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I'm just dabbling here.
#1
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?
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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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