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It's only semi autobiographical
Thursday, December 08, 2005
Crisps
Here I present to the world a proof that crisps are by no means the ultimate snack, and I use MATHEMATICS, wich as we all know is the purest form of, and thus outranks, SCIENCE!
We prove by contradiction. Let us assume that crisps are the ultimate snack. Ultimate can be defined to mean "the finest or most superior quality of its kind"
Let S be a finite, unbounded set of all snacks, and let C be a subset of Snacks. Let C be closed and finite (but not necessarily bounded.)
Now let all flavours of crisps be members of C. We have that for all flavours f in C, f>=s
Where s is a member of S. This is because for crisps to be ultimate ALL crisps must be superior to all other snacks
So, for all x, y members of C, x>=s and y>=s.
But, x and y are in S, so x>=y and y>=s Thus we have x=y, for all x, y in C contained within S
Now, without loss of generality, let us take x=Cheese and onion and y=Salt and vinegar. Clearly these two flavors are not equal, so our initial assumption must have been false.
As required, crisps are NOT the ultimate snack.
Now I'm bored, and I'm going to do stats.
Comments:
I'm not sure you can have an unbounded finite set....
The problem with your proof (poor fool) is thus. We claim that crisps are the greatest snacks. Therefore f>=s for all s member of S WITHOUT C, as crisps are better than all other snacks...
So ha. Gonna have to get up earlier than that to better SCIENCE.
The problem with your proof (poor fool) is thus. We claim that crisps are the greatest snacks. Therefore f>=s for all s member of S WITHOUT C, as crisps are better than all other snacks...
So ha. Gonna have to get up earlier than that to better SCIENCE.
I had, in fact considered this, but decided the definition of Ultimate seemed to suggest my method.
However, if f>=s for all s in S\C, then we simply have to find a single s >c for ANY c and ANY s.
Trivially we find that a chocolate hobnob mini-pack is superior to cheese and onion crisps and we are done.
Not as elegant, but it will suffice.
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However, if f>=s for all s in S\C, then we simply have to find a single s >c for ANY c and ANY s.
Trivially we find that a chocolate hobnob mini-pack is superior to cheese and onion crisps and we are done.
Not as elegant, but it will suffice.
