I assume most fo you are familiar with the demonstration by recursion.
Here is a nice example:
(i) In all boxes that contains 1 pen, all the pen(s) are of the same color.
(ii) Let us assume H(N):In all boxes containing N pen, all the pen(s) are of the same color.
Lets consider a box of N+1 pen with N>=1 a natural integer. If I remove a pen from the box, i am left with a box with N pen, and according to H(N), they are all of the same color.
Now lets replace the pen i just took in the box and take another one. The box still contains N pens, so following H(N), again all pen are of the same color, including the first one we removed. So we can put back hte pen we just removed, and all the pen in hte box have the same color ; so H(N+1) is true.
As H(1) is true ANd H(N+1) is true if H(N) is, H(N) is true for all (finite)N. [note than trying to remove a pen from a box containg an infiinte number of pens would still left us with an infinite number of them, so the argument does not hold in that case
]Hence, any box with a finite number of pen can only contains pens of the same color.
Unless you can spot an error in the demo ?
the best audience for this enigma is someone who just learn about recursion , it usually throw them for a loop
or have them digging for a fault in the wrong place i once explained this demonstration to someone not particuraly savvy in math, and he took it as a demonstration that math has nothing to do with the real world
