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...
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}

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

\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

x \cdot 2x - 7 &
\qquad\mbox{is not the ...
...qquad &
x(2x-7),  \\
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}

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Thomas L. Scofield 2003-09-04