In Puzzle 42, you were asked to find the value of an endless fraction. There are an infinitude of such fractions, even if we limit ourselves to those that have rational values. One source of such fractions is the formula for creating a quadratic equation given the sum (S) and product (P) of the roots: x² – Sx + P = 0. From this we get x² – Sx = -P, x(x – S) = -P, x = -P/(x – S), x = P/(S – x). If you want to get any two particular numbers as the value of an endless fraction, insert the sum and product of those two numbers into the template: . For example: If we want the value of the fraction to be 2 and 3, then S = 5 and P = 6 and the puzzle would be: What is the value of the fraction ? And the procedure to determine the value would be, as indicated in the solution of Puzzle 42, to let x equal the fraction and solve for x. If x = , then x = 6/(5 – x), 6 = 5x – x², x² – 5x + 6 = 0, and by factoring, we get x = 2, x = 3, so the value of the fraction is 2 and 3. (We’ll deal with that “and” momentarily.)
It seems to me that this situation is very peculiar.
There is nothing peculiar about two solutions to a quadratic equation. All we are saying there is that it is possible to substitute two different values for x in the quadratic expression that will make the equation a true statement. And that is not astonishing, because there is nothing inherently odd about being able to apply two different sets of operations to the same numbers to get the same result (1•2•3 = 1 + 2 + 3, 4•3 ÷2 – 1 = 4(3 – 2) + 1, 9(9 ÷ )² – 9 = (9 – 9 ÷ 9)9 ) or to apply the same set of operations to two or more different numbers to get the same result (6 ÷ 2 + 1 • 3 – 5 = 9 ÷ 3 + 2 • 1 – 4, most polynomial equations of degree ≥ 2).
What we are dealing with here, however, is a sequence of operations with a set of numbers and without changing any numbers or any operations, getting two different results. It’s like saying a + b = 2 and a + b = 3, which clearly implies 2 = 3, so we reject the conjunction of those two statements. To the objection that the quadratic equation solution is not a conjunction, but a disjunction, x = 2 or x = 3, we reply that the disjunctive quality is inherent in the situation; (1) (x – 2)(x -3) = 0 implies x – 2 = 0 or x – 3 = 0 and (2) we can’t substitute two different numbers for x in x² – 5x + 6 = 0 at the same time.
But the endless fraction contains no variables, only univocal operations with univocal numbers. It is the method of calculation itself that introduces a variable. And sometimes, as in the case of absolute value equations ( /x – 6/ = 2x ) or rational equations ( = ), the solution method itself introduces extraneous roots that do not check, so we reject them. In this case, however, there can be no extraneous roots in the sense of values to be substituted for a variable that result in a contradiction. The contradiction in this case is in the given itself; “x” is ambiguous, but “6”, “5”, “-” and the fraction bar are not.
The problem is that we can’t do the calculation without injecting the result of the calculation into the calculation, and our insertion of “x” into the calculation highlights this fact. This is a self-referential calculation. Part of the calculation that must be done in order to get the result is itself the result of the calculation. And like other self-referential concepts, it has a strong potentiality to yield a contradiction.
It is similar to the Liar’s Paradox, which concerns the attempt to determine the truth value of a statement such as: The sentence you are now reading is false. We can’t evaluate the truth value of this sentence independently of the result of our evaluation and when we assume either truth value, the evaluation yields the opposite truth value.
In the endless fraction, if we assume either numerical value obtained by the method we used, that value is confirmed, the same way both solutions to a correctly solved quadratic equation are confirmed when you check them. The difference in the endless fraction case is that the result is a conjunction and there is no reason to reject the result of either calculation rather than the other except to say that when an infinitely long calculation is made, there is the potential to yield a contradiction, and in the event that it does, we must reject both parts of the contradiction and assert that there is a hidden ambiguity or some other flaw that renders the calculation meaningless.
Hmmm… a very dissatisfying conclusion. TO BE CONTINUED