Content-Type: text/shitpost


Subject: Comparing heaps of beans
Path: your-brain!your-host!ihnp4!kremvax!plovergw!shitpost!mjd
Date: 2017-12-22T16:15:41
Newsgroup: talk.mjd.math-se-shitposting
Message-ID: <4b8dab90744a625e@shitpost.plover.com>
Content-Type: text/shitpost

An earlier shitty post on this shitty blog offered a simple method for proving that !!2^{50} < 3^{33}!! without calculating the values:

Take a heap of red beans of size !!2^{50}!! and a heap of navy beans of size !!3^{33}!!. Repeatedly remove one bean from each pile until the red pile is exhausted. At that point some navy beans will remain and the claim is proved.

Katara very correctly observed that this is incomplete, because it may not be clear how to carry out the instruction “take a heap of red beans of size…”.

But there is a solution! Start with a heap of 32 red beans. Now double the size of the heap by adding one bean for each bean already in the heap; this can be done without any calculation or even counting:

  while any beans remain in the heap:
      transfer one bean to a secondary heap
      add a new bean to the secondary heap

You now have a heap of !!2^6!! beans. Repeat this doubling process 44 more times. One can similarly construct a heap of !!3^{33}!! navy beans without calculation.

This assumes that we can count as far as 44, which I think is not too unreasonable; I myself can usually get as far as 50 or even 65 on a good day. But if not, workarounds are available.

I will continue to resist the temptation to suggest that there is a deep and subtle point lurking here.