Top Algebra Errors Made by Calculus Students

by Thomas L. Scofield
Assistant Professor of Mathematics
Calvin College
Sept. 5, 2003




Preface, for the reasons behind this document

Linear Function Behavior (LFB)

Lines through the origin are peculiar in that they have an expression of the form $f(x) = mx$, where $m$ is a constant (the slope). This formula makes possible the following ``additive" property:

\begin{displaymath}f(x_1 + x_2) = m(x_1 + x_2) = mx_1 + mx_2 = f(x_1) + f(x_2). \end{displaymath}

For the particular choices of $x_1 = x^2$ and $x_2 = 2x$, we would have
\begin{displaymath}f(x^2 + 2x) = f(x^2) + f(2x) = f(x^2) + 2f(x). \end{displaymath}

Despite students' tendancies to treat every function as additive, other functions just do not have this property. Typical mistakes made include:
\begin{displaymath}\sqrt{3x^2 + 2x} = \sqrt{3x^2} + \sqrt{2x}
\mbox{\hspace{0.6in} \epsfig{0.5in}{lfb.eps}} \end{displaymath}

\begin{displaymath}(x + 2)^3 = x^3 + 8
\mbox{\hspace{0.6in} \epsfig{0.5in}{lfb.eps}} \end{displaymath}

\begin{displaymath}\ln(2x-1) = \ln(2x) - \ln 1
\mbox{\hspace{0.6in} \epsfig{0.5in}{lfb.eps}} \end{displaymath}

\begin{displaymath}\frac{1}{x+5} = \frac{1}{x} + \frac{1}{5}
\mbox{\hspace{0.6in} \epsfig{0.5in}{lfb.eps}} \end{displaymath}

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 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, $(3x-1)$ and $(x+1)$ -- and those of the denominator -- $(x)$ and $(2-x)$. Factoring, in this case, did not lead to any cancelling, as is often the case.

Confusing Negative and Fractional Exponents (CNFE)

Students can make a variety of mistakes when it comes to working with exponents. Two of the most common are Multiplying Exponents that should be Added (MEA), and Adding Exponents that should be Multiplied (AEM). This section does not deal with either of these, but rather with a problem that some students have applying two basic rules about exponents, the ones concerning reciprocals and roots. Specifically, these are

\begin{displaymath}\frac{1}{x^m} \;=\; x^{-m} \qquad\mbox{and}\qquad
\sqrt[q]{x^p} \;=\; x^{p/q}, \end{displaymath}

respectively, where the understanding is that a square root ( $\sqrt{\phantom{x}}$) is to be taken as ( $\sqrt[2]{\phantom{x}}$).

The first of these says that a factor of the denominator (see the discussion on CES) raised to a power (be it positive or negative) may be written as a factor to the oppositite power of the numerator (i.e., a $(-2)$ power becomes $(+2)$, a $(3/4)$ power becomes $(-3/4)$). The only change is to the sign of the exponent. An example of a valid application of this rule is

\begin{displaymath}\frac{3}{2x^3} \;=\; \frac{3}{2}x^{-3} \qquad\mbox{or}
\qquad 3(2)^{-1}x^{-3}. \end{displaymath}

The second rule shows how to write a root as a power, which can be especially helpful in calculus when a derivative is desired. Things like

\begin{displaymath}\begin{array}{lll}
\ds{\sqrt[3]{3x^2}} &
\mbox{may be writt...
... as} &
\ds{\frac{\sqrt{5}x^{1/2}}{(x-2)^{1/2}}}.
\end{array} \end{displaymath}

Some students seem to confuse these two rules. The main errors seem to come from students trying to reciprocate the wrong thing
\begin{displaymath}\frac{1}{x^2} = x^{1/2},
\mbox{\hspace{0.6in} \epsfig{0.5in}{cnfe.eps}} \end{displaymath}

\begin{displaymath}\frac{2}{x^{1/2}} = 2x^2,
\mbox{\hspace{0.6in} \epsfig{0.5in}{cnfe.eps}} \end{displaymath}

or from students putting a minus in when none is required
\begin{displaymath}\sqrt[3]{x^2} = x^{-2/3},
\mbox{\hspace{0.6in} \epsfig{0.5in}{cnfe.eps}} \end{displaymath}

\begin{displaymath}\frac{3}{\sqrt{2x-1}} \;=\; \frac{3}{(2x-1)^{-1/2}}
\;=\; 3(2x-1)^{1/2}.
\mbox{\hspace{0.6in} \epsfig{0.5in}{cnfe.eps}} \end{displaymath}

