Explain how and why the cancellation of $ 6 $’s in $ \dfrac{16}{64} $ to get $ \dfrac{1}{4} $ is a fallacious statement.
By John Thompson
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?
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$\begingroup$The way someone might have justified that:
"Just remember that
$$\require {cancel}\frac {10}{20} = \frac{1\cancel {0}}{2\cancel {0}} = \frac{1}{2} $$
Therefore
$$\require {cancel}\frac{16}{64} = \frac{1\cancel {6}}{\cancel {6}4} = \frac{1}{4} $$
$\blacksquare$"
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:
$$\require {cancel}\frac{k}{j} = \frac{k'\cdot10}{j'\cdot10} = \frac{k'\cdot\cancel {10}}{j'\cdot\cancel {10}} = \frac{k'}{j'}$$
$\endgroup$ 3 $\begingroup$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.
$\endgroup$ 3 $\begingroup$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.
$\endgroup$ $\begingroup$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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