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 hyper 0 dantheman163 Junior Fellow Posts: 13 Threads: 3 Joined: Oct 2009 10/25/2009, 11:29 PM I have a couple of ideas about the hyper 0 operator. I have see a couple of different definitions of it that don't agree with each other and nobody seems to have a definite solution. Some people say that a[0]b= b+1. Others have sort of a piece wise approach that is discontinuous and frankly doesn't make much sense. I have found a different approach (that may be completely wrong). The solution that i have found is that a[0]b= (a+b)/a + a The initial conditions that i worked with were as follows: 1. a[0]a=a+2 2. 2[0]4=5 (2[x]4 = 2[x+1]3) 3. This operation should be something that is fundamental (this is not really an initial condition but rather something i figured to be true) Now what is operation really means is something that may not be inherently obvious from the the definition a[0]b= (a+b)/a + a. I am going to explain using fingers. To start you would put a fingers in one hand and b fingers in the other. Next you would figure out how many groups of a fingers you had total. Then you would perform the sum (# of groups) + (# in each group ) which is the same as (# of groups) + a. For example 3[0]3 You have 6 fingers total. So you have 2 groups of 3 fingers. So the answer is 2+3 = 5 3[0]2 You have 5 fingers total.That is 1 and 2/3s groups of 3. So the answer is 5/3+3 = 14/3 An error that some people may see is that a[0](a[0]a) DOES NOT= a+3 however, I feel that a[0]a[0]a = a+3. a[0]b[0]c using "the finger method" would equal (a+b+c)/a + a. I hope that you guys will not have to struggle too hard to understand what I am saying and I also hope that all of this is not completely wrong Thanks. andydude Long Time Fellow Posts: 509 Threads: 44 Joined: Aug 2007 10/26/2009, 01:43 AM I see another error. I have 10 fingers, not 6. « Next Oldest | Next Newest »

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