spot the error

[perceval]

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

[HermanHiddema]

The recursive step involves boxes of N+1 pens for N>= 1, so N+1 is at least 2 and the recursion does not reach H(1)

[emeraldemon]

I think this is the error:

The box still contains N pens, so following H(N), again all pen are of the same color, including the first one we removed.

Consider the N=1 case: all pens are the same color (say red). We have one red pen in the box. We take our N+1 pen, say it is blue. We swap the two pens: now we have a blue pen in the box: H(N) is still true. But it isn't true that the pen in our hand is the same color as the pen in the box, and adding our red pen back in doesn't make H(N+1) true.

[perceval]

The recursive step involves boxes of N+1 pens for N>= 1, so N+1 is at least 2 and the recursion does not reach H(1)

I do not see teh issue: I assume H(N) is true, and i want to prove it for N+1 so indeed we can reach N = 1 ? or i misunderstood something ?

[perceval]

I think this is the error:

The box still contains N pens, so following H(N), again all pen are of the same color, including the first one we removed.

Consider the N=1 case: all pens are the same color (say red). We have one red pen in the box. We take our N+1 pen, say it is blue. We swap the two pens: now we have a blue pen in the box: H(N) is still true. But it isn't true that the pen in our hand is the same color as the pen in the box, and adding our red pen back in doesn't make H(N+1) true.

that's it : the operation described doesnt work when you start with 2 pens because you need at least a third pen to say the the first and the second you removed have the same color as some other pen ,and thud are the same color

[HermanHiddema]

The recursive step involves boxes of N+1 pens for N>= 1, so N+1 is at least 2 and the recursion does not reach H(1)

I do not see teh issue: I assume H(N) is true, and i want to prove it for N+1 so indeed we can reach N = 1 ? or i misunderstood something ?

From your proof (ii):
"Lets consider a box of N+1 pen with N>=1 a natural integer."

So your step (ii) poses as a prerequisite that a box contains at least 2 pens (N>=1 and N+1 pens in the box).

Therefore it is entirely unrelated to step (i), which has a box with 1 pen.

Your proof instead reduces to the case of a box with two pens. And indeed, if it was a given that any box containing two pens would necessarily have both pens the same color, then your proof would be valid.

[drmwc]

Consider the inductive bit of the "proof" applied to a box of two pens, a black and a blue pen.

Remove the black pen from the box. We have a box of 1 pens, and the pen is the same colour as itself (blue). Now put the pen back, and remove the other pen. We again have a box of 1 pen of the same colour (black this time). However, it is a different colour than in the first iteration and so we have not proved that black equals blue.

At no point does the argument show that the two pens need to be the same colour - we merely show that a single pen is the same colour as itself. H(1) does not consider the two pens together at any stage.

If we can show that any box of two pens are necessarily the same colour, then the induction may work. But we can't assume that, because it isn't true.

[schawipp]

...
(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.
...

In your H(N+1) example, I feel free to assume that the pen, which you initially removed from the box to get back to H(N), is from a different color. So, your induction step seems somehow flawed...