The Prodigies

HARVARD SENIOR DOES IMPOSSIBLE, headlined the Boston Globe last week.

The senior in question was Richard Friedberg of Manhattan, a modest 20-year-old who plays the piano, sings in the glee club, majors in mathematics. Last week not only the Globe but mathematicians across the U.S. were buzzing about what has turned out to be a major mathematical achievement.

Friedberg's feat was to solve a problem, first posed by the late Emil Post of New York's City College in 1944, that has been baffling mathematicians ever since. The problem: Does each non-recursive, recursively enumerable set have the property that every recursively enumerable set is recursive in it? Post himself thought not, but it was not until young Friedberg came along that anyone had the proof.

To be recursive, a set of integers must be subject to some sort of description that will tell whether any particular integer is in the set or not. The set 2, 4, 6, 8, ... is recursive because all its integers can be described as divisible by two. Sometimes, however, a set is recursive only in terms of another. Set A, for instance may have no regularity of its own, but if it contains just those integers whose squares are in Set B, it can be considered recursive in B. A recursively enumerable set is one which can be produced by a specific method, e.g., by squaring any integer to infinity, but which may or may not be recursive. Friedberg's problem was to find a recursively enumerable set which is not recursive, but in which some other recursively enumerable set is also not recursive.

The two sets Friedberg was after had to possess an infinite number of distinct properties. But in the construction of the sets, the assigning of one property often interfered with the assigning of another. Friedberg overcame the difficulty by arranging the properties in a priority list and resolving all interferences in favor of the properties standing higher on the list. After four months, he found the two sets he was looking for. Though laymen could hardly be expected to appreciate the virtues of his solution, his seniors in the field were dazzled. One professor at the University of California asked to include his work in a new book he is writing. M.I.T. has heard him lecture, and so has the Institute for Advanced Study in Princeton. But Dick Friedberg himself has proved something of a disappointment to U.S. mathematicians. "Unfortunately," says Czech-born Kurt Godel of the institute, "he wants to study medicine. An achievement like this at his age comes only once in a lifetime."

Friedberg was not the only student that Harvard was boasting about. There was also Sophomore French Anderson, 18, who as a student at the Tulsa, Okla. Central High School produced the first theory of how an ancient Roman did his multiplying, dividing and square-rooting with Roman numerals. Says Anderson, "It's really pretty easy."

The trick is to remember that while the value of a number in Arabic numerals depends on its position (e.g., 2 in 126 is worth 20), an X, V, L or C is still ten, five, 50 or 100 no matter where it appears. Thus, in finding the answer to "What is XXVIII multiplied by XII?", a Roman might have multiplied from left to right the top number by each numeral in the bottom number. Taking X (10) times XXVIII, he would get 20X5, ten Vs and 30 Is, which would become CCLXXX. After adding XXVIII and XXVIII to this, he would get CCLXXXXXXXV-VIIIIII, or CCCXXXVI--i.e., 336.

Last week, French Anderson announced that he had moved over to the Linear B numerals of ancient Crete--with such symbols as a perpendicular stroke for one, a dash for ten, a circle for 100, a circle with four spikes for 1,000.

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