Calculations with transfinite ordinals

Before to read the post, I would like to show a list of the most import rules of calculations. Have a look the list below:

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

The reader is invited to verify, by easy direct calculation or easy induction, the following identities:

{ (\omega +m) }^{ n }={ \omega }^{ n }+{ \omega }^{ n-1 }m+{ \omega }^{ n-2 }m+\cdot\cdot\cdot\omega m+m,                m,n\in \omega
{ (\omega +1) }^{ n }={ \omega }^{ n }+{ \omega }^{ n-1 }+{ \omega }^{ n-2 }+\cdot\cdot\cdot\omega +1,                n\in \omega  the binomial theorem is not true any more !!
In particular:
{ (\omega +1) }^{ 2 }={ \omega }^{ 2 }+\omega +1,         no double product !!!
{ (\omega +1) }^{ 3 }={ \omega }^{ 3 }+{ \omega }^{ 2 }+\omega +1\neq ({ \omega }^{ 2 }+1)(\omega +1)=({ \omega }^{ 2 }+1)\omega+({ \omega }^{ 2 }+1)1={ \omega }^{ 3 }+{ \omega }^{ 2 }+1, classical decompositions don’t work any more !!
{ ({\omega}^{2} +1) }^{ 2 }={ \omega }^{ 4 }+{\omega}^{2} +1    substitution of \omega by {\omega}^{2} here is true !!
{ ({\omega}^{2} +1) }^{ 3 }={ \omega }^{ 6 }+{ \omega }^{ 4 }+{ \omega }^{ 2 }+1
{ ({\omega}^{k} +1) }^{ n }={ \omega }^{ kn }+{ \omega }^{ k(n-1) }+{ \omega }^{ k(n-2) }+\cdot\cdot\cdot {\omega}^{k} +1,                k,n\in \omega
{ (\omega p +m) }^{ n }={ \omega }^{ n }p+{ \omega }^{ n-1 }pm+{ \omega }^{ n-2 }pm+\cdot\cdot\cdot\omega pm+m,                p,m,n\in \omega
Recall that     (\omega p +m) =m(\omega + 1)p but observe that {m}^{n}{(\omega +1)}^{n}{p}^{n}={m}^{n}({ \omega }^{ n }+{ \omega }^{ n-1 }+{ \omega }^{ n-2 }+\cdot\cdot\cdot\omega +1){p}^{n}={ \omega }^{ n }{p}^{n}+{ \omega }^{ n-1 }+{ \omega }^{ n-2 }+\cdot\cdot\cdot\omega +{m}^{n} is not the same as before.
More generally:
{ ({\omega}^{k} p +m) }^{ n }={ \omega }^{kn }p+{ \omega }^{ k(n-1) }pm+{ \omega }^{ k(n-2) }pm+\cdot\cdot\cdot {\omega}^{k} pm+m,                k,p,m,n\in \omega

As a consequence of these results we obtain:
\sqrt [ n ]{ { \omega }^{ n }{ a }_{ n }+{ \omega }^{ n-1 }{ a }_{ n-1 }+{ \omega }^{ n-2 }{ a }_{ n-2 }+\cdot\cdot\cdot\omega { a }_{ 1 }+{ a }_{ 0 }}=\begin{cases} w{ a }_{ n }+{ a }_{ 0 }\quad if\quad { a }_{ n-1 }={ a }_{ n-2 }=\cdot\cdot\cdot ={ a }_{ 1 }={ a }_{ n }{ a }_{ 0 } \\ no\quad solution\quad otherwise \end{cases}

Decomposition into irreducible (unzerlegbar) multiplicative factors, an example explains better than unreadable  formules:
{\omega}^{\delta + \alpha + \beta + \gamma}c+{\omega}^{\delta + \alpha + \beta}b+{\omega}^{\delta + \alpha}a+{\omega}^{\delta}d={\omega}^{\delta}({\omega}^{\alpha + \beta + \gamma}c+{\omega}^{\alpha + \beta}b+{\omega}^{\alpha}a+d)={\omega}^{\delta}({\omega}^{\alpha + \beta}b + {\omega}^{\alpha}a +d)({\omega}^{\gamma}c + 1)={\omega}^{\delta}({\omega}^{\alpha}a + d)({\omega}^{\beta}b + 1)({\omega}^{\gamma}c + 1)={\omega}^{\delta}d({\omega}^{\alpha}a + 1)({\omega}^{\beta} + 1)b({\omega}^{\gamma} + 1)c =\\ {\omega}^{\delta}d({\omega}^{\alpha} + 1)a({\omega}^{\beta} + 1)b({\omega}^{\gamma} + 1)c

