Multiplying Infinite Cardinals (by Zero Specifically) On the Wikipedia page on Cardinal Numbers, Cardinal Arithmetic including multiplication is defined. For finite cardinals there is multiplication by zero, but for infinite cardinals only defines multiplication for nonzero cardinals. Is multiplication of an infinite cardinal by zero undefined? If so, why is it? 
Also does $\kappa\cdot\mu= \max\{\kappa,\mu\}$ simply means that the multiplication of the two is simply the cardinality of the higher cardinal? Why is this?
 A: Suppose $A,B$ are sets, we have a good sense about $|A|,|B|$ which are the cardinality of these sets.
The arithmetics of cardinals is quite simple:


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*$|A|+|B| = |(\{0\}\times A)\cup(\{1\}\times B)|$ (That is the disjoint union of $A$ and $B$)

*$|A|\cdot|B| = |A\times B|$

*$|A|^{|B|}=|\{f:B\to A\mid f\text{ is a function}\}|$


It is easy to see that these operations are well defined (that is, $|A|=|A'|$ and $|B|=|B'|$ then $|A|+|B|=|A'|+|B'|$ etc.) and it is also quite simple to see that the usual addition, multiplication and exponentiation correspond to the same operations (with the "exception" that $0^0=1$).
If $|B|=0$ then $B=\varnothing$ and we know that $A\times\varnothing=\varnothing$, for every set $A$. Therefore $\kappa\cdot 0=0$.

As for your second question, assuming the axiom of choice holds, every two cardinalities are comparable and the result is that multiplication and addition is the same thing as taking the maximum. This is a result of the well ordering theorem, which is equivalent to the axiom of choice. 
Such assumption is widely accepted today and it's quite rare to find people working in contexts where it is false (like yours truly). In situations like that it is usually the case that $|A|\cdot|B|\neq\max\{|A|,|B|\}$.
A: For any cardinal $\kappa$ whatsoever, $0\cdot\kappa=\kappa\cdot 0=0$. This is an immediate consequence of the definition and the fact that for any set $X$, $\varnothing\times X=\varnothing$.
Yes, if one assumes the axiom of choice, the product of two infinite cardinals is simply the larger of them; so is their sum. The product of a non-zero finite cardinal and an infinite cardinal is that infinite cardinal, so it’s also simply the larger of the two. This fails when the finite cardinal is $0$, because then the product is $0$.
Even without the axiom of choice it’s true that if $\kappa$ and $\mu$ are well-orderable cardinals, $\kappa\cdot\mu=\max\{\kappa,\mu\}$. This is proved by constructing a bijection between $\kappa\times\mu$ and $\max\{\kappa,\mu\}$.
