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## Unique Ordinal Representation

Edited: Late Feb 2016 to align to changes to my blog on Fundamental Sequences.

This blog will demonstrate how to create a unique representation of ordinals with arbitrary complexity using the definitions from my blog on Extended Normal Form. In turn this blog will be referred to in my other blogs for the J Function and in particular by Sandpit $$J_8$$.

## Extended Normal Form

An arbitrary ordinal can be defined as follows using the definitions of Extended Normal Form.

$$\gamma = (\lambda\uparrow\uparrow t)^{\gamma_e}.\gamma_c + \gamma_a = (\varphi(\gamma_{[q]})\uparrow\uparrow t)^{\gamma_e}.\gamma_c + \gamma_a$$

Where $$\gamma$$ is a arbitrary transfinite ordinal, $$\lambda$$ is an arbitrary limit ordinal, and $$t, q$$ are finite integers.

## Unique Representation

A unique representation for any such ordinal $$\gamma$$ can be demonstrated as a sequence of finite integers defined as:

$$<g> = <q, <L>, t, <g_e>, <g_c>, <g_a>>$$

The sequence is constructed in the following order:

Let $$q$$ be any finite integer

Recursively define $$<L> = <<g>_{[q]},p,m>$$

$$= <<g_1>,<g_2>,<g_3>,...,<g_q>,p,m>$$ where $$0 < p <= q$$ and $$0 < m$$

Let $$t$$ be any finite integer

Recursively define $$<g_e>, <g_c>$$, and $$<g_a>$$

## Unique Representation and Monotonically Ascending

The resulting sequence of finite integers $$<g>$$ will be unique for any ordinal. The sequence will also impose monotonically ascending representations, such that if $$<x> > <y>$$ then $$\gamma_x > \gamma_y$$.

For one sequence $$<x>$$ to be greater than another sequence $$<y>$$, then the first non-identical integer (from left to right) between the sequences must be greater in sequence $$<x>$$, as shown here:

Let

$$<x> = <x_{[i-1]},x_i,x_{[j]}>$$

$$<y> = <y_{[i-1]},y_i,y_{[k]}>$$

Then

$$<x> > <y>$$

Only If

$$x_{[i-1]} = y_{[i-1]}$$ and $$x_i > y_i$$ for some value of $$i$$

## Unique Big Number Representation

A further extension to the above can be made to create a sequence of finite integers $$<f>$$ as follows:

Assign $$<f> = <<g>,n,p>$$ where $$n, p$$ are finite integers and $$p > max(max(<g>)+1,n)$$

This sequence will represent any Big Number (Googolism) of the form:

$$f_g^n(p)$$ where $$g < SVO$$ the Small Veblen Ordinal

## Ruleset

Ruleset for creating $$<g>$$ sequences for ordinals is:

$$<g> = <q, <L>, t, <g_e>, <g_c>, <g_a>>$$

• when $$q = 0$$ then $$<L> = t = <g_e> = <g_c> =$$ Null and $$<g_a>$$ collapses to $$a$$ i.e. a finite integer
• and when $$q > 0$$ then
• $$<L> == \lambda = \varphi(1) = \omega$$ when $$q = 1$$
• $$<L> = <<g>_{[q]},p,m> == \lambda = \varphi^m(\gamma_{[p-1]},(\gamma_p)_*,\gamma_{[q-p]})$$ when $$q > 1$$ and
• $$0 < <g_1> == \gamma_1 < \varphi(1,0_{[q]})$$ or
• $$0 < <g_1> < < q >$$
• The sequence $$<L>$$ follows the construction procedure for $$\lambda$$, starting with:
• $$<L_0> = < u >_{[q]} == \lambda_0 = \varphi(\mu_{[q]})$$ where $$\mu < \varphi(1,0)$$
• $$<L_1> = <<L_0>,1,m_0> == \lambda_1 = \varphi^{m_0}(\mu_{[1]}*,\mu_{[q-1]})$$
• and
• $$<L_2> = <<L_1> + a_1,<g>_{[1]},<0,0>_{[q-2]},2,m_1>$$
• $$<L_3> = <<L_2> + a_2,<g>_{[2]},<0,0>_{[q-3]},3,m_2>$$
• $$<L_q> = <<L_{q-1}> + a_{q-1},<g>_{[q-1]},q,m_{q-1}>$$
• then
• $$<L_q+1> = <<L_q> + a_q,<g>_{[q]},q,m_q>$$
• until finally $$< L > = < L_{q+j} >$$
• $$t$$ is a finite integer
• $$0 < <g_e> < <q,<L>,2>$$
• $$0 < <g_c> < <q,<L>,t>$$
• $$<g_a> < <q,<L>,t,<g_e>>$$

