A Problem with the Order of OperationsAuthor(s): Arthur GoodmanSource: The Mathematics Teacher, Vol. 72, No. 2 (FEBRUARY 1979), pp. 92-93Published by: National Council of Teachers of MathematicsStable URL: http://www.jstor.org/stable/27961546 .

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sharing teaching ideas

A Problem with the Order of Operations A common experience shared by many

mathematics teachers at all levels is the wonderful ingenuity that students show at

obtaining the "right answer"?but for all the wrong reasons. The usual teacher reac tion to such a situation is to provide a

counterexample that shows the student that the "wrong" method will not work in gen eral.

Students, however, are frequently skepti cal about mathematics and continue to wonder how they could manage to be so

lucky and get the right answer by an incor rect method. Regardless of how many counterexamples are given, this mystical feeling is not dispelled.

The following reports an attempt to dis

pel a student's misgivings. As a result, the class was led into a discussion that yielded a nice solution that justified the "incorrect" method in particular cases. An added bene fit was that the discussion provided the class with an opportunity to apply factor ing techniques as well.

During a lesson with a precalculus class on the order of the arithmetic operations and the use of grouping symbols, the class was asked to evaluate the following ex

pression.

7 - 4[7 -

4(7 -

4)] Several students simplified the expression as follows.

7 - 4[7 -

4(7 -

4)] = (3)[(3)(3)] = 27, which is the correct answer obtained by an incorrect method. After pointing out the

errors, I naturally proceeded to give what I

thought was a good counterexample.

5 - 3[5 -

3(5 -

3)]

However, when this is evaluated by the in correct method it gives

(2)[(2)(2)] = 8, the correct answer.

Now even my curiosity was aroused at the coincidence. After showing the class a

counterexample demonstrating that the "incorrect" method fails in general, we de cided to try to unravel the coincidence.

The problem was formulated: when will it be true that

(1) a - b[a- b(a -

b)] =

(a- b)[(a -b)(a -

b)] =

(a -

6)3? Now if the operations are performed cor

rectly on the left side of equation (1) while at the same time multiplying out (a

? b)3

equation (1) yields

(2) a - ab + ab2 - b* = a3 - 3a2b + 3ab2 - b3

or

(3) a3 - 3a2b + lab2 + ab - a = 0.

Now these equations are true if a = 0. Therefore, assume a 0, and divide both sides of equation (3) by a, which yields

(4) a2 - 3ab + 2b2 + b - 1 = 0.

An equation like (4) with integral coeffi cients that is known to have more than one

pair of integral solutions is a second degree Diophantine equation?something totally

Sharing Teaching Ideas offers practical tips on the teaching of topics related to the secondary school curriculum. We hope to include classroom-tested approaches that offer new slants on familiar subjects for the beginning and the experienced teacher. Please send an original and four copies of your ideas to the managing editor for review.

92 Mathematics Teacher

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new for a class at this level. The equation, however, can be solved using only the tools of intermediate algebra.

The class recognized that if we could fac tor the left-hand side of equation (4), we had a chance for a solution. This example demonstrates that factoring by grouping can be used in a quite natural way. The

expression on the left side of equation (4) can be factored as follows:

a2 - 3ab + 2b2 + b - 1 = 0

(5) (a -

2b)(a- b) + b -1 = 0

Now if we think of

a ?

2b =

and a -

b = y,

then y - =

(a -

b) -

(a -

2b) =

b,

which is the middle term of equation (5). So

equation (5) is of the form

(6) xy + (y -

) - 1 = 0.

Factoring the left side of equation (6) gives

( + 1)0 -

1) = 0.

Substituting = a ? 2b and y

= a ? 6, we

get

(a-2b+ \)(a - * - 1)

= 0.

So either

a-2b + \= 0ora-b-l=0

gives the two solutions

0 = 26-1 or = 6 + 1.

The second solution a = b + 1 means a ?

6=1, which is an obvious solution to equa tion (1). It was also noted that our class

examples were of the first type, that is, 7 =

2(4) - 1 and 5 = 2(3)

- 1. An alternative method of factoring (4) is

to use the method of completing the square. In order to produce a perfect square tri

9 nomial from a2 ? 3ab, the term

^ b2 must be

added. So from equation (4) we get

a2 - 3ab + ^b2 -\b2

+ 262 + b - 1 = 0

a2 - 3ab + |

b2 - \

b2 + b - 1 = 0

a2 - 3ab + |

62 = ^b2

- b + 1.

Multiplying this last equation by 4, we get

4a2 - \2ab + 962 = b2 - 4b + 4.

Factoring both sides of the equation, we get

(2a -

3b)2 =

(b -

2)2.

Then either 2a - 3b = b - 2 or 2a - 3b = -

(b -

2) gives the two solutions a = 2b ?

1 or a = b + 1. As a consequence of this discussion, the

class seemed to be more satisfied that the "incorrect" method was valid only in spe cial cases.

To err is human, to forgive, Diophan tine-

Arthur Goodman

Queens College City University of New York

Flushing, NY 11367

Teaching Fractions

The concept of fractional unit is some

times difficult for students to comprehend. I have found the following game and its variations to be very helpful in conveying such concepts as (1) fractional units of the

whole; (2) the relationship between proper fractions, improper fractions, and mixed

numbers; and (3) the addition and sub traction of fractions.

The game board and spinner are shown in figure 1. All the materials are drawn on

regular paper, cut out, and mounted on

stiff cardboard. Tokens for each player are made. The arrow is mounted to the playing board with a pin, fastener, or rivet.

To play the game, all players place their

tokens on start. The first player spins the arrow and must move his or her token the

February 1979 93

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