Correstness of the mathematical expression I have a question regarding some strange mathematical expression defined in the set of positive integers - N :


*

*g(N) - greatest power of 2 that divides N.
e.g. g(24) = 8 because 8 is divided by 2,4,8 but is not divided by 16

*g(0) = ∞
Is point 2. represents correct mathematical expression? Can we use ∞ notation without using limits?
 A: The function $g$ is a function $\Bbb N \to \Bbb N \cup \{\infty\}$. This is a well-defined thing to do, but you have to be careful about what you do with the values of $g(x)$.
On one level, $\infty$ is just a symbol; it would be equally valid to define $g(0) = \clubsuit$ at which point $g$ is a function $\Bbb N \to \Bbb N \cup \{\clubsuit\}$. 
However, this is only considering $\Bbb N \cup \{\infty\}$ as a set. Sets are crucial to mathematics, but a set on its own doesn't have much structure. The natural numbers have lots more structure than a set: you can add them, multiply them, work out whether one divides another, work out which of two is bigger, and so on. (And the real numbers have lots more structure!) These operations aren't defined for $\infty$, and while we can sometimes write down plausible definitions for what those operations should be, we have to be careful, because any theorems we proved about the natural numbers don't necessarily hold for $\Bbb N \cup \{\infty\}$.
(I talk about an example of this, admittedly talking about real numbers rather than natural numbers, in my answer at What is the limit of zero times x, as x approaches infinity?)
However, there is one bit of structure that definitely does extend nicely, which is the comparison relations on $\Bbb N$. For natural numbers $n$ and $m$, we know how to define whether $n < m$ or not. We can extend this definition to $\Bbb N \cup \{\infty\}$ by saying $n < \infty$ for all $n \in \Bbb N$. ("Infinity is bigger than any natural number"). 
This also has the advantage of making the two parts of the definition consistent: if you define
$$h(x) = \sup \{2^i: i \in \Bbb N, \exists r \in \Bbb N\text{ s.t. }2^ir = x\}$$
then $h(0) = \sup \{2^i: i \in \Bbb N\} = \infty$, and so $h(x) = g(x)$ for all $x \in \Bbb N$.
(Also, I disagree with @MPW; $0$ is a positive integer, and certainly $0 \in \Bbb N$, though I would normally call $\Bbb N$ the natural numbers rather than the positive integers, partly because of the confusion as to whether "positive" means "$> 0$" or "$\geq 0$".)
A: No, for two reasons:


*

*$0$ is not a positive integer

*$\infty$ is not a power of $2$
My point is that $g$ is ill-defined as stated.
