If $ \in$ Card where Card is the class of all infinite cardinals, and if $0 \neq \lambda \in$ Cd where Cd is the class of all cardinals, how can I prove the followings:
(but without possible usage of Godel's pairing function)
$a)$ $\kappa \times \kappa = \kappa$
$b)$ $\kappa \times \lambda =$ max {$\kappa , \lambda$}
$c)$ $\kappa + \lambda =$ max {$\kappa , \lambda$}
Remark: I use the following definition of cardinals:
card$(x):=$ min {$\alpha \in$ Ord | $\exists$ bijection $f:\alpha \to x$}
We'll first assume that we proved a) and deduce b) and c). Then we'll prove a) (obviously without using our newly acquired results).
So let $\kappa, \lambda$ be two infinite cardinals. Assume $\lambda\leq \kappa$. We let $x\leq y$ denote "there is an injection $x\to y$". We have $\kappa \leq \kappa +\lambda \leq \kappa\times\lambda \leq \kappa\times\kappa\leq \kappa$; which allows us to conclude for b) and c) using Cantor-Bernstein's theorem. The first inequality is trivial, the second one is obtained by $\kappa +\lambda \sim \kappa\times\{0\}\cup\lambda\times\{1\}$ and then using the inclusion map (here we use $\lambda\neq 0$); the third one comes from our assumption that $\lambda\leq \kappa$, and the final one comes from a). In other words, it suffices to prove a).
The proof for a) is a bit more delicate, as it uses induction, but I'm sure you must be familiar with this. The result is obviously true for $\omega$ (there are many explicit bijections $\omega\times\omega\to\omega$). Therefore let $\kappa$ be any infinite cardinal and assume the result holds for all cardinals $<\kappa$.
Define the following order on $\kappa\times\kappa$: $(x,y)\preceq (x', y')$ if and only if : either $max(x,y) < max(x',y') $ or ($max(x,y)=max(x',y')$ and $(x,y)\leq (x', y')$ ) (in the last condition, $\leq$ denotes the lexicographic ordering). It's quite easy to see that this is a well ordering on $\kappa\times\kappa$ (I'll leave the details to you)
Therefore, $(\kappa\times\kappa,\preceq)$ is isomorphic (with a unique isomorphism $f$) to a unique ordinal $\delta$. What we want to show is that $\delta$ cannot be bigger than $\kappa$, which will then show that it is $\kappa$.
So assume $\kappa < \delta$. Then let $(\lambda_0, \lambda_1)= f^{-1}(\kappa)$. Both $\lambda_i <\kappa$, so if we let $\lambda = sup(\lambda_0,\lambda_1)$, we have $\lambda<\kappa$.
Since $f$ is an isomorphism, $\kappa \subset Im(f_{\mid \lambda\times \lambda})$. But using our induction hypothesis, $\lambda\times\lambda$ has a cardinal $<\kappa$ : indeed it is either finite, in which case it's obvious, or it is infinite and then its cardinal is the cardinal of $\lambda <\kappa$. This is a contradiction, since $Im(f_{\mid \lambda \times \lambda})$ has cardinal at most $card(\lambda\times\lambda)$.
Therefore our assumption was wrong, $\delta\leq \kappa$, and so $\kappa\times\kappa \leq \kappa$. The reverse inequality being trivial, Cantor-Bernstein's theorem shows that $\kappa\times \kappa \sim\kappa$.
This gives us a), and thus b) and c)