User blog:Rgetar/Ordinal arithmetic

Hi!

This summer I derived rules of ordinal arithmetic. I used Wikipedia article Ordinal arithmetic. Some of this I rediscovered independently, including Cantor normal form, addition and multiplication rules, which are contained in the same article in Cantor normal form section. Rules for my [X]a function need verification.

Addition basic rules
Let α, β, γ, δ - ordinals. β (as any ordinal) is 0, or a successor ordinal, or a limit ordinal.

Let α+1 is the successor of α.

Definition of addition (α+β):

1. α+0 = α  (rule for β = 0) 

2. α+(β+1) = (α+β)+1  (rule for β is a successor ordinal) 

3. α+β = limit of the α+δ for all δ<β (if β is a limit ordinal)  (rule for β is a limit ordinal) 

(that is in the 3rd rule of the definition α+β is least ordinal larger than α+δ for all δ<β)

Some properties:

α+0 = 0+α = α

(α+β)+γ = α+(β+γ)

Addition is non-commutative: α+β ≠ β+α in general case

Multiplication basic rules
Let α, β, γ, δ - ordinals.

Definition of multiplication (α·β):

1. α·0 = 0  (rule for β = 0) 

2. α·(β+1) = (α·β)+α  (rule for β is a successor ordinal) 

3. α·β = limit of the α·δ for all δ<β (if β is a limit ordinal)  (rule for β is a limit ordinal) 

(that is in the 3rd rule of the definition α·β is least ordinal larger than α·δ for all δ<β)

Some properties:

α·0 = 0·α = 0

α·1 = 1·α = α

(α·β)·γ = α·(β·γ)

Multiplication is non-commutative: α·β ≠ β·α in general case

α·(β+γ) = α·β+α·γ  (note: (α+β)·γ) ≠ α·γ+β·γ in general case) 

Middle dot often omitted: α·β = αβ

Exponentiation basic rules
Let α, β, γ, δ - ordinals, n - natural number

ω - least countable ordinal

Definition of exponentiation (α·β):

1. α0 = 1  (rule for β = 0) 

2. αβ+1 = (αβ)·α  (rule for β is a successor ordinal) 

3. αβ = limit of the αδ for all δ<β (if β is a limit ordinal)  (rule for β is a limit ordinal) 

(that is in the 3rd rule of the definition α·β is least ordinal larger than α·δ for all δ<β)

Some properties:

0α = 0, α>0

1α = 1

α1 = α

αβ·αγ = α(β+γ)

(αβ)γ = αβ·γ

nω = ω (in particular 2ω = ω)

lp, fp, of, zp, rt, zc, vs
I made up number of functions:

lp means "limit part"

fp means "finite part"

of means "omega factor"

zp means "zero part"

rt means "rest"

zc means "zero coefficient"

vs means "Veblen sector"

Cantor normal form
Cantor normal form: an ordinal α can be uniquely represented as linear combination of degrees of ω with decreasing exponents:

α = ωα0·n0 + ωα1·n1 + ωα2·n2 + ...

α0, α1, α2, ... - ordinals

α0 > α1 > α2 > ...

n0, n1, n2, ... - natural numbers

finite part (fp)
If last αi = 0 then fp(α) = ni else fp(α) = 0

limit part (lp)
Rest of Cantor normal form of α is lp(α)

So,

α = lp(α) + fp(α)

If α - limit ordinal then α = lp(α) ≠ 0, fp(α) = 0

If α - successor ordinal then fp(α) ≠ 0

If α - finite ordinal then lp(α) = 0

omega factor (of)
lp(α) = ω·of(α)

α = ω·of(α) + fp(α)

zero part (zp)
zp(α) = ωα0·n0 (first term of Cantor normal form)

rest (rt)
rt(α) is rest of α:

rt(α) = ωα1·n1 + ωα2·n2 + ...

So,

α = zp(α) + rt(α)

zero coefficient (zc)
zc(α) = n0

Veblen sector (vs)
vs(X(α)) is function. Its argument is function X(α) of ordinal α to sequence of ordinals (see Ordinals array function blog, section Sequence X and its separators). Its value is function of ordinal α to ordinal.

Definition of vs:

φ(X(vs(X(α)))) ≤ α < φ(X(vs(X(α))+1))

if α < φ(X(0)) then vs(X(α)) = -1

φ - Veblen function

Examples:

X(α) = α

φ(vs(α)) ≤ α < φ(vs(α)+1)

φ(α) = ωα

ωvs(α) ≤ α < ωvs(α)+1

vs(α) = -1 for α = 0

vs(α) = 0 for 1 ≤ α < ω

vs(α) = 1 for ω ≤ α < ω2

vs(α) = 2 for ω2 ≤ α < ω3

zp(α) = ωvs(α)·zc(α)

So,

α = ωvs(α)·zc(α) + rt(α)

X(α) = 1,α

φ(1,vs(1,α)) ≤ α < φ(1,vs(1,α)+1)

φ(1,α) = εα

εvs(1,α) ≤ α < εvs(1,α)+1

vs(1,α) = -1 for α < ε0

vs(1,α) = 0 for ε0 ≤ α < ε1

vs(1,α) = 1 for ε1 ≤ α < ε2

vs(1,α) = 2 for ε2 ≤ α < ε3

X(α) = 2,α

φ(2,vs(2,α)) ≤ α < φ(2,vs(2,α)+1)

φ(2,α) = ζα

ζvs(2,α) ≤ α < ζvs(2,α)+1

vs(2,α) = -1 for α < ζ0

vs(2,α) = 0 for ζ0 ≤ α < ζ1

vs(2,α) = 1 for ζ1 ≤ α < ζ2

vs(2,α) = 2 for ζ2 ≤ α < ζ3

X(α) = 1,α,0

φ(1,vs(1,α,0),0) ≤ α < φ(1,vs(1,α,0),0+1)

vs(1,α,0) = -1 for α < φ(1,0,0)

vs(1,α,0) = 0 for φ(1,0,0) ≤ α < φ(1,1,0)

vs(1,α,0) = 1 for φ(1,1,0) ≤ α < φ(1,2,0)

vs(1,α,0) = 2 for φ(1,2,0) ≤ α < φ(1,3,0)

How to add two ordinals
Let α, β - ordinals.

