Predicate logic statements with one and multi parameters

Question

I'm new to the Math stack exchange community so my apologies in advance in making mistakes for putting my question into non-fitting categories and such.

Ultimate Question: When are you supposed to use n-ary predicate statements for proving any sort of proof-based questions? (In my context, any of the three inductions) In case there may be any confusion due to many different theories and notation for predicate logic n-ary as in... i.e. P(x, y) or P(x, y, z) or n-ary P(x_0, x_1, ... , x_n-1)

Sub-Question #1: I thought about this question myself and came up with sub-question like what the difference is between the two following predicates and I was not able to find the answer what the difference is searched online however no really specific and detailed answer but here it is:

Sub-Question #2: Does every variable that is used in the predicate sentence need to be bounded by a quantifier?

Consider, the two following question:

1. Prove that for all natural numbers $n, k*n \ge n$ for any natural number $k \ge 2.$

I thought of two predicates and two separate claims that I have made for the above question as follows:

Define a set $S =\{s \in \mathbb {N}\}.$

Also, define set $Y = \{y \in \mathbb {N} \; | \; y\ge 2\}.$

Predicate and Claim #1

Define the predicate $P:S \rightarrow \{True, False\}$

and is defined as $P(n): "k*n \ge n"$

Claim #1: $ \forall \; n\in S, P(n)$ holds.

NOTE: The bounded variable in the predicate would be $n$ and the free variable would be the $k$ because it is not bounded by some quantifier. So a natural question to ask is, do I have to/neccessary to bound the free variable $k$ in proving this question?

Claim #2: $ \forall \; n \in S , \forall \; k \in Y, P(n)$ holds.

Predicate and Claim #2

Define the predicate $P: (S \times S) \rightarrow \{True, False\}$

and is defined as $P(n, k): \; "k*n \ge n"$

Claim: $\forall n \in S, \; \forall k \in Y, \; P(n, k)$ holds.

Sub-Question #3: (I just thought of this question while writing this post...) If I have some arbitrary predicate statement (and ONLY THE predicate statement) $P(n)$, is the $n$ considered to be a free variable?

So out all the this attempts are any of them correct and if not can anyone provide with a proper rigorously defined predicate for the above question? Also, it will be very appreciated if someone can point out what basic background I might lack in to have this problem and also suggest any sort of textbooks or any online material that can possibly help me strengthen my knowledge in this field! I personally find that defining a predicate is the basis of learning how to rigorously prove any mathematical or non-mathematical statements.

Thank you so much in advance! :)

EDIT #1: Thank you for both of your answers but I did not state explicitly what the question I had in my mind. (My apologies, it was quite late last night wasn't able to think very straight)

I want to ask what the difference between my Claim #1 and Claim #2 and the difference between both of the two predicates that I defined. From my perspective the only difference I can spot in the predicate is the very explicit view which is my first predicate is unary and second predicate is binary BUT do not understand what the mathematical difference is.

Answer 1

We have that :

"for all natural numbers $n, k n ≥ n$ for any natural number $k ≥ 2$"

is:

$\forall n \forall k \ (k \ge 2 \to kn \ge n)$.

Thus, both variables $n$ and $k$ are (universally) quantified.

We assume that the domain is $\mathbb N$ : thus we can omit the "specification" : $\in \mathbb N$.

If we want to set-up a typical inductive proof, we can consider the binary predicate $P(n,k) := kn \ge n$ and apply induction on $k$ :

(i) Basis : $k=2$.

We have that $2n = n+n \ge n$.

(ii) Induction step : assume that the property holds for $k \ge 2$ and prove for $k+1$.

If $k+1 \ge 2$ (and thus $k \ge 1$, ensuring that $kn \ne 0$), we have that $(k+1)n=kn+n \ge n+n \ge n$.

In conclusion, we have $P(n,2) \text{ and } \forall k (k \ge 2 \to (P(n,k) \to P(n,k+1)))$. Thus, we can apply induction to conclude with:

$\forall k \ (k \ge 2 \to P(n,k))$.

But the result holds for $n$ whatever; thus we can generalize it to:

$\forall n \ \forall k \ (k \ge 2 \to kn \ge n)$.