I checked my dictionary, but apparently there is no english word for the dutch word ‘tangconstructie’. Anyway, remember how I, rhetorically, asked ‘how else could it be?’ yesterday? Well, I am more than delighted to say that I am able to give a counterexample to the metaconjecture that every proof of a counterexample to the question of Erdös and Graham I talked about in the previous post, works via a case-by-case-argument.

Consider the set and let be the least common multiple of all members of , for a set of positive integers. Then . Now, assume we have a partition of in two sets and with and . Then in particular we have:

And boom goes the dynamite.

### Like this:

Like Loading...

*Related*

This entry was posted on November 3, 2010 at 19:05 and is filed under Problem Solving. You can follow any responses to this entry through the RSS 2.0 feed.
You can leave a response, or trackback from your own site.

## Leave a Reply