Based on what we know from elementary and middle school teachers, most of us know that 16/64 correctly equals 1/4 because 16/64 is simplified with a common divisibility of 16. However, there is another way to prove that 16/64 equals 1/4 without dividing the numerator and denominator by 16. Who can explain how and why that method leads to a fallacious statement?

  • $\begingroup$ $$\require{cancel} \frac{12}{24} = \frac{1\cancel2}{\cancel24} = \frac14, \qquad \frac{13}{39} = \frac{1\cancel3}{\cancel39} = \frac19, \qquad \frac{14}{42} = \frac{1\cancel4}{\cancel42} = \frac12, \qquad \frac{15}{45} = \frac{1\cancel5}{4\cancel5} = \frac14, \qquad \cdots$$ $\endgroup$ – Rahul Nov 12 '16 at 23:29
  • $\begingroup$ en.wikipedia.org/wiki/Mathematical_fallacy#Howlers ...and this even has it's own Wiki page Anomalous cancellation $\endgroup$ – Winther Nov 12 '16 at 23:37
  • $\begingroup$ @Rahul Okay. So what is fallacious about these four other true statements? $\endgroup$ – Mindozas Nov 12 '16 at 23:39
  • $\begingroup$ @Winther Thanks, Winther. Glad you figured that out. $\endgroup$ – Mindozas Nov 12 '16 at 23:40
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    $\begingroup$ Hint: None of them are true statements. $\endgroup$ – Rahul Nov 12 '16 at 23:47

The way someone might have justified that:

"Just remember that

$$\require {cancel}\frac {10}{20} = \frac{1\cancel {0}}{2\cancel {0}} = \frac{1}{2} $$


$$\require {cancel}\frac{16}{64} = \frac{1\cancel {6}}{\cancel {6}4} = \frac{1}{4} $$


There is actually a problem on project euler regarding this type of fractions. Those that can be fallaciously simplified to something that holds as true.

Why it does not work:

There is a widely-used simplification that is

$$\require {cancel}\frac{a\cdot b}{a\cdot d} = \frac{\cancel {a}b}{\cancel {a}d} $$

That works because we have a product. The above fraction is just syntatic sugar for

$$a\cdot b \cdot \frac{1}{a} \cdot \frac{1}{d}$$

But the product is commutative and therefore we have

$$\require{cancel} a\cdot b \cdot \frac{1}{a} \cdot \frac{1}{d} = \cancel {a}\cdot b \cdot \cancel {\frac{1}{a}} \cdot \frac{1}{d} = \frac{b}{d}$$

The problem with the digits is that $16$ is not $1\cdot6$ just as $64 \not= 6\cdot4$. That means $\frac{1}{64} $ is not syntatic sugar for $\frac{1}{6}\cdot\frac{1}{4} $ and the 6s won't cancel. It only works when the numbers end in 0 because if $k $ and $j $ end in 0, then $k $ is the product of $k'$ with $10$ and $j $ is the product of $j'$ with $10$. Then we have:

$$\require {cancel}\frac{k}{j} = \frac{k'\cdot10}{j'\cdot10} = \frac{k'\cdot\cancel {10}}{j'\cdot\cancel {10}} = \frac{k'}{j'}$$

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    $\begingroup$ There's a difference of squares "identity" like this and I just saw someone use it not 5 minutes ago grading; $$\require{cancel}\dfrac{a \pm b}{a^2 - b^2} = \dfrac{a \pm b}{(a + b)(a - b)} = \dfrac{1}{a \mp b} ``=" \dfrac{\cancel{a} \pm \cancel{b}}{a^\cancel{2} - b^\cancel{2}};$$ it drives me crazy that this "works"... $\endgroup$ – pjs36 Nov 12 '16 at 23:26
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    $\begingroup$ To be honest, RSerrao, to the OP asking this question in the first place, I find this answer to be more confusing than helpful to the OP. You never explicitly address what is fallacious: i.e., the incorrect procedure you're employing. Indeed, your answer can be seen by newcomers as endorsing that tactic. Anyway, you're obviously "singing to the choir", hence the upvotes. For me, I'd rather "sing to the OP". $\endgroup$ – amWhy Nov 13 '16 at 0:05
  • $\begingroup$ @amWhy thanks for the critique. I have edited the answer to be a "better" answer as to what the OP is concerned. Please review it as well and let me know if there is more room for improvement. $\endgroup$ – RGS Nov 13 '16 at 0:25

The wrong proof is more of a joke than a serious fallacy: $$ \frac{16}{64} = \frac{16\llap{/}}{\rlap{/}64} = \frac 14 $$ This joke exploits the notational ambiguity that writing two symbols next to each other can either mean multiplication or -- if the symbols happen to be digits -- be part of the usual decimal notation for numbers, in which case it means something quite different from multiplying the digits together.

In the joke proof we pretend that $16$ and $64$ mean $1\cdot 6$ and $6\cdot 4$ (which of course they don't) and then "cancel the common factor" of $6$.

This doesn't really work because the $6$ is not a factor.

  • $\begingroup$ And that other method I was asking about you showed is considered a howler. Am I right? $\endgroup$ – Mindozas Nov 12 '16 at 23:34
  • $\begingroup$ @Mindozas: That's in the eye of the beholder. I don't find it particularly funny myself. $\endgroup$ – hmakholm left over Monica Nov 12 '16 at 23:36
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    $\begingroup$ I don't find it funny either. $\endgroup$ – Mindozas Nov 12 '16 at 23:43

It's not fallacious. $\frac{16}{64}$ is equal to $\frac 14$.

But the method is unsound, because it only works for a few special fractions. For example, if you try it with $\frac{12}{24}$ you get $$\require{cancel} \frac{12}{24} = \frac{1\cancel 2}{\cancel 2 4} = \frac14$$ which is completely wrong.


They may well know that

$$\require {cancel}\frac {25}{50} = \frac{1}{2} $$

If their logic was always correct, this would also be, which is not:

$$\require {cancel}\frac {25}{50} = \frac{2\cancel {5}}{\cancel{5} {0}} = \frac{2}{0} $$

So their cancellation method is not bullet proof. Once it does not work in all cases, they learn to focus better in the underlying logic of the cancellation.


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