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Do It Wrong, Get It Right Author(s): David J. C. Elgin Source: Mathematics in School, Vol. 33, No. 1 (Jan., 2004), pp. 14-15 Published by: The Mathematical Association Stable URL: http://www.jstor.org/stable/30215650 . Accessed: 06/10/2013 09:33 Your use of the JSTOR archive indicates your acceptance of the Terms & Conditions of Use, available at . http://www.jstor.org/page/info/about/policies/terms.jsp . JSTOR is a not-for-profit service that helps scholars, researchers, and students discover, use, and build upon a wide range of content in a trusted digital archive. We use information technology and tools to increase productivity and facilitate new forms of scholarship. For more information about JSTOR, please contact [email protected]. . The Mathematical Association is collaborating with JSTOR to digitize, preserve and extend access to Mathematics in School. http://www.jstor.org This content downloaded from 129.64.99.141 on Sun, 6 Oct 2013 09:33:56 AM All use subject to JSTOR Terms and Conditions

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Page 1: Do It Wrong, Get It Right

Do It Wrong, Get It RightAuthor(s): David J. C. ElginSource: Mathematics in School, Vol. 33, No. 1 (Jan., 2004), pp. 14-15Published by: The Mathematical AssociationStable URL: http://www.jstor.org/stable/30215650 .

Accessed: 06/10/2013 09:33

Your use of the JSTOR archive indicates your acceptance of the Terms & Conditions of Use, available at .http://www.jstor.org/page/info/about/policies/terms.jsp

.JSTOR is a not-for-profit service that helps scholars, researchers, and students discover, use, and build upon a wide range ofcontent in a trusted digital archive. We use information technology and tools to increase productivity and facilitate new formsof scholarship. For more information about JSTOR, please contact [email protected].

.

The Mathematical Association is collaborating with JSTOR to digitize, preserve and extend access toMathematics in School.

http://www.jstor.org

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Page 2: Do It Wrong, Get It Right

P& it wrong, get it right

by David J. C. Elgin

A pupil was sitting in my class the other day doing fractions. He was not doing too well because in attempting to do the subtraction

5 9 4 12

what he did, was to subtract the numbers on the numerator and then subtract the numbers on the denominator, simplify and get an answer

5 9 -4 4 1 i.e. 2 - 8

- 8

4 12 -8 8 2

(He did actually know that a negative divided by a negative was a positive.)

I tried to explain to him that to do such problems you had to make the denominators the same and therefore you would

5 15 change 4 to 5. You could then subtract the fractions

4 12

because the denominators were the same and after cancelling, you would get the correct answer

15 9 6 1

12 12 12 2

"There you go", he said, "same answer as I got but my method was much simpler".

"This must be a fluke" I thought.

I tried other examples to show him that his method produced the correct answer only very rarely and that it was really luck which had actually given him the right answer.

It was a while before we found another example which produced the same type of result, i.e. do the wrong steps and

7 16 get the right answer. This example was7

1 3 12

5 9 A third producing the same effect was found.

2 6 We then got interested in looking for examples where this effect was occurring. It was time to turn to a bit of algebra.

Can I find numbers a, b, m and n so that the following result occurs?

a b a-b m n m-n

an - bm a- b Common denominator on LHS

- bm a - b )

mn m - n

Cross-multiplying gives

(an - bm) (m - n) = mn (a - b).

Removing the brackets and simplifying gives

an2 + bm2 - 2bmn = 0. (1)

Dividing through by am2, a 0, m 0 O, gives

n 2 + 0. -2 ') +

(a b n

Letting R =- and r = a, we obtain r2-2Rr + R = 0 (2) a m

which is a quadratic in r.

This quadratic has solutions

2R a - 4R2 - 4R r -

2

which simplifies to

r = Ra -\R2 -R, r = R aR (R - 1).

We need R < 0 or R> 1 for real roots.

If R > 1 then b > a, and if R < 0 then 'a' and 'b' must be of opposite sign which would lead to the case of the sum of two fractions.

From (2) we see that

-2 R - .

(3) 2r- 1

n Now r is rational since r - , and hence R is rational also.

m

A procedure therefore for obtaining examples could be

(i) Choose a value for r.

(ii) This fixes the ratio n/m.

(iii) Use (3) to find R and this fixes the ratio b/a.

(iv) Select values for a, b, m and n which fit the two ratios.

You therefore have values for a, b, m and n which will satisfy the criteria for (M) to be true.

14 Mathematics in School, January 2004 The MA web site www.m-a.org.uk

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Page 3: Do It Wrong, Get It Right

Example

Take r = 3, then n/m = 3/1.

This gives R = 9/5 so b/a = 9/5.

If you take a = 5, b = 9, m = 4 and n = 12 say then you get the non-trivial example

5 9

4 12

and even if a pupil did all the wrong things they would still end up with the right answer.

Some Possible Examples 9 12 3 4 6 8 9 12 5 10' 6 12' 7 14' 7 14

5 9 5 9 3 4 2-6'4 12 5 10

Keywords: Fractions; Quadratic equation.

Author David J. C. Elgin, High School of Dundee, Mathematics Department, Euclid Crescent, Dundee DD1 1HU. e-mail: [email protected]

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Mathematics in School, January 2004 The MA web site www.m-a.org.uk 15

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