Teachers discover new math theoremAuthor(s): HELEN G. RENZI and GEORGE C. CROSSSource: The Arithmetic Teacher, Vol. 12, No. 8 (DECEMBER 1965), pp. 625-626Published by: National Council of Teachers of MathematicsStable URL: http://www.jstor.org/stable/41185271 .

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Teachers discover new math theorem

HELEN G. RENZI and GEORGE C. CROSS Williamstown Public Schools, Williamstown, Massachusetts Mrs. Renzi and Mr. Cross are fourth-grade teachers at Grant School. Also, Mr. Cross is the head teacher of Grant School.

Who's afraid of modern mathematics? Not the teachers of the Williamstown, Mass., elementary schools who recently stumbled upon what we believe to be a new mathematics theorem. While the theorem itself will not change the course of the world, the circumstances in which it was discovered help to prove the value of the new approach to the teaching of mathematics.

This new development occurred as a re- sult of a math workshop taught by George F. Feeman, Professor of Mathematics at Williams College. Mr. Feeman, in leading up to a discussion of Euclid's Algorithm, wrote a series of numbers on the board to encourage the group to do some intuitive thinking on a certain relation involving the lowest common multiple and the greatest common divisor of two natural numbers.

The table of numbers, which the instruc- tor chose at random, looked like this :

a b lem gcd (1) 18 24 72 6 (2) 6 8 24 2 (3) 6 9 18 3 (4) 10 12 60 2 (5) 10 15 30 5 (6) 8 12 24 4

Mr. Feeman tried to lead the teachers to discover the fact that aX6=(gcd) X(lcm). Some members of the group did arrive at that conjecture, but in so doing one of us realized that in certain examples, a+6+gcd = lcm is true, for instance, in cases 3, 5, and 6. This, in turn, inspired her fellow author to do some intuitive

thinking and come to the conclusion that this occurs when a and b have a 2-3 ratio.

From this fortuitous combination of one elementary school workshop, sparked by one mathematics professor plus a ran- dom choice of numbers, we obtained the following theorem:

Let a and b be natural numbers. We de- note by [a, b] the lowest common multiple of a and ò, by (a, b) the greatest common divisor of a and b, and use the known re- sult that

аХЬ=[а, Ь]х(а, Ь).

Theorem. Let a, о, с, d be natural numbers where с and d are relatively prime and where

с b = - X a.

d

Then a+b= [a, b] - (a, b) if and only if

с г 2 - = - or - • d 2 3

Proof: (a) Suppose

- C = - 3 and A ' = - 3 - = - and A o = - a. d 2 2

Since b is a natural number, a must be even. So a = 2& for к some natural num- ber and therefore b = 3k. Thus, a + b = 5k. In addition, (a, b) = (3fc, 2k) = к and

. . aXb 2/сХЗ/с . [a,b]=--

. = = 6*. (a, 6) к

Hence, [a, b] - (a, b) - 5k and the result is established, i.e., a+b= [a, 6] - (a, b).

December I960 625

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The proof is quite similar for

с 2

7~ÌT '

(b) To prove the converse we let

a

where с and d are relatively prime. Since b is a natural number, a = dl f or I a natural number. Then ò = cZ and a+6 = (c+d)l by the distributive law. In addition, (a, b) = (cZ, dl) = l, since с and d are relatively prime, and

[a,b'=-- = cdl. (a, b)

Thus, [a, b]-(a, b)=cdl-l=(cd-l)l. Now [a, 6] - (a, ò) = a+ò implies (c+d)l = (cd- 1)1 which, by the cancellation law, yields c-'-d = cd - 1 or

d+1 °~d-l

'

We seek positive integral solutions of this equation so need only consider d>l. For d=2, c = 3, and

с 3

7~~2" '

For d=3, c = 2, and

с 2

d" 3

For d>3, we observe that Kc<2, since

d+1 2 c=- =1+

d-1 d-1

Hence, these are the only possible integral solutions for с and d, i.e., с and d are in the ratio 3 : 2. The converse is proved.

Mr. Feeman then suggested that we change the ratio of с to d to see if we would get other theorems of a similar nature. And we did !

For example, suppose the ratio eld is 3:5, so that & = fa. Then we get 2(a+b) = [a, &] + (a, b). The proof is similar to the preceding proof and is as follows:

If ò = fXa, and b is a natural number, then a = 5fc, ò = 3fc, (a, b) = k, and [a, b] = Ш, showing that 2(a+ř>) = 16fc = (a, b) + [a, Ò].

Conversely, if с

b =-Xa d

and ò is a natural number, then a = dl, b = clj (a, b) = l, and [a, b] = cdl, assuming that с and d are relatively prime. The condition 2(a+b)= [a, &] + (a, b) leads to the equation 2(c+d) = cd+l. From this we get

2d-l c = 1

d-2

and again we seek positive integral solu- tions. When d = 3, c = 5 and when d = 5, c = 3. Since

2d-l 3 c = = 2-' ,

d-2 d-2

it is clear that for d>5, 2<c<3. Hence, с and d in ratio 3 to 5 are the only solu- tions.

In this fashion, a whole class of theorems can be developed. We think it would be fun for the reader to find some others for himself.

A reminder NCTM members have previously had a choice of The Arithmetic Teacher, or The Mathematics Teacher, or both. Now, with any of these three options, members may elect to receive The Math-

ematics Student Journal for an addi- tional fee of 50 cents. This choice is avail- able now to new members and to present members at the time they renew their memberships.

626 The Arithmetic Teacher

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