Multiplication Ignoring Powers (MIP)

Another law of exponents frequently misunderstood by students is

\begin{displaymath}(a^m)(b^m) = (ab)^m. \end{displaymath}

This means that such statements as
\begin{eqnarray*}
4x^2 & = & 2^2 x^2 = (2x)^2, \qquad\mbox{and} \\
3\sqrt{x} & = & 9^{1/2}x^{1/2} = (9x)^{1/2} = \sqrt{9x}.
\end{eqnarray*}

are correct. But many students ignore the significance of having identical powers in these multiplications. They make statements like the following, all of which are incorrect:
\begin{displaymath}2x^{1/2} = \sqrt{2x},
\mbox{\hspace{0.6in} \epsfig{0.4in}{mip.eps}} \end{displaymath}

\begin{displaymath}-(3x)^2 = 9x^2,
\mbox{\hspace{0.6in} \epsfig{0.4in}{mip.eps}} \end{displaymath}

\begin{displaymath}3(x+1)^2 = (3x+3)^2,
\mbox{\hspace{0.6in} \epsfig{0.4in}{mip.eps}} \end{displaymath}

\begin{displaymath}\frac{3}{2x^2} = 3(2x)^{-2}.
\mbox{\hspace{0.6in} \epsfig{0.4in}{mip.eps}} \end{displaymath}

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...
...p}{1}
\;=\; \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 $\ds{\frac{\sqrt{2+x+h} - \sqrt{2+x}}{h}}$.
\begin{eqnarray*}
\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}},
\end{eqnarray*}

which cannot be further simplified.



Multiplication Without Parentheses (MWP)

The discussion here necessarily must begin with an appeal to the order of algebraic operations (OO). These are rules of hierarchy as to which operations to perform 1st, 2nd, etc. when an algebraic expression requires more than one operation be performed. There are three levels of hierarchy:

  1. powers,
  2. multiplication and division, and
  3. addition and subtraction.
When faced with an expression like the one below that has both an addition and a multiplication in it, the order of operations dictates that the multiplication be performed first:
\begin{displaymath}2 + 3\cdot5 \quad\mbox{is}\quad 17, \quad\mbox{not}\quad 30. \end{displaymath}

The levels above do not give the whole story, however. For instance, what if an expression has both an addition and a subtraction, operations which appear at the same level? The answer here is that operations appearing on the same level are always performed left-to-right:
\begin{displaymath}2 + 3 - 5 \quad\mbox{is}\quad 0, \qquad
2 - 3 + 5 \quad\mbox...
...ox{and}\qquad
2 / 3 \cdot 5 \quad\mbox{is}\quad \frac{10}{3}. \end{displaymath}

Also, one may use parentheses to override these rules. Things in parentheses are performed before things outside of those parentheses, starting from the inside and working out. So
\begin{displaymath}2 - (3 - (2 - 6)) \quad\mbox{is}\quad -5, \end{displaymath}

while
\begin{displaymath}2 - (3 - 2 - 6) \quad\mbox{is}\quad 7 \end{displaymath}

and
\begin{displaymath}2 - 3 - 2 - 6 \quad\mbox{is}\quad -13. \end{displaymath}

These order of operations apply to expressions involving variables as well. Thus

\begin{displaymath}\begin{array}{lll}
x \cdot 2x - 7 &
\qquad\mbox{is not the ...
...qquad &
x(2x-7),  \\
3-x/x^2 & &
(3-x)/x^2.
\end{array} \end{displaymath}