Squares and square roots:
given \alpha = {\omega}^{\gamma}{n}_{1}({\omega}^{{\beta}_{1}}+1){n}_{2}({\omega}^{{\beta}_{2}}+1)\cdot\cdot{n}_{s-1}({\omega}^{{\beta}_{s}}+1){n}_{s}, we distinguish two cases: \gamma = 0 that is \alpha successor or \gamma \neq 0 that is \alpha limit.
In the first case
{\alpha}^{2}=\alpha\alpha ={n}_{1}({\omega}^{{\beta}_{1}}+1){n}_{2}({\omega}^{{\beta}_{2}}+1)\cdot\cdot{n}_{s-1}({\omega}^{{\beta}_{s}}+1){n}_{s} {n}_{1}({\omega}^{{\beta}_{1}}+1){n}_{2}({\omega}^{{\beta}_{2}}+1)\cdot\cdot{n}_{s-1}({\omega}^{{\beta}_{s}}+1){n}_{s}
and since this “double form” is in canonical form it is the canonical decomposition of {\alpha}^{2}: let’s call it form a)
In the second case it’s easily seen that .{\alpha}^{2}={\omega}^{\gamma +{\beta}_{1}+\cdot\cdot {\beta}_{s}+\alpha}{n}_{1}({\omega}^{{\beta}_{1}}+1){n}_{2}({\omega}^{{\beta}_{2}}+1)\cdot\cdot{n}_{s-1}({\omega}^{{\beta}_{s}}+1){n}_{s} that we call form b).
In summary: a successor is a square iff of form a) a limit number is a square iff of form b)

examples:
\sqrt{{\omega}^{3}+{\omega}^{2}+1}=\sqrt{({\omega}^{2}+1)(\omega+1)}, the radicand is a successor not in double form a) \Rightarrow we have no root.
\sqrt{{\omega}^{4}+{\omega}^{3}}=\sqrt{{\omega}^{3}(\omega+1)}, we search \alpha such that \alpha +1+\alpha = 3 that gives \alpha = 1 and hence the solution \omega (\omega +1)
\sqrt{{\omega}^{4}2+{\omega}^{2}6+3}=\sqrt{3({\omega}^{2}+1)6({\omega}^{2}+1)2}=\sqrt{3({\omega}^{2}+1)2\cdot 3({\omega}^{2}+1)2}=3({\omega}^{2}+1)2={\omega}^{2}2+3

Powers and roots:
Similarly {\alpha}^{n} has a form a):
{\alpha}^{n}={n}_{1}({\omega}^{{\beta}_{1}}+1){n}_{2}({\omega}^{{\beta}_{2}}+1)\cdot\cdot{n}_{s-1}({\omega}^{{\beta}_{s}}+1){n}_{s} {n}_{1}({\omega}^{{\beta}_{1}}+1){n}_{2}({\omega}^{{\beta}_{2}}+1)\cdot\cdot{n}_{s-1}({\omega}^{{\beta}_{s}}+1){n}_{s}\cdot\cdot{n}_{1}({\omega}^{{\beta}_{1}}+1){n}_{2}({\omega}^{{\beta}_{2}}+1)\cdot\cdot{n}_{s-1}({\omega}^{{\beta}_{s}}+1){n}_{s}
or a form b):
{\alpha}^{n}={\omega}^{\gamma +{\beta}_{1}+\cdot\cdot {\beta}_{s}+\gamma (n-1)}{n}_{1}({\omega}^{{\beta}_{1}}+1){n}_{2}({\omega}^{{\beta}_{2}}+1)\cdot\cdot{n}_{s-1}({\omega}^{{\beta}_{s}}+1){n}_{s}

