Equation Properties for Expressions (EPE)

Early on in one's high school algebra courses one learns several properties of equality -- namely

Notice that both of these properties pre-suppose that we start with an equation, usually one we are supposed to solve (say, for $x$). These properties are helpful in achieving that goal, as in: Solve $3x - 1 = 7$:

Add 1 to both sides: $3x = 8$
Divide both sides by 3: $x = \ds{\frac{8}{3}}$
or, solve $\ds{\frac{4x + 5}{3x^2 + 1}} = 0$:

Multiply both sides by the never-zero quantity $(3x^2+1)$: $4x + 5 = 0$
Subtract 5 from both sides: $4x = -5$
Divide both sides by 4: $x = -5/4$.

In contrast, these are not, generally speaking, properties one uses when trying to simplify an expression. (There are exceptions to this, such as in the simplifying of $\int e^x \sin x dx$ and $\int e^x \cos x dx$ using integration by parts, but these are relatively rare.) Students asked to find the derivative of

\begin{displaymath}f(x) := \frac{3x}{2x-1} \end{displaymath}

may find it easier to work with the function
\begin{displaymath}\frac{3x}{2x-1}\cdot(2x-1), \qquad\mbox{or}\qquad 3x, \end{displaymath}

but they shouldn't be under any illusions that $3x$ and $3x/(2x-1)$ are the same functions, nor that they have derivatives that are equal. If one is simplifying an expression like
\begin{displaymath}\frac{12/p - 5}{3p}, \end{displaymath}

it may be tempting to multiply by $p$, which gives
\begin{displaymath}\frac{12/p - 5}{3p}\cdot p
\;=\; \frac{12/p - 5}{3p}\cdot \f...
\;=\; \frac{(12/p - 5)p}{3p}
\;=\; \frac{12 - 5p}{3p}, \end{displaymath}

but, of course, multiplying by $p$ changed the expression. One must both multiply and divide by $p$ (equivalent to saying that we're multiplying by $1$) if the expression is going to remain the same (but hopefully simplified):
\begin{displaymath}\frac{12/p - 5}{3p}\cdot\frac{p}{p}
\;=\; \frac{12 - 5p}{3p^2}. \end{displaymath}

Another example is in simplifying the difference quotient $\ds{\frac{\sqrt{2+x+h} - \sqrt{2+x}}{h}}$.
\frac{\sqrt{2+x+h} - \sqrt{2+x}}{h}
& = & \frac{\sqrt{2+x+h}...
... + \sqrt{2+x})} \\
& = & \frac{1}{\sqrt{2+x+h} + \sqrt{2+x}},

which cannot be further simplified.

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Thomas L. Scofield 2003-09-04