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Vol. 10 No 6 - Pi Mu Epsilon

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498 PI MU EPSILON JOURNAL<br />

PROBLEMS AND SOLUTIONS 499<br />

Solution by Heath Schulterman, student, Fort Smith <strong>No</strong>rthside High<br />

School, Barling, Arkansas.<br />

From the quadratic formula we know that the solutions to the above<br />

equation are<br />

We cube these roots to see that<br />

a± Ja 2 - 4<br />

X= •<br />

2<br />

3 a 3 - 3a a 2 1 ~-=---<br />

x = ± - ya 2 - 4.<br />

2 2<br />

If a is odd, then both 03 - 3a and a 2 - 1 are even numbers. Thus we may<br />

take p = (03 - 3a)/2, q = (a 2 - 1)/2, and r = a 2 - 4, and p, q, and rare all<br />

integers. If a is even, then p = (03 - 3a)/2 is still an integer. <strong>No</strong>w a 2 is<br />

divisible by 4, so we take r = (a2 - 4)/4 and q = a 2 - 1, and both q and r<br />

are integers. In either case we have shown that X3 can be written in the<br />

desired form.<br />

Also solved by Avraham Adler, Miguel Amengual Covas, Frank P.<br />

Battles, Paul S. Bruckman, James Campbell, William Cbau, Lisa M. Croft,<br />

Charles R. Diminnie, Russell Euler, George P. Evanovicb, Mark Evans,<br />

Victor G. Feser, Jayanthi Ganapatby, Robert C. Gebhardt, S. Gendler,<br />

Richard I. Hess, Joe Howard, Thomas C. Leong, Peter A. Lindstrom,<br />

David E. Manes, Yosbinobu <strong>Mu</strong>rayoshi, William H. Peirce, Bob Prielipp,<br />

John F. Putz, H.-J. Seiffert, Skidmore College Problems Group, Kenneth<br />

M. Wilke, Rex H. Wu, Monte J. Zerger, and the Proposer.<br />

879. [Spring 1996} Proposed by Barton L. Willis, University of<br />

Nebraska at Kearney, Kearney, Nebraska.<br />

A Mystery Space. LetS be a set of ordered pairs of elements. Define<br />

binary operations + , *, and + on S by<br />

and<br />

(a, b) + (c, d) = (a + c, b + d),<br />

(a, b)*(c, d) = (ac, ad + be),<br />

(a, b) + (c, d) = (a + c, b + c - ad+ ~).<br />

Although it might be fun to deduce properties of space S (commutativity,<br />

associativity, etc.), the problem is to find an application for S.<br />

I. Solution by PaulS. Bruclanan, Highwood, lllinois.<br />

Let the ordered pair (a, b) denote the fraction bla. A problem occurs if<br />

a = O, which may be circumvented by including a point at infinity [(0, b)<br />

= oo if b ¢. 0, and (0, 0) is undefined} or by requiring that a be a positive<br />

integer.<br />

One more relation needs to be defined:<br />

(ka, kb) = (a, b) for all k '#- 0.<br />

The binary operation "+" is then defined as the operation whereby the<br />

numerators and denominators of the two fractions, respectively, are added.<br />

The result is called the mediant of the two given fractions, one application<br />

of which is found in baseball batting averages. A batter with <strong>10</strong> hits in 40<br />

times at bat bas a batting average of <strong>10</strong>00 X <strong>10</strong>/40 = 250. If in the next<br />

game that batter gets 2 bits in 4 times at bat, his or her average for the<br />

game is <strong>10</strong>00 x 214 = 500 and the new accumulative batting ave~ge<br />

becomes the mediant <strong>10</strong>00(<strong>10</strong> + 2)/(40 + 4) = 273 of the two prevtous<br />

fractions.<br />

The binary operations "*" and "+" represent respectively the sum and<br />

difference of the two fractions. That is,<br />

and<br />

(a, b)"*" (c, d) = bla + die = (ad + bc)lac = (ac, ad+ be)<br />

(a, b)"+" (c, d) = bla -die = (be- ad)lac = (b/c - adl~)l(alc)<br />

= (ale, b/c- ad/~).<br />

Of course, there is no unit under "+" (unless we allow (0, 0)), the unit _<br />

under "*" is (1, 0), and since we have (a, b) "+" (1, 0) = (a, b), then (1, 0) •<br />

is the right hand unit under "+ ".

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