Search All of the Math Forum:
Views expressed in these public forums are not endorsed by
NCTM or The Math Forum.


Paul
Posts:
729
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 kequivalent 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 1element sequences x = 2 and y = 3 kequivalent? This is completely unclear to me, and it seems that "yes" or "no" could be argued with equal force. The yescampaigners might say: "Yes, they are 5equivalent. No occurrence of 5 occurs in either sequence, so vacuously, all the terms of x and all the terms of y occur after the nonexistent occurrences of 5. Since the only disagreements between x and y occur after these nonexistent occurrences, x and y are 5equivalent." However, a nocampaigner might argue: "No, they are not 5equivalent. No occurrence of 5 occurs in either sequence, so vacuously, all the terms of x and all the terms of y occur before the nonexistent occurrences of 5. Since x and y disagree before we see the final (nonexistent) occurrences of 5, x and y are not 5equivalent." From context, it's clear that the authors are siding with the yescampaigners but I don't see how that's clear from the logic. Is this a genuine ambiguity? If not, why are 2 and 3 5equivalent, and why is the noargument wrong?
Thank You,
Paul Epstein