## Unique Representation for $$\omega^{\omega} + \omega.3 + 2$$

Here is a unique representation for this arbitrary ordinal:

$$\gamma = \omega^{\omega} + \omega.3 + 2 = (\varphi(\gamma_{[q]})\uparrow\uparrow t)^{\gamma_e}.\gamma_c + \gamma_a$$

• Let $$q = 1$$
• This will collapse $$\lambda = \varphi(\gamma_{[1]}) = \varphi(\gamma_1) = \varphi(1) = \omega$$
• Let $$t = 2$$
• Recursively define $$\gamma_e == <g_e>$$
• $$\gamma_e = (\varphi(\gamma_{[q]})\uparrow\uparrow t)^{\gamma_{e_2}}.\gamma_c + \gamma_a$$
• Let $$q = 0$$
• This will collapse $$\gamma_e = \gamma_a = e$$ (i.e. a finite number) by definition
• Let $$e = 1$$
• Assign $$<g_e> = <0,1>$$
• Recursively define $$\gamma_c == <g_c>$$
• $$\gamma_c = (\varphi(\gamma_{[q]})\uparrow\uparrow t)^{\gamma_e}.\gamma_{c_2} + \gamma_a$$
• Let $$q = 0$$
• This will collapse $$\gamma_c = \gamma_a = c$$ (i.e. a finite number) by definition
• Let $$c = 1$$
• Assign $$<g_c> = <0,1>$$
• Recursively define $$\gamma_a == <g_a>$$
• $$\gamma_a = (\varphi(\gamma_{[q]})\uparrow\uparrow t)^{\gamma_e}.\gamma_c + \gamma_{a_2}$$
• Let $$q = 1$$
• This will collapse $$\lambda = \varphi(\gamma_{[1]}) = \varphi(\gamma_1) = \varphi(1) = \omega$$
• Let $$t = 1$$
• Recursively define $$\gamma_e == <g_e>$$
• $$\gamma_e = (\varphi(\gamma_{[q]})\uparrow\uparrow t)^{\gamma_{e_2}}.\gamma_c + \gamma_a$$
• Let $$q = 0$$
• This will collapse $$\gamma_e = \gamma_a = e$$ (i.e. a finite number) by definition
• Let $$e = 1$$
• Assign $$<g_e> = <0,1>$$
• Recursively define $$\gamma_c == <g_c>$$
• $$\gamma_c = (\varphi(\gamma_{[q]})\uparrow\uparrow t)^{\gamma_e}.\gamma_{c_2} + \gamma_a$$
• Let $$q = 0$$
• This will collapse $$\gamma_c = \gamma_a = c$$ (i.e. a finite number) by definition
• Let $$c = 3$$
• Assign $$<g_c> = <0,3>$$
• Recursively define $$\gamma_{a_2} == <g_{a_2}>$$
• $$\gamma_{a_2} = (\varphi(\gamma_{[q]})\uparrow\uparrow t)^{\gamma_e}.\gamma_c + \gamma_{a_3}$$
• Let $$q = 0$$
• This will collapse $$\gamma_{a_2} = \gamma_{a_3} = a$$ (i.e. a finite number) by definition
• Let $$a = 2$$
• Assign $$<g_{a_2}> = <0,2>$$
• Assign $$<g_a> = <q,<L>,t,<g_e>,<g_{c_2}>,<g_a>> = <1,null,1,<0,1>,<0,3>,<0,2>>$$

Assign $$<g> = <q,<L>,t,<g_e>,<g_c>,<g_a>>$$

$$= <1,null,2,<0,1>,<0,1>,<1,null,1,<0,1>,<0,3>,<0,2>>>$$

$$= <1,2,<0,1>,<0,1>,<1,1,<0,1>,<0,3>,<0,2>>>$$

$$= <1,2,0,1,0,1,1,1,0,1,0,3,0,2>$$

Then

$$\gamma = \omega^{\omega} + \omega.3 + 2 == <1,2,0,1,0,1,1,1,0,1,0,3,0,2>$$

## Unique Representation for $$f_{\omega^{\omega} + \omega.3 + 2}^2(4)$$

$$f_{\omega^{\omega} + \omega.3 + 2}^2(4) == <<1,2,0,1,0,1,1,1,0,1,0,3,0,2>,n,p>$$

$$= <<1,2,0,1,0,1,1,1,0,1,0,3,0,2>,2,4>$$

$$= <1,2,0,1,0,1,1,1,0,1,0,3,0,2,2,4>$$