# I Notation question

Tags:
1. Sep 20, 2016

### Math_QED

Hello everyone. I have read a proof but I have a question concerning the notation. To give some context, I will write down this proof as written in the book.

Theorem: There is a unique binary operation $+: \mathbb{N} \times \mathbb{N} \rightarrow \mathbb{N}$ that satisfies the following two properties for all $n,m \in \mathbb{N}$
1) n + 1 = s(n)
2) n + s(m) = s(n + m)

(s is the successor function as described in the Peano Postulates)

Proof: Uniqueness: I'm going to skip this here as it is bot important for my question.

Existence:

For $p \in \mathbb{N}$, we can apply the recursiob theorem to the set $\mathbb{N}$, the element $s(p) \in \mathbb{N}$ and the function $s: \mathbb{N} \rightarrow \mathbb{N}$ to deduce that there is a unique function $f_p: \mathbb{N} \rightarrow \mathbb{N}$ such that $f_p(1) = s(p)$ and $f_p \circ s = s \circ f_p$. Let $+: \mathbb{N} \times \mathbb{N} \rightarrow \mathbb{N}$ be defined by $c + d = f_c(d)$ for all $(c,d) \in \mathbb{N} \times \mathbb{N}$. Let $n,m \in \mathbb{N}$. Then $n + 1 = f_n(1) = s(n)$, which is part 1) and $n + s(m) = f_n(s(m)) = s(f_n(m)) = s(n + m)$, which is part 2).

Now, here comes this silly question. Why does the author use the notation $f_c(d)$? It seems that he's 'hiding' that $f_c$ depends on 2 variables $c,d$ instead of 1. Although I do understand the proof, I feel uncomfortable with this notation.

Last edited by a moderator: Sep 20, 2016
2. Sep 20, 2016

### Simon Bridge

He wants to consider c as an identifying parameter while d is the variable, in order to be consistent with the definition earlier of $f_p$ ... $f_c$ satisfies the same definition with $p=c$.

It can be valid to use f(p,d) instead, with the modified notation in the defnition.

Consider the analogous situation:
$f(x)=\sum_{n=0}^N a_ng_n(x)$ vs $f(x)=\sum_{n=0}^N a(n)g(n,x)$ ...

3. Sep 20, 2016

### Math_QED

Thanks. But if I would write $f(p,d)$ instead, this would indicate that $f$ has as domain $\mathbb{N} \times \mathbb{N}$ where the domain is in fact $\mathbb{N}$, wouldn't it?

4. Sep 20, 2016

### Simon Bridge

Well, by that argument, $g_n(x)$, in the analogy, has domain $\mathbb N \times \mathbb R$ right?
Are you unhappy with the subscript notation there too?

Consider the set of polynomials ... if y is a polynomial of degree in in x, then we can write $y = p_n(x)$ right?
But p still maps one dimension onto one dimension even though I need another number to specify the degree.

An example in physics would be the single atomic state wavefunction, which would be: $\Psi_{nlms}(x,y,z,t)$ ... so now we have 8 variables, four of them are subscripts specifying the state and four are arguments. What do we gain from writing $\Psi(n,l,m,s,x,y,z,t)$?

An advantage of using the subscript notation over including it as an argument of the function is that you can talk about $g_n$ (etc) as a particular function, and discussing properties of, without referring to the argument explicitly. This is, in fact, what the author does.

Perhaps it would help to think of $f_p(s)$ as holding the value of p constant and varying s - but, at the same time, recognising the c may take more than that one value in general. This is an implication that notation $f(p,s)$ does not provide. Note: if p is a parameter rather than an argument, then the domain is still 1D.

The author's use is proper and reasonable and consistent and to the purpose of the proof.
What is the problem?

5. Sep 20, 2016

### Math_QED

I did not cover the thing with the sums, yet, but your explanation here helped a lot and now I understand it. I just ended high school and I am not familiar with such notations so most likely that's where the confusion started. Once I will start at the university, I will get used to it. Now, there is not a problem anymore. Thank you for helping me out.