One of the joys of hosting the PoW is the chance to show some really beautiful new ideas to you and to my students. #864 is such an example. Enjoy!

Problem of the Week 864

The Ten Implications

Suppose P is a property of natural numbers such that

1. P[0] is true
2. Whenever P[n] is true, then P[n+1] is true.

   You may use the general law of implication [X and X => Y yield Y] ten times only, but -- the catch -- you are NOT allowed to use induction. What is the largest n for which you can prove P[n]?

   Two examples should clarify the rules: Here is a proof of P[10].

   P[0] -> P[1] -> P[2] -> P[3] -> P[4] -> P[5] -> P[6] -> P[7] -> P[8] -> P[9] -> P[10]

   And here is how to get to P[25]: (B) is used five times to prove an auxiliary assertion, (C), which is then used five times.

3. For all n, P[n] ==> P[n+5]:

   Proof:

   P[n] -> P[n+1] -> P[n+2] -> P[n+3] -> P[n+4] -> P[n+5]

   Now use (C) five times:

   P[0] -> P[5] -> P[10] -> P[15] -> P[20] -> P[25]

How high a value can you get? Note that the auxiliary assertions are allowed to be complicated!

If you tell me your top value I will tell you if a higher value is possible.

Source: Michael Sheard, St. Lawrence Univ.
Thanks to Sheard and Dan Velleman and John Guilford for their comments on drafts.

© Copyright 1998 Stan Wagon. Reproduced with permission.