Do Any Other Countries Hospitals Get...
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Another Maths Problem.
23 Answers
Since people seem to be in the mood for mathematical oddities today, take a look at this:
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How many regions would you get from 6 dots all joined together?
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How many regions would you get from 6 dots all joined together?
Answers
I think I've completely solved the problem and have some understandin
20:16 Wed 29th Oct 2014
It's one of those where induction will lead you astray, at least. It's quite cute to see the function F(x) - 2^(x-1) plotted -- within the region [1,5] they appear to agree quite well, but the difference is only zero at precisely the points (1 and, I think?) 2, 3, 4 and 5 and otherwise it oscillates between a difference of about +/- 0.05 or so.
Another interesting problem in dodgy induction is the following "proof":
Theorem: Everyone shares the same birthday.
Proof (by induction): Clearly one person shares his own birthday. Let us suppose that P(k) is true, ie that in any group of k people all have the same birthday. Now consider adding a new person to the group so that we have ({k}, {1}) people. However, we can remove one person from the first group and replace him by the new person, creating a partition of ({k}-1+{1},1) = ({k},{1}) again. Since any group of k people shares the same birthday by assumption, the new person shares his birthday with the other k-1 people, but the person we removed from the first group also shares that birthday with the other k-1 people. So, in fact, all (k+1) people share a birthday and the inductive step is complete.
Hence P(1) and P(k) implies P(k+1), and everyone has the same birthday by induction. QED... right?
Another interesting problem in dodgy induction is the following "proof":
Theorem: Everyone shares the same birthday.
Proof (by induction): Clearly one person shares his own birthday. Let us suppose that P(k) is true, ie that in any group of k people all have the same birthday. Now consider adding a new person to the group so that we have ({k}, {1}) people. However, we can remove one person from the first group and replace him by the new person, creating a partition of ({k}-1+{1},1) = ({k},{1}) again. Since any group of k people shares the same birthday by assumption, the new person shares his birthday with the other k-1 people, but the person we removed from the first group also shares that birthday with the other k-1 people. So, in fact, all (k+1) people share a birthday and the inductive step is complete.
Hence P(1) and P(k) implies P(k+1), and everyone has the same birthday by induction. QED... right?