α+β = ?

Represent α and β in Cantor normal form. We get sum of

ωαi·ni

terms which are terms of Cantor normal forms of both α and β (now αi may be not decreasing).

We can use (α+β)+γ = α+(β+γ) and add terms one by one.

Rules of adding two ωαi·ni terms:

1. ωα1·n1+ωα2·n2 = ωα1·(n1+n2) (if α1 = α2)

2. ωα1·n1+ωα2·n2 = ωα1·n1+ωα2·n2 (if α1 > α2) (the sum is not changed)

3. ωα1·n1+ωα2·n2 = ωα2·n2 (if α1 < α2)

So, ωαi·ni "eats" all terms leftward until it faces a term with larger αi, and if it faces a term with equal αi, its ni is added with its own ni.

Examples:

α = ω9·2 + ω7 + ω4·3 + 1

β = ω5 + ω3·5 + ω2·2 + ω

α+β = ω9·2 + ω7 + ω4·3 + 1 + ω5 + ω3·5 + ω2·2 + ω

ω5 "eats" 1 and ω4·3

α+β = ω9·2 + ω7 + ω5 + ω3·5 + ω2·2 + ω

We get α+β in Cantor normal form.

α = ω10 + ω3·6 + ω·5

β = ω3·5 + ω·7 + 8

α+β = ω10 + ω3·6 + ω·5 + ω3·5 + ω·7 + 8

ω3·5 "eats" ω·5 and added with ω3·6

α+β = ω10 + ω3·11 + ω·7 + 8

If vs(α) < vs(β) then zp(β) "eats" all α, so, α+β = β

Example:

α = ω + 1

β = ε0

α+β = ω + 1 + ε0 = ω + 1 + ωε0 = ωε0 = ε0

How to multiple two ordinals
Let α, β - ordinals.

α·β = ?

Represent α and β in Cantor normal form. We get product of two sums.

We can use α·(β+γ) = α·β+α·γ and we get sum of products of α and terms ωαi·ni of Cantor normal form of β:

sum of α·ωαi·ni

How to add ordinals, we already know. Now we should find out how to multiple α and ωαi·ni.

Rules:

1. α·0 = 0

2. α·ni = zp(α)·ni + rt(α) = ωvs(α)·(zc(α)·ni) + rt(α) (if ni > 0)

(zc(α) and ni are multiplied as natural numbers)

3. α·ωαi·ni = ωvs(α)+αi·ni (if αi > 0, ni > 0)

Examples.

α = ω9·2 + ω7 + ω4·3 + 1

zp(α) = ω9·2

rt(α) = ω7 + ω4·3 + 1

vs(α) = 9

zc(α) = 2

Let β = 5

From Rule 2

α·β = ωvs(α)·(zc(α)·β) + rt(α) = ω9·10 + ω7 + ω4·3 + 1

Let β = ω·6

vs(β) = 1

zc(β) = 6

From Rule 3

α·β = ωvs(α)+vs(β)·zc(β) = ω10·6

How to raise a natural number to the power of an ordinal
Let n - natural number, α - ordinal.

nα = ?

nα = ωof(α)·nfp(α)

Example.

n = 2

α = ω9·2 + ω7 + ω4·3 + 1

fp(α) = 1

of(a) = ω8·2 + ω6 + ω3·3

nα = ωof(α)·nfp(α) = ωω 8·2 + ω6 + ω3·3 ·n

[β]α
Let X in my function [X]α is β.

[β]α = α+1+β

Special case for lp(β) ≠ 0:

[β]α = α+β

[1,β]α
[1,β]α = -1+(1+α)·21+β

Special case for lp(α) ≠ 0:

[1,β]α = α·21+β

Special case for lp(α) ≠ 0 and lp(β) = 0:

[1,β]α = α·21+β = ωvs(α)·(zc(α)·21+β) + rt(α)

Special case for lp(α) ≠ 0 and lp(β) ≠ 0:

[1,β]α = α·2β = ωvs(α)+of(β)·2fp(β)

[2,0]α
[2,0]α = [1,α]α = -1+(1+α)·21+α

Special case for lp(α) ≠ 0:

[2,0]α = ωvs(α)+of(α)·2fp(α)

[n,β]α
Let α, β - ordinals, lp(α) ≠ 0, of(β) ≠ 0, n - natural number.

[n,β]α = φ(n-1,vs(n-1,α)+of(β)) (if fp(β) = 0)

[n,β]α = [n,fp(β)-1]φ(n-1,vs(n-1,α)+of(β)) (if fp(β) ≠ 0)

[n,0]α
Let α, β - ordinals, of(α) ≠ 0, n - natural number, n > 1.

[n,0]α = φ(n-2,vs(n-2,α)+of(α)) (if fp(α) = 0)

[n,0]α = [n-1,fp(α)-1]φ(n-2,vs(n-2,α)+of(α)) (if fp(α) = 0)

Next step may be other functions of ordinals including [X]α with other X such as γ,β and with more than two elements.