I am sorry for the two quick posts, but this is a bit too long to just stick in the previous one as an edit.
A few days ago we proved that any chocolate bar of n squares can be broken up into 1x1 squares by n-1 breaks. We didn't consider the case n=0. I claimed that P(n) was vacuously true for n=0 because there are no chocolate bars with 0 squares. Prof. Heap countered this claim, but I am not fully convinced and this matter has been bothering me since then.
How can there be a chocolate bar with 0 squares? If we reduce the size of a chocolate bar to 0 rows and 0 column, we're left with nothing. So there's nothing there, so there is no chocolate bar at all.
Right?
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Once you're done eating, there's a chocolate bar with zero squares (zero triangles, for that matter).
We have empty strings, empty arrays, so empty chocolate bars are a natural extension. They're very easy to carry around, but not very filling.
I see. So we are looking at a chocolate bar as if it was a set of squares. Then the 0x0 chocolate bar is the empty set, and so the set of chocolate bars of 0 squares has cardinality 1.
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