Is (a/b)/c = a/(b/c)?
This is part of a series on common misconceptions.
True or False?
For non-zero numbers \(a\), \(b,\) and \(c\),
\[ \frac{\left(\frac{a}{b}\right)}{c} = \frac{a}{\left(\frac{b}{c}\right)}.\]
Why some people say it's true: Just like with multiplication, the order that you choose to evaluate the two division operations doesn't matter.
Why some people say it's false: It's almost never true. Maybe there are a few very special cases for choosing \(a\), \(b,\) and \(c\) so that it's true.
This statement is \( \color{red} {\textbf{false}}\).
In general, \(\frac{\left(\frac{a}{b}\right)}{c} \neq \frac{a}{\left(\frac{b}{c}\right)}\). Consider the following example:
Does \(\frac{\left(\frac{36}{6}\right)}{2} = \frac{36}{\left(\frac{6}{2}\right)}\)?
We have \(\frac{\left(\frac{36}{6}\right)}{2} = \frac{6}{2} = 3.\)
We also have \( \frac{36}{\left(\frac{6}{2}\right)} = \frac{36}{3} =12.\)
Clearly, \(3 \neq 12\). Therefore, the given equation is false. \(_\square\)
The special cases when \(\frac{\left(\frac{a}{b}\right)}{c} = \frac{a}{\left(\frac{b}{c}\right)}\):
If \(a\), \(b,\) and \(c\) all equal 1, then this equation is true. But there are many other examples.
If \(\frac{\left(\frac{a}{b}\right)}{c} = \frac{a}{\left(\frac{b}{c}\right)}\), then by cross-multiplying, we can simplify to
\[\begin{align} \frac{a}{b} \times \frac{b}{c} &= a \times c\\ \frac{a \times b}{b \times c} &= a \times c\\
\frac{a}{c} &= a \times c\\
\frac{1}{c} &= c. \end{align}\]The only two values that satisfy the equation \(\frac1c = c\) are \(c=1\) and \(c=-1\). Therefore, the set of all cases when the original equation is true are those where \(c = 1 \text{ or } -1\). The other two variables can take on any values, so long as \(c\) is so chosen.
Example:
Does \(\frac{\left(\frac{36}{6}\right)}{-1} = \frac{36}{\left(\frac{6}{-1}\right)}\)?
We have \(\frac{\left(\frac{36}{6}\right)}{-1} = \frac{6}{-1} = -6.\)
We also have \( \frac{36}{\left(\frac{6}{-1}\right)} = \frac{36}{-6} =-6.\)
Therefore, \((a,b,c)=(36,6,-1)\) is one special case when the given equation is true. \(\square\)
Rebuttal: But then, how do you know what to do first if you have an expression like\[36 \div 6 \div 3 =\, ?\]
Reply: The chosen convention for such circumstances is to evaluate from left to right. Therefore,
\[36 \div 6 \div 3 = (36 \div 6) \div 3 = 6 \div 3 = 2.\]
Rebuttal: I guess I believe that it's false now, but my intuition for why is all messed up.
Reply: Here's an intuitive way to think about why \(\frac{\left(\frac{a}{b}\right)}{c}\) is not equivalent to \(\frac{a}{\left(\frac{b}{c}\right)}.\)
Think about the numbers, \(a\), \(b,\) and \(c\) as troops of soldiers in a video game. Division by a number larger than 1 is like a sneak attack of one group on another that cuts down the attacked force's number. Our troop is troop \(a\) and we want to know what's left of them at the end of evaluating these fractions in each of the two given situations
\[\frac{\left(\frac{a}{b}\right)}{c}\ \text{ and }\ \frac{a}{\left(\frac{b}{c}\right)}.\]
The first option \(\frac{\left(\frac{a}{b}\right)}{c}\) corresponds with the following situation: First, \(b\) sneak attacks our group \(a\), cutting \(a\) down to a smaller size. Then \(c\) comes and attacks \(a\) as well, cutting our number down even further.
On the other hand, \(\frac{a}{\left(\frac{b}{c}\right)}\) corresponds to the situation where, first, \(c\) attacks \(b\) . Then b attacks \(a\), but having been previously cut down by \(c,\) the attack is much less effective. Clearly, troop \(a\) fairs much better in the second scenario. In fact, in the second scenario, \(c\) is effectively an ally to \(a\), whereas in the first scenario, it was an attacking force.
This intuition can even be used to identify the special cases for which the given equation is true. The only case when an enemy force is effectively the same as an ally is when that force is inactive or neutral. As either 1 or -1, \(c\) doesn't affect the magnitudes of the other values in the equation. Therefore, those are the only special cases when the equation is true. (See the proof above for details.)
Want to make sure you've got this concept down? Try this problem:
See Also
