How do we prove or visualize $[x+(x-2)]=[2+(x-2)]$ the same way we prove or visualize $0+5\mathbb Z = 5 + 5\mathbb Z$? Denote $\langle x-2\rangle$ as the principal ideal generated by $x-2$ in the polynomial ring $\mathbb R[x]$.


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*$[x+\langle x-2\rangle]$ and $[2+\langle x-2\rangle]$ are elements of the quotient ring $\mathbb R[x]/\langle x-2\rangle$, which happens to be a field because $x-2$ is monic irreducible in $\mathbb R[x]$ (and the Proposition here).


Here is what I tried:
Let us take an element in one side and show it is in the other side. One of the elements in $[x+\langle x-2\rangle]$ is $x+x(x-2)$. Now we must find $q \in \mathbb R[x]$ such that
$$x+x(x-2) = 2+q(x-2)$$
And it is $q(x)=x+1$ by solving for $q$.
In general for $x+r(x-2)$ and $r \in \mathbb R[x]$, $q=r+1$.
Right to left is similar.
Is that correct?


*And then in general, to show


$$[a+\langle x-2\rangle] = [b+\langle x-2\rangle]$$ for elements $\overline a=\overline b$ in $\mathbb R[x]/\langle x-2\rangle$, to show an element on the left hand side is on the right hand side, we are given and $r$ and must find $q$ such that
$$a+r(x-2)=b+q(x-2)$$
and we solve for $q$:
$$a+r(x-2)=b+q(x-2)$$
$$\iff a-b+r(x-2)= q(x-2)$$
$$\iff c(x-2)+r(x-2)= q(x-2), c \in \mathbb R[x]$$
$$\iff (c+r)(x-2)= q(x-2), c \in \mathbb R[x]$$
Therefore $q=c+r$ where $c$ exists as a polynomial with coefficients in $\mathbb R$ by definition of "$\overline a=\overline b$ in $\mathbb R[x]/\langle x-2\rangle$", which is that as
$\overline a=\overline b$ in $\mathbb Z/\langle n \rangle$" means that $a-b=cn$ for some $c \in \mathbb Z$,
$\overline a=\overline b$ in $\mathbb R[x]/\langle p \rangle$" means that $a-b=cp$ for some $c \in \mathbb R[x]$.
Is that correct?

The analogy should be $5+5\mathbb Z=0+5\mathbb Z$ and $[x-2+\langle x-2 \rangle] = [0+\langle x-2 \rangle]$:
$$5+5\mathbb Z = 5+\{...,-5,0,5,...\} = 0+\{...,0,5,10,...\}=0+5\mathbb Z$$
or
$$5+5\mathbb Z = 5+\{5m\} = 0+\{5+5m\}=0+\{5(m+1)\}=0+\{5(n)\}$$
Similarly,
$$x-2+\langle x-2 \rangle = x-2+\{(x-2)(p)\} = 0+\{(x-2)(p+1)\} = 0+\{(x-2)(q)\}$$
