# 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= $\omega$a+b and y= $\omega$c+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