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 I'm just dabbling here. ChaoticMC Newbie Posts: 1 Threads: 1 Joined: Jul 2018 07/07/2018, 02:44 AM (This post was last modified: 07/07/2018, 02:46 AM by ChaoticMC. Edit Reason: Changed "Can" to "Could" at the very last line, to sound less ''pushy". ) I'm only a new user, so I don't know what to do here. I'm only an amateur too. Anyways, here's how we define hyperoperations. H(b,x,0) = S(b) H(b,0,1) = b H(b,1,r≥2) = b H(b,x+1,r+1) = H(H(b,x,r+1),b,r) Alright, so we want the negative hyperoperations. So let's denote S(b) as S(b,x). This time with 2 inputs. So, it has to obey the rule that H(b,b,r) = H(b,2,r+1)...right? Well, if S(b,b) did equal b+2, then would there something wrong? S(b,b) = b+2 = S(b,b+1), nothing wrong with that. However, 0+1 = S(0+0,0) = S(0,0) = 2. This is a contradiction. This is the only instance, however, where this is wrong. Why? Because 0+0=0. It's the identity of addition. In fact, even if the identity was 2, then 2+3 = S(2+2,2) = S(2,2) = 2. If the identity was 3, then 3+4 = S(3+3,3) = S(3,3) = 2. -Wait! If the identity was 1... Then 1+2 = 2...which would actually be true. So, in order for S(b,x) to be functional, then the identity of addition has to be 1. But wait! [Note: Let's denote the inverse of S as P] 0+(3-1) = P((0+3),(0+1)) = P(3,0) = 2 0+(3-1) = 2 0+(3-1) = 0+3 = 3 2=3 And this is when the identity of addition is 1. So, no matter which identity we choose, there will always be a contradiction. So, that's it. It's just 0. Meaning that S(b,b) ≠ b+2. Meaning that S(b,x) = S(b). Alright, well this does change a bit about the negative hyperoperations, but not completely. Anyways. Let's denote H(b,x,-1) as ®. So, S(b,x) just equals S(b) for all x. So, S(b,S(x)) = S(b,x)®b = S(b)®b S(b)®b = S(b) And...this is just it. This is the solution to the hyperoperation rank -1. We can't go any farther. Could somebody check my work? « Next Oldest | Next Newest »

 Messages In This Thread I'm just dabbling here. - by ChaoticMC - 07/07/2018, 02:44 AM RE: I'm just dabbling here. - by sheldonison - 07/09/2018, 02:07 PM RE: I'm just dabbling here. - by Xorter - 07/09/2018, 08:34 PM RE: I'm just dabbling here. - by sheldonison - 07/10/2018, 10:34 PM

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