In this light, acceptable notation for the product of two expressions like $(3)$ and $(-5x^2)$ is
\begin{displaymath}(3)(-5x^2) \qquad\mbox{or, more simply}\qquad -15x^2, \end{displaymath}

not, as so many students write,
\begin{displaymath}3\cdot -5x^2.
\mbox{\hspace{0.6in} \epsfig{0.5in}{mwp.eps}}\end{displaymath}

Frivolous Parentheses (FP)

There really isn't an error, per se, with using too many parentheses. Nevertheless, students who consistently employ more parentheses than needed are demonstrating as much of a lack of understanding of the order of algebraic operations as those who use too few. Expressions such as

\begin{displaymath}\begin{array}{lll}
\ds{\left(\frac{(2x^2+3x)}{(x-1)}\right)}...
...{x^3}}
\;\;\;\mbox{or}\;\;\;
(3-x)/x^3.  \\
\end{array}. \end{displaymath}



Undo Multiplication with Division (UMD); also, Undo Addition with Subtraction (UAS)

The properties of equality that were mentioned earlier are, by some students, implemented incorrectly even when the situation calls for their use. For instance, when solving an equation like

\begin{displaymath}3x + 7 = 4, \end{displaymath}

two steps are called for:
\begin{displaymath}3x = -3 \qquad\mbox{({\bf {subtraction prop. of equality}}; $7$
subtracted from both sides)} \end{displaymath}

and
\begin{displaymath}x = -1 \qquad\mbox{({\bf {division prop. of equality}};
both sides divided by $3$)}. \end{displaymath}

Notice that, in the expression $(3x + 7)$, the order of operations dictates that the multiplication by 3 comes before the addition of 7, and the ``undoings" of these processes -- the subtraction of 7 and the division by 3 -- were carried out in reverse order. That is not to say that we could not have undone things in a different order, but students who do so often make the following error. Dividing by 3, they often neglect the fact that all terms on both sides are to be divided by 3. In other words, after dividing by 3 they write
\begin{displaymath}x + 7 = \frac{4}{3} \qquad\mbox{instead of}\qquad
x + \frac{7}{3} = \frac{4}{3}. \end{displaymath}

They are too set on the idea that they will be subtracting 7 from both sides to realize that, having divided by 3 first, it is not 7, but rather $7/3$, which must be subtracted, giving the same answer $x=-1$ as before. One other note is in order here. If parentheses appear in an equation such as
\begin{displaymath}3(x - 1) = 5, \end{displaymath}

then the order of operations are preempted (the subtraction within the parentheses comes before the multiplication by 3). In solving for $x$, we may of course, distribute the 3, thereby eliminating the parentheses and making the problem appear like the last one discussed. Even fewer steps are required if one just ``undoes" the multiplication and subtraction in their opposite order:
\begin{displaymath}x - 1 = \frac{5}{3} \qquad (\mbox{{\bf {division prop. of equality}};
dividing both sides by 3}), \end{displaymath}

and then
\begin{displaymath}x = \frac{8}{3} \qquad (\mbox{{\bf {addition prop. of equality}};
adding 1 to both sides}). \end{displaymath}

Now let us return to the equation

\begin{displaymath}3x + 7 = 4, \end{displaymath}

and investigate the more telling errors that gave the titles UMD and UAS to this section. Some students recognize the need for two steps (like those carried out when this equation was being considered above) to isolate $x$, but have little feel for which operations will achieve this. For instance, realizing that, like the $4$ on the right-hand side of the equation, $7$ is a ``non-$x$" term, a student may write
\begin{displaymath}3x = \frac{4}{7},
\mbox{\hspace{0.6in} \epsfig{0.5in}{uas.eps}}\end{displaymath}

misunderstanding that she has subtracted 7 on the left side, but divided by 7 on the other side. The original equation and the new one no longer have the same solutions as a result. The same student may then recognize that she needs to move the $3$ over to the other side. Since the 3 is multiplied by the $x$, she should ``undo" this by dividing both sides by 3. But she may (wrongly) write
\begin{displaymath}x = \frac{4}{7} - 3,
\mbox{\hspace{0.6in} \epsfig{0.5in}{umd.eps}}\end{displaymath}

having divided on the left but subtracted on the right. Again, the solution $x = -17/7$ is different from the one that solved the original equation $3x+7=4$, namely $x=-1$.

