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Which is the fastest method to compare the below fractions with minimum calculation possible and finding which is greatest and which the smallest?? $$\frac{26}{686},\quad \frac{48}{874},\quad \frac{80}{892},\quad \frac{27}{865}$$

J. M. ain't a mathematician
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Jay
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5 Answers5

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The denominators of the last three fractions are within $27$ of one another; $3\cdot27=81$, so $27$ out of $865$ is less than $1$ part in $30$. The numerators of those fractions differ by much more than $1$ part in $30$, so we can safely rank those three fractions in the order $$\frac{27}{865}<\frac{48}{874}<\frac{80}{892}\;.$$ The only real question is where the first fraction fits, but that’s easy: $26$ and $27$ are very close, while $686$ is much less than $865$, so $$\frac{26}{686}>\frac{27}{865}\;.$$ On the other hand, it’s pretty obvious that $$\frac{26}{686}<\frac{80}{892}\;,$$ so the smallest and largest must be $\dfrac{27}{865}$ and $\dfrac{80}{892}$, respectively.

Brian M. Scott
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  • @Jay: $892-865=27$, and $27/865<27/810=1/30$. Thus, both $874$ and $892$ are less than $\frac{31}{30}\cdot 865$. But the numerators $48$ and $80$ are much more than $\frac{31}{30}\cdot 27$, so the fractions $48/874$ and $80/892$ must be bigger than $27/865$. Similar reasoning shows that $48/874<80/892$. The key is that if $\frac{a}b=\frac{c}{bk}$, then $c=ak$. Thus, if $k<\frac{31}{30}$, and $c$ is obviously more than $\frac{31}{30}\cdot a$, then $\frac{a}b$ must be less than $\frac{c}{bk}$. – Brian M. Scott Jul 08 '12 at 19:57
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As I mentioned in detail here, a general quick algorithm for comparing fractions (or reals) is to employ continued fractions, i.e. do the comparison in parallel with the continued fraction expansion, computing the integer parts, then recursing on the reciprocals of the fractionals parts (see the link for an algorithm). Here, all fractions have integer part $= 0$, so comparing reciprocals:

$$\bigg\lfloor \frac{865}{27}\bigg\rfloor = 32\ >\ \bigg\lfloor \frac{686}{26}\bigg\rfloor = 26\ >\ \bigg\lfloor \frac{874}{48}\bigg\rfloor = 18\ >\ \bigg\lfloor \dfrac{892}{80}\bigg\rfloor = 11$$

Since reciprocating reverses inequalities, reversing the above yields the sought equalities.

Community
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Bill Dubuque
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$\dfrac{1}{30} < \dfrac{26}{686} < \dfrac{1}{20}$ because $30 \times 26 = 780 > 686$ while $20 \times 26 = 520 < 686$.

$\dfrac{1}{20} < \dfrac{48}{874} < \dfrac{1}{12}$ because $20 \times 48 = 960 > 874$ while $12 \times 48 = 144 \times 4 = 576 < 874$

$\dfrac{1}{12} < \dfrac{80}{892}$ because $12 \times 80 = 960 > 892$.

$\dfrac{27}{865} < \dfrac{1}{30}$ because $30 \times 27 = 810 < 865$.

So $\dfrac{27}{865} < \dfrac{26}{686} < \dfrac{48}{874}< \dfrac{80}{892}$.

Robert Israel
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You can order the fractions using following method. For any two fractions $A= \frac{a}{b}, B= \frac{c}{d}$ take $\frac{\frac{a}{b}}{\frac{c}{d}} =C$. Then if $C>1$ you have $A>B$ if $C<1$ then $B<A$ follows, finally if $C=1$ obviously $A=B$.

borg
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its enough compare 1/100 multiply each fraction! so we have 26/6.86, 48/8.74, 80/8.92, 27/8.65 so we have approximation 4, 6, 10, 3 respectively for this fractions, and obviously 27/865 is smallest and 80/892 is the biggest fraction.

M.K. Dadsetani
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