# Explain how and why the cancellation of 6’s in 16/64 to get 1/4 is a fallacious statement. 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?

Explain how and why the cancellation of $6$’s in $\frac{16}{64}$ to get $\frac{1}{4}$ is a fallacious statement.
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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escobamesmo
The way someone might have justified that:
"Just remember that
$\frac{10}{20}=\frac{1\overline{)0}}{2\overline{)0}}=\frac{1}{2}$
Therefore
$\frac{16}{64}=\frac{1\overline{)6}}{\overline{)6}4}=\frac{1}{4}$
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
$\frac{a\cdot b}{a\cdot d}=\frac{\overline{)a}b}{\overline{)a}d}$
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
$a\cdot b\cdot \frac{1}{a}\cdot \frac{1}{d}=\overline{)a}\cdot b\cdot \overline{)\frac{1}{a}}\cdot \frac{1}{d}=\frac{b}{d}$
The problem with the digits is that $16$ is not $1\cdot 6$ just as $64\ne 6\cdot 4$. 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 ${k}^{\prime }$ with $10$ and $j$ is the product of ${j}^{\prime }$ with $10$. Then we have:
$\frac{k}{j}=\frac{{k}^{\prime }\cdot 10}{{j}^{\prime }\cdot 10}=\frac{{k}^{\prime }\cdot \overline{)10}}{{j}^{\prime }\cdot \overline{)10}}=\frac{{k}^{\prime }}{{j}^{\prime }}$

anudoneddbv
The wrong proof is more of a joke than a serious fallacy:
$\frac{16}{64}=\frac{16/}{/64}=\frac{1}{4}$
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.