Can someone quickly tell me what the set $\mathbb R[X]$ is defined as, where $\mathbb R$ is the set of real numbers?
Is it $$\mathbb R[X]=\{ a_nX^n+.........+a_1X+a_0\mid a_n,...,a_0 \in\mathbb R\}$$? but I do not know what n is? is it a natural number or integer or............?
What can n be?
I need to write this own because mathematics stack is not letting me post this question because of the lack of description in this question, but there is nothing more to say.
A common informal definition of $\mathbb{R}[X]$ is
the set of all expressions $a_0+a_1X+\dots+a_nX^n$, where $n$ is a nonnegative integer and $a_0,a_1,\dots,a_n\in\mathbb{R}$; addition is defined by reducing alike terms, multiplication is defined by the usual rule, applying the distributive property.
However, this lacks mathematical rigor: while we “informally” know what an expression is, this is not what's expected from a precise mathematical definition.
A more rigorous definition is as follows. Consider the set $\mathcal{S}$ of all sequences of real numbers, that is, all maps $\mathbb{N}\to\mathbb{R}$. Among these, consider $\mathcal{S}_\omega$ consisting of all sequences $s$ such that there exists $n$ with $s_m=0$, for all $m>n$ (in the context of sequences it's customary to write $s_0$, $s_1$ and so on instead of $s(0)$, $s(1)$ and so on).
A polynomial with real coefficients is then an element of $\mathcal{S}_\omega$.
We need to define operations on this set so that, if $a\in\mathcal{S}_\omega$, we can rewrite it as $a_0+a_1X+\dots+a_nX^n$, with suitable conventions, where $a_0$ $a_1$ and so on are exactly the terms of the sequence.
It's easier if we assume one knows about vector spaces. The set $S_\omega$ is clearly a vector space under addition defined by $$ a+b\colon n\mapsto a_n+b_n\qquad \alpha a\colon n\mapsto \alpha a_n $$ for $\alpha\in\mathbb{R}$ and $a,b\in\mathcal{S}_\omega$.
Define, for natural $m$, $e^{(m)}$ as the sequence $$ e^{(m)}\colon n \mapsto \begin{cases}1&\text{if $n=m$},\\0&\text{if $n\ne m$}.\end{cases} $$ It easy enough to show (by induction) that any set $$ \mathcal{B}_n=\{\,e^{(0)},e^{(1)},\dots,e^{(n)}\,\} $$ is linearly independent and that any $a\in\mathcal{S}_\omega$ can be written as a linear combination of one of these sets: precisely, if $a_m=0$, for all $m>n$, then $$ a=a_0e^{(0)}+a_1e^{(1)}+\dots+a_ne^{(n)} $$
Now the big step: we define a multiplication on polynomials. If $a,b\in\mathcal{S}_{\omega}$, define $ab=c$, where $$ c_n=\sum_{k=0}^na_kb_{n-k} $$
It's quite tedious, but basically trivial, to verify that this multiplication is associative and distributes over addition.
That's it! Note that $e^{(0)}$ is the unity for this multiplication and that $$ e^{(n)}=\underbrace{e^{(1)}\cdot\dots\cdot e^{(1)}}_{\text{$n$ times}} $$ so, if we set $X=e^{(1)}$, we can write $e^{(n)}=X^n$. Thus the above expression for $a\in\mathcal{S}_\omega$ can be also written $$ a_0X^0+a_1X^1+a_2X^2+\dots+a_nX^n $$ and of course it's customary to omit $X^0$ and abbreviate $X^1$ into $X$.
Try your hand with the multiplication between the sequences $$ a\colon n\mapsto \begin{cases} 3 & \text{if $n=0$}\\ -2 & \text{if $n=1$}\\ 1 & \text{if $n=2$}\\ 0 & \text{if $n>2$} \end{cases} \qquad\text{and}\qquad b\colon n\mapsto \begin{cases} -2 & \text{if $n=0$}\\ 1 & \text{if $n=1$}\\ 0 & \text{if $n>1$} \end{cases} $$ with the above definition corresponds to the usual multiplication between the polynomials written explicitly as $$ 3-2X+X^2\qquad\text{and}\qquad -2+X $$
Do we really need this formal definition? Yes, we do. Or, at least, we need to know how the set of polynomials can be realized “concretely”.
Higher level descriptions are possible, which shed more light on the concept (or perhaps make it more obscure ;-)).
$\mathbb{R}[x]$ is the set of all polynomials with coefficients from $\mathbb{R}$.
All polynomials mean that polynomial of each degree is present in $\mathbb{R}[x]$. And also degree of a (non-zero) polynomial is $\ge 0$. So $n=0,1,2,....$
