1
$\begingroup$

Building on this question on SO, when comparing positive fraction to determine which is the largest/smallest, is it safe to consider the absolute difference between the Numerator and Denominator, assuming the fractions are in the lowest forms?

Although the attached SO question mentions it false, is it feasible if the fractions are in their lowest forms?

For example, consider the fractions:

a) 33/128

b) 45/138

c) 53/216

d) 83/324

e) 15/59

If we want to determine the smallest of these, we proceed as:

a) |128 - 33| = 95

b) |138 - 45| = 93

c) |216 - 53| = 153

d) |324 - 83| = 241

e) |59 - 15| = 44

Using this procedure can we safely assume 83/324 is the smallest fraction? (considering all of the fractions are in the lowest form)

$\endgroup$
4
  • $\begingroup$ Based on the answer in your link, it should rather be the lowest absolute difference, hence $15/59$. $\endgroup$
    – nicomezi
    Nov 6 '19 at 7:18
  • $\begingroup$ but should't the difference between the numerator and denominator be highest for the smallest fraction? $\endgroup$
    – Abrar
    Nov 6 '19 at 7:26
  • $\begingroup$ The answer in your link showed that, if $a<b$ and $c<d$, then $|a - b| < |c - d| \iff \frac{a}{b} < \frac{c}{d}$. Since $15 <59$ and $83<324$, then $15/56 < 83/324$ (which can be verified by a calculator). $\endgroup$
    – nicomezi
    Nov 6 '19 at 7:29
  • $\begingroup$ Note that 15/59 is not the largest despite having the smallest difference $\endgroup$
    – David
    Nov 6 '19 at 12:42
2
$\begingroup$

Even in lowest terms, what you're asking is still not necessarily correct. Consider just the $2$ fractions $\frac{1}{3}$ and $\frac{7}{11}$. The first is less than $\frac{1}{2}$ while the second is greater, so the first is the smallest. However, the absolute value of the difference between the numerator & denominator is $|1-3| = 2$ for the first one and $|7-11| = 4$ for the second one. Using the logic you suggest would indicate the second fraction, i.e., $\frac{7}{11}$, was smaller instead.

Note that $\frac{2}{3}$ and $\frac{3}{11}$ shows the reverse, i.e., the fraction with the smallest absolute difference is not necessarily the smallest value either.

The answers in the Math SE question you linked to explain in more detail why this checking of the absolute value of difference between the numerator and denominator doesn't always work, with their logic not depending on whether or not the fraction is in lowest terms. Instead, as the various answers indicate (e.g., the Edit section of the accepted answer), there are several conditions which must be satisfied to be able to use the differences of the numerator & denominator to compare the values of fractions.

$\endgroup$
0
1
$\begingroup$

That's not correct.

Let $A=\frac{1}{2}$ and $B=\frac{997}{1000}$ Note that $B$ is almost twice as big as $A$ but

$2-1=1$

$1000-997=3$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.