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#21 |
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FFR Simfile Author
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There is a contradiction from lines 3 to 4.
Root (1/-1) does not equal root(1)/root(-1), hence the rest is all incorrect. i^2= -1 -1=-1
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Last edited by Reach; 10-29-2006 at 08:58 PM.. |
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#22 |
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FFR Player
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I thought that too, but it is true.
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#23 |
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FFR Simfile Author
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Uh no it isn't. You can't take the root of the denominator there unless it's positive because the result does not exist.
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#24 | |
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FFR Player
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EDIT, crap. reach is right.
The calculation rule (a^.5) * (b^.5) = (a*b)^.5 is only valid for real, non-negative values of a and b
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Quote:
Last edited by spyke252; 10-29-2006 at 09:32 PM.. Reason: Stupid Reach. |
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#25 |
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Away from Computer
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Proof: Everyone in the world is the same age
Let n = number of people in the world (a group) Let s(n) = the maximum number of different ages existing in any group of n s(1) = 1, (with one person in the world, everyone in the world is the same age) assume s(k) = 1: by showing that s(k+1) = 1, we show thateveryone in the world has the same age by induction Take two people in the group of k+1. Call these people Al and Bob. Take the group of everyone except Al. This is a group of k people. By the assumption, everyone in this group has the same age. Take the group of everyone except Bob. This is a group of k people. By the assumption, everyone in this group has the same age. Take a third person, Carol. Carol is the same age as Bob and is also the same age as Al by the two statements above, therefore Bob and Al are the same age. Since Al and Bob are generic people in the group, by showing that they are the same age, I have shown that any two people in the group are the same age, therefore everyone in the group of k+1 is the same age, or s(k+1) = 1. By induction, everyone in the world is the same age.
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#26 |
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Private College
Join Date: Feb 2006
Location: Lol badger
Posts: 536
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Inductive hypothesis fails for k=2, since there's no overlap.
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<img src="Bent Lines" /> |
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#27 |
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Banned
Join Date: Aug 2006
Posts: 212
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true but like i said I DIDNT DO ****!
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