Worse still is when a student thinks he can solve in one step (that is, take care both of the multiplication by 3 and the addition of 7 via one operation). Such a student may write something like

\begin{displaymath}3x + 7 = 4 \quad\Rightarrow\quad x = \frac{4}{3+7}.
\mbox{\hspace{0.6in} \epsfig{0.5in}{uas.eps}/\epsfig{0.5in}{umd.eps}}\end{displaymath}

Again, the answer this student gets, $x = \frac{2}{5}$, is different than the correct one $x=-1$.



Misunderstood Relationship between Roots and Zeros (MRRZ)

Much of one's mathematical experience prior to the calculus is spent in solving equations. There are the kind of equations, known as identities, where every number in the domain is a solution. The equation

\begin{displaymath}(x - 2)(2x + 3) = 2x^2 - x - 6, \end{displaymath}

is such an identity. It is not this, but the other type of equation, known as a conditional equality, that one learns to solve, precisely because solutions, also known as roots, of conditional equalities are not everywhere to be found. Often there are very few numbers, perhaps even none at all, which make a conditional equality true.

Another fact about conditional equalities is that comparatively few of them may be solved exactly. Leaps in technology have made it commonplace for students, with the purchase of a handheld calculator, to have at their fingertips powerful graphing capability and numerical methods for finding approximate solutions to many, perhaps even most conditional equalities. This does not mean that one should forego learning the algebraic techniques which lead to exact solutions, thinking that deftness in pushing the right pair of buttons is an appropriate substitute for the thinking processes such algebraic methods introduce. Still, there is added value in the knowledge one gets by investigating graphical methods. By these methods one comes to think of the solutions of, say,

\begin{displaymath}3x^2 - 2x = 2x^3 + x^2 - 1 \end{displaymath}

as the $x$-values of points of intersection between the graphs of the two functions
\begin{displaymath}y = 3x^2 - 2x \qquad\mbox{and}\qquad y = 2x^3 + x^2 - 1. \end{displaymath}

As another example, solutions of the equation
\begin{displaymath}x^2 + 5x = -6 \end{displaymath}

would be found at points of intersection between the graphs of
\begin{displaymath}y = x^2 + 5x \quad\mbox{(a parabola)}\qquad\mbox{and}\qquad
y = -6 \quad\mbox{(a horizontal line)}. \end{displaymath}

It is in solving equations like this latter one that students become confused. What some students do is the following:
\begin{eqnarray*}
x^2 + 5x = -6 & \quad\Rightarrow\quad & x(x+5) = -6
\quad\mb...
...or}\;\; x = -5.
\mbox{\hspace{0.6in} \epsfig{0.6in}{mrrz.eps}}
\end{eqnarray*}
In mathematical terms, the student who does these steps has found the zeros of $f(x) = x^2 + 5x$; that is, the values for $x$ which make the output of $f$ be zero. There are several ways to see that this work is wrong. One way to see it is that, in the equation, we want values of $x$ whose output value is $(-6)$, not zero. Another angle which reveals the errors is the one that notes that, while there are a lot of pairs of numbers which may be multiplied to give (-6) -- (-1) and 6, 12 and (-1/2), 55 and (-6/55) are three such pairs -- one thing which we can say for certain is that neither of the numbers in the pair is zero, which is quite counter to the idea of setting the factors equal to zero. (Of course, neither is it enough to set the factors equal to (-6), as in
\begin{eqnarray*}
x(x+5) = -6 & \qquad\Rightarrow\qquad & x = -6 \;\;\mbox{or}
...
... x = -11, \mbox{\hspace{0.6in}
\epsfig{0.6in}{mrrzPrime.eps}},
\end{eqnarray*}
since it is not enough for either one of these conditions to hold by itself; that is, if $x=-6$ then we would need the other factor $(x+5)$ to be equal to 1 in order for their product to be $(-6)$, and clearly these things cannot occur at the same time.)

