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Topic: Parsing the logic of an elementary concept.
Replies: 2   Last Post: Feb 2, 2014 5:50 AM

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 Paul Posts: 764 Registered: 7/12/10
Parsing the logic of an elementary concept.
Posted: Feb 1, 2014 10:18 AM

The URL: http://www.math.ucsd.edu/~ronspubs/74_01_van_der_waerden.pdf contains an elementary proof of Van der Waerden's theorem that is particularly simple.
However, even this proof contains readability issues as far as I'm concerned.
For example: BEGIN QUOTE Let [a, b] denote the set of integers x with a <= x <= b. Let (x_1, ..., x_m), (y_1, ..., y_m) be members of [0, k] ^ m. Then (x_1, ..., x_m), (y_1, ..., y_m) are called k-equivalent if they agree up through their last occurrences of k. END QUOTE
Suppose k = 5 and m = 1. Suppose also that we are working with the logic of the definition, without reading ahead to get more context. Are the 1-element sequences x = 2 and y = 3 k-equivalent? This is completely unclear to me, and it seems that "yes" or "no" could be argued with equal force. The yes-campaigners might say: "Yes, they are 5-equivalent. No occurrence of 5 occurs in either sequence, so vacuously, all the terms of x and all the terms of y occur after the non-existent occurrences of 5. Since the only disagreements between x and y occur after these non-existent occurrences, x and y are 5-equivalent." However, a no-campaigner might argue: "No, they are not 5-equivalent. No occurrence of 5 occurs in either sequence, so vacuously, all the terms of x and all the terms of y occur before the non-existent occurrences of 5. Since x and y disagree before we see the final (non-existent) occurrences of 5, x and y are not 5-equivalent."
From context, it's clear that the authors are siding with the yes-campaigners but I don't see how that's clear from the logic. Is this a genuine ambiguity? If not, why are 2 and 3 5-equivalent, and why is the no-argument wrong?

Thank You,

Paul Epstein

Date Subject Author
2/1/14 Paul
2/1/14 William Elliot
2/2/14 Paul