Cancelling Everything in Sight

Cancelling Everything in Sight (CES)

Seeing a complicated fraction become less ugly as elements are cancelled from both the numerator and denominator can be something of an enjoyable experience. One's first exposure to this magical process usually comes in grade school when reducing fractions, such as

$\begin{displaymath}\frac{52}{30} \;=\; \frac{\cancel{2}\cdot 26}{\cancel{2}\cdot 15} \;=\; \frac{26}{15}. \end{displaymath}$

High school algebra classes build upon this, showing us that we may also cancel expressions involving variables, as in

$\begin{displaymath}\frac{(x+2)(x+4)}{(2x-1)(x+2)} \;=\; \frac{\cancel{(x+2)}(x+4)}{(2x-1)\cancel{(x+2)}} \;=\; \frac{x+4}{2x-1}. \end{displaymath}$

What some students do not notice is that these cancellations only are performed once the numerator and denominator are factored. Factoring a numerator (or denominator) turns it into an expression which is, at its top level, held together by multiplication. For instance, in the expressions

$\begin{displaymath}\begin{array}{ll} \ds{\frac{x(3-x)}{5(x+2)}} & \mbox{numera... ...ither numerator nor denominator is factored.} \\ \end{array} \end{displaymath}$

To be sure that one performs valid cancellations only, it is necessary to

be patient, making sure to factor numerator and denominator first, and cancelling only those factors common to both, and
accept that many times no factorization is possible, at least none that leads to a common, cancellable factor.

With this in mind, cancellations such as those below may only be labelled instances of someone ``cancelling everything in sight", with no attention given to the discussion above, and having no validity whatsoever.

$\begin{displaymath}\frac{3x^2 + 2x - 1}{2x - x^2} \;=\; \frac{3\cancel{x^2} + 2... ...}{- 1} \;=\; -2. \mbox{\hspace{0.4in} \epsfig{0.5in}{ces.eps}}\end{displaymath}$

Any attempt to simplify the original fraction (rational expression) should start with factoring:

$\begin{displaymath}\frac{3x^2 + 2x - 1}{2x - x^2} \;=\; \frac{(3x-1)(x+1)}{x(2-x)}, \end{displaymath}$

at which stage we see that there is no matching factor between those of the numerator -- namely,

and

-- and those of the denominator --

and

. Factoring, in this case, did not lead to any cancelling, as is often the case.

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Thomas L. Scofield 2003-09-04