In summary, the error occurs in finding the zeros of a function and taking to be the roots of the equation, when the two concepts do not coincide. There is a simple way to make them coincide. We simply make one side zero (using a valid algebraic step, of course).

\begin{eqnarray*}
x^2 + 5x = -6 & \quad\Rightarrow\quad & x^2 + 5x + 6 = 0 \\
...
...x + 3 = 0 \\
& \Rightarrow & x = -2 \;\;\mbox{or}\;\; x = -3.
\end{eqnarray*}
The zeros of $g(x) = x^2 + 5x + 6$ are the numbers which make $g$ equal zero, and that is exactly what we want in a solution of the equation $x^2 + 5x + 6 = 0$, so the two concepts coincide. Why do students mess this up? The most likely answer is that many are looking to do as little work as possible, and bringing the $(6)$ over makes factoring a more difficult job (it is harder to factor $x^2 + 5x + 6$ than to factor $x^2 + 5x$); of course, the quadratic formula is an option for this case. What may help to avoid this confusion is remembering this graphical interpretation of what one is doing (still applied to the example above):



Multiplication Not Distributive (MND)

In precalculus/algebra we become familiar with the distributive laws that address interactions between multiplication and addition/subtraction. Specifically, these laws say

\begin{displaymath}a(b+c) = ab + ac \qquad\mbox{and}\qquad (a + b)c = ac + bc. \end{displaymath}

We use these laws all the time, both in expanding
\begin{eqnarray*}
3x^2(x - 2y) & = & 3x^3 - 6x^2y \qquad\mbox{and} \\
(x - 3)...
...x-3)x + (x-3)(7) \;=\; x^2 - 3x + 7x - 21
\;=\; x^2 + 4x - 21,
\end{eqnarray*}
and in factoring
\begin{displaymath}30x^2y - 12xy^2 + 3xy = 3xy(10x - 4y + 1). \end{displaymath}

We even use it (although we don't often think about it this way and usually don't include the middle step below) when combining like terms, as in
\begin{displaymath}4xy - 15xy \;=\; (4 - 15)xy \;=\; -11xy. \end{displaymath}

The problem is when students misinterpret these laws, thinking they also say something about interactions between more than one multiplication; that is, they ``invent" for themselves a law that looks something like:
\begin{displaymath}a(b\cdot c) = (ab)(ac).
\mbox{\hspace{0.6in} \epsfig{0.5in}{mnd.eps}}\end{displaymath}

This clearly is false, as most would see if these were all numbers -- few (though I cannot go so far as to say no one) would assert, say, that
\begin{displaymath}7(3\cdot 10) = (21)(70) = 1470.
\mbox{\hspace{0.6in} \epsfig{0.5in}{mnd.eps}}\end{displaymath}

But when the objects involved are expressions involving variables, the error is frequently made, such as in this case:
\begin{displaymath}5(2x^2 y^3) = (10x^2)(5y^3) = 50x^2 y^3,
\mbox{\hspace{0.6in} \epsfig{0.5in}{mnd.eps}}\end{displaymath}

or
\begin{displaymath}3[(x-1)(7x)] = (3x-3)(21x).
\mbox{\hspace{0.6in} \epsfig{0.5in}{mnd.eps}}\end{displaymath}



Poor Use of Mathematical Language (PUML)

A prerequisite skill to writing good mathematics is the ability to write well in one's native tongue. People who cannot write a complete English sentence should take remediation in English composition before reading on.

What may surprise some students is that good writing using mathematical symbols (even in the write-up of homework problems) consists of using complete sentences, setting up one's ideas clearly and then following through on the details, much as one expects from a good English essay. The language and symbols of mathematics are used just like regular English words and phrases to express ideas, albeit ideas which one would often struggle to use any other means of expressing.

Nobody studies mathematical writing as a subject. Your mathematics professor(s) got to be good writers of mathematics, if good they be, by reading papers and books by other mathematicians, not by reading a treatise such as this one. If a book on good mathematical writing does exist (and there are probably a number of such books), they will say much more than I say here. I will only describe the most common example of poor mathematical writing I see when grading students' work: Using Equals as a Conjunction (UEC).