For the following exercises the following rules are useful:
{({\omega}^{\alpha}m+ something\quad less\quad important)}^{\omega}={\omega}^{\alpha \omega}
{n}^{\omega}=\omega   n\in\omega
{\epsilon}_{\alpha}={\omega}^{{\epsilon}_{\alpha}}           definitory property of \epsilon-number
{{\epsilon}_{\alpha}}^{{{\epsilon}_{\alpha}}}={\omega}^{({\omega}^{({\epsilon}_{\alpha}2) } )}{{\epsilon}_{\alpha}}^{{{\epsilon}_{\alpha}}}={({\omega}^{{\epsilon}_{\alpha}})}^{{\epsilon}_{\alpha}}={\omega}^{{\epsilon}_{\alpha}{\epsilon}_{\alpha}}={\omega}^{{\omega}^{{\epsilon}_{\alpha}}{\omega}^{{\epsilon}_{\alpha}}}={\omega}^{{\omega}^{{\epsilon}_{\alpha}+{\epsilon}_{\alpha}}}={\omega}^{{\omega}^{{\epsilon}_{\alpha}2}}
{\epsilon}^{({\epsilon}^{\epsilon})}={\omega}^{({\omega}^{({\omega}^{\epsilon 2})})} : {\epsilon}^{({\epsilon}^{\epsilon})}={({\omega}^{\epsilon})}^{({\epsilon}^{\epsilon})}={\omega}^{\epsilon{\epsilon}^{\epsilon}}={\omega}^{({\epsilon}^{1+\epsilon})}={\omega}^{({\epsilon}^{\epsilon})}={\omega}^{({({\omega}^{\epsilon})}^{\epsilon})}={\omega}^{({\omega}^{\epsilon\epsilon})}={\omega}^{({\omega}^{({\omega}^{\epsilon}{\omega}^{\epsilon})})}={\omega}^{({\omega}^{({\omega}^{\epsilon 2})})}
in general a tower of n \epsilon is a tower of n \omega with on the top \epsilon 2
{\epsilon}^{2}={\omega}^{\epsilon}{\omega}^{\epsilon}={\omega}^{\epsilon 2} and in general {\epsilon}^{n}={\omega}^{\epsilon n}
{\epsilon}_{\alpha}{\epsilon}_{\beta}={\epsilon}_{\beta} if \alpha <\beta because every \epsilon={\omega}^{\epsilon}={\omega}^{{\omega}^{\epsilon }} is a delta number!
{{\epsilon}_{\alpha}}^{{\epsilon}_{\beta}}={\epsilon}_{\beta} se \alpha <\beta in fact {{\epsilon}_{\alpha}}^{{\epsilon}_{\beta}}={({\omega}^{{\epsilon}_{\alpha}})}^{{\epsilon}_{\beta}}={\omega}^{{\epsilon}_{\alpha}{\epsilon}_{\beta}}={\omega}^{{\epsilon}_{\beta}}={\epsilon}_{\beta}
{\epsilon}^{({\epsilon}^{\omega})}={\omega}^{({\epsilon}^{\omega})}   in fact   {\epsilon}^{({\epsilon}^{\omega})}={({\omega}^{\epsilon})}^{({\epsilon}^{\omega})}={\omega}^{(\epsilon {\epsilon}^{\omega})}={\omega}^{({\epsilon}^{1+\omega})}={\omega}^{({\epsilon}^{\omega})}

Put in normal Cantor form the following numbers:
{ 3 }^{ { \omega }^{ 2 }5+\omega 2+4 }
sol: { 3 }^{ { \omega }^{ 2 }5+\omega 2+4 }={ 3 }^{ { \omega }^{ 2 }5}{ 3 }^{ \omega 2 }{ 3 }^{ 4}={({3}^{\omega)}}^{\omega 5}{({3}^{\omega})}^{2}81={\omega}^{\omega 5}{\omega}^{2}81={\omega}^{\omega 5+2}81
{\epsilon}_{0}
sol: {\omega}^{{\omega}^{{\omega}^{{\cdot}^{\cdot}}}}   by construction of the first fixed point after \omega
{\epsilon}_{1}
Solve the following equations:
{5}^{\omega + x}= \omega 5: sol.   {5}^{\omega}{5}^{x}= \omega 5,    \omega {5}^{x}= \omega 5,   {5}^{x} = 5  (right multiplicative injectivity)   x = 1
Solve the following sistems:
\begin{cases}x+y=\omega 3+1 \\ xy={\omega }^{ 2 }+\omega 2 \end{cases}
sol: the first equation bounds the possible solution to x=\omegaa+b and y=\omegac+d with a,b,c,d in \omega and with a,b \leq 3, substituting these expressions and equating coefficients we have \begin{cases}x=\omega 2 \\ y=\omega + 1 \end{cases}

Divisions with rest:
{\omega}^{\omega}5+\omega 7+4   divided by   \omega 3 +2
sol: we reproduce elementary division: {\omega}^{\omega}5 = (\omega 3){\omega}^{\omega}5(\omega 3 +2){\omega}^{\omega}5 ={\omega}^{\omega}5 =; if we subtract this from the dividend we obtain the rest \omega 7+4\omega 7 = (\omega 3)1+\omega 2 so we have the natural 2 as second term of the quotient; we multiply \omega 3 +2 by this second term quotient and obtain \omega 6 +2; we subtract this from \omega 7+4 and obtain \omega +4; this is the rest because it’s less than \omega 3+2. Summing up: {\omega}^{\omega}5+\omega 7+4 =(\omega 3 +2)({\omega}^{\omega}5 +2)+(\omega + 4)

References
H. Bachmann       Transfinite Zahlen    Springer Verlag 1967
J. Donald Monk   Introduction to set theory   Mc Graw Hill  New York 1969
Lectures on Set Theory  euclid.colorado.edu/~monkd/setth.pdf
M. Di Nasso        people.dm.umipi.it/dinasso/~Mauro Di Nasso