integer-valued in the numerator and denominator. –  Clayton Apr 11 '13 at 19:53
    
So being an integer will it satisfy this rule ? . Is this rule a standard accepted rule ?.Also ,we can make it an integer by multiplying by power of 10 –  Harish Kayarohanam Apr 11 '13 at 19:55
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Your 'complex' example is in error as the textbook requires that the numerator is smaller than the denominator, and with regards to your 20/30 I would assume the textbook intends the fractions to be in simplest form. However, 1/2 is smaller than 14/17 so unless there are other clauses you haven't mentioned, it is incorrect. –  Mala Apr 11 '13 at 19:56
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4 Answers

up vote 7 down vote accepted

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.

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Hey thank you very much for the example with a proof which challenges what the book stated . Thanks . –  Harish Kayarohanam Apr 11 '13 at 20:10
    
@HarishKayarohanam: You're welcome. BTW, just out of curiosity, what book is this, so that I can avoid it? :-) –  ShreevatsaR Apr 11 '13 at 20:23
    
"T.I.M.E" considered to be Top most institute for CAT coaching having produced many IIM entries . It's the material created by this institute . Even that video I have posted here is done by the person in this link " elceducation.com/founder.php " .He also seems to have a good reputation . –  Harish Kayarohanam Apr 11 '13 at 20:30
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The statement is incorrect, even without considering possibly special cases. Take for example $$\frac{901}{1000}<\frac{9011}{10000}.$$

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Hey thank you very much for the example with integer which gainsays the statement the book proposed . Thanks –  Harish Kayarohanam Apr 11 '13 at 20:09
    
You're welcome @HarishKayarohanam! –  Clayton Apr 11 '13 at 20:34
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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

$$\rm\displaystyle \frac{A}B\, >\, \frac{a}b \iff \frac{b}a\, >\, \frac{B}A\ \stackrel{X\to\ X-1\!} {\iff}\ \frac{b-a}a\, >\, \frac{B-A}A \iff b\!-\!a\, >\, \frac{a}A\, (B-A)$$

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The test is as follows:

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

  2. 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).

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The test is accurate as long as the conditions are met. –  user81201 Jun 6 '13 at 10:40
    
Note: Most people mistakenly assume that if the test is not answered, that the converse is true.i.e, for example in the case of proper fractions if (D-N)>(d-n)people assume that it means that n/d>N/D, which is not true. Hence my clarification that the conditions are sufficient and not necessary. –  user81201 Jun 6 '13 at 12:37
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