The word ``equals" has a very specific meaning. It requires two objects, and it asserts that these two objects are the same. In mathematics, the two objects are usually quantities, like the mathematical expression $(3x + 5)$, or the number 7. Even within this tight definition, mathematical equations, as I mentioned earlier, come in two varieties: identities and conditional equations. A conditional equation is one such as the equation

\begin{displaymath}3x + 5 = 7, \end{displaymath}

which is only for particular values of $x$ (in this case one particular value). In algebra courses one often sees conditional equalities in homework problems accompanied by the instruction ``Solve the equation". There are some quantities that are the same regardless of the value of the variable. A familiar example is the identity
\begin{displaymath}\sin^2 x + \cos^2 x = 1, \end{displaymath}

which is true no matter what real value $x$ takes. These two types of equations encompass the two most common (and only?) valid ways to use an `equals' ($=$) sign.

Consider the typical calculus problem of evaluating a limit like

\begin{displaymath}\Lim{x}{2} \frac{x^2 - x - 2}{x^2 - 4}. \end{displaymath}

What we are given here is not an equation, but an expression. If we begin writing a series of equalities to simplify/evaluate this expression, we will want them to be identities, as in
\begin{eqnarray*}
\Lim{x}{2} \frac{x^2 - x - 2}{x^2 - 4} & = &
\Lim{x}{2} \fra...
...
& = & \Lim{x}{2} \frac{x + 1}{x + 2} \\
& = & \frac{3}{4}.
\end{eqnarray*}
The original expression, along with each of the ensuing expressions, as it turns out, are all equal to the number $3/4$.

In contrast, suppose we begin with a (conditional) equation like

\begin{displaymath}3x + 5 = 7, \end{displaymath}

which we are asked to solve. If a student who understands very well the discussion of UAS and UMD (found earlier in this piece) makes a mistake, she is most likely to do so writing something like
\begin{displaymath}3x + 5 \;=\; 7 - 5 \;=\; \frac{2}{3}. \end{displaymath}              UEC

Such a string of equalities asserts three things:
  1. that $3x + 5 = 2$,
  2. that $3x + 5 = 2/3$,
  3. and that $2 = 2/3$.
(i) and (ii) are conditional equations in their own right, but it should be clear that they do not have the same solutions as the original equation $3x + 5 = 7$ (nor does (i) have the same solution as (ii)). And (iii) has no solution at all, for it is never true. What I am really saying is that the string of equations
\begin{displaymath}
3x + 5 \;=\; 7 - 5 \;=\; \frac{2}{3}
\end{displaymath} (1)

is really three equations, and there is no common solution between them (and, even if there had been, such a solution would have no relevance to the original problem, that of solving $3x + 5 = 7$). The student most likely never intended to assert these three equations in place of the original; she simply began writing out her ideas, and used an equals sign to join them together whenever it seemed some sort of conjunction was required.

The student who writes (1) actually appears to have some facility in the techniques for solving linear equations, but lacks the ability to put her ideas onto paper in a meaningful fashion. One good way to express the solution of the previous equation is

$\displaystyle 3x + 5 \;=\; 7$ $\textstyle \qquad\Rightarrow\qquad 3x \;=\;2$    
  $\textstyle \qquad\Rightarrow\qquad x \;=\; \ds{\frac{2}{3}}.$   (2)

The symbol $\Rightarrow$ can be translated here as ``which implies". Yes, (2) is more writing than (1), much in the same way the complete sentence ``I am taking the train to Chicago this weekend" requires more writing than the three words ``weekend, Chicago, train". A more favorable comparison is between (2) and the same ideas expressed in English words: If the sum of three times $x$ and five is seven, then subtracting five from both sides and dividing by three yields the value of two-thirds for $x$, or, perhaps more literally, Assuming that the sum of three times $x$ and five is seven, this implies that three times $x$ is two, and that $x$ is two-thirds.



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Thomas L. Scofield 2003-09-04