Equivalence classes are sets, a collection, not a singular element like some arbitrary integer. You are comparing apples to oranges. How do you solve this equation: $3+\mathbb{Q}$? What does $+$ even mean in this context?

Thank you @BertrandWittgenstein'sGhost . I explained how I moved away from equivalence classes and opted for $f(5m+n)$. The thing is I wanted something with similar properties as modular arithmetic without it being so. The letters $A$, $B$, and so on are symbols. They aren't meant to be solved for.

Sorry, about that. I just fixed it.

Your question is very - very - confusing. What do you mean by $f(5m+1)$? What does it map to? $f(5m+1)=[1]=\{6,11,16,....\}$? This doesn't make sense because, again, this is not a function but a set. The same idea applies here. You can't add sets unless you define what $+$ means.

$5m+n=k$. While $k,m,n∈ℤ$, $0≤n<5$. $N∈\left\{ z, A, B, C, D \right\}$. $f:ℤ→N$. So $f(5m+1)=A$. I'm merely mapping integers to valueless letters. My mistake was using modular arithmetic because $-3A≠-A-A-A$ when $A=[1]$. What I'm hoping is that $-3A=-A-A-A$ when $A=f(5m+1)$.

I think you may meant $f:\mathbb{Z}\to\{z, A, B, C, D\}$. Besides, why do you think that $-3A\neq -A-A-A$ using usual modular arithmetic?

Regardless of how you are phrase it, the end result is that you are trying to define a binary operation on set of set, it doesn't work unless you give it some nice properties. Call it $-3A=-A-A-A$, or call it $-3f(Dom(f))=-f(Dom(f))-f(Dom(f))-f(Dom(f))$ these are all sets, they are not integers.

Thank you for the clarification. Is their a term or topic that I can research that deals with providing such formal properties? I meant to ask you about what you meant by properly "defining addition." but I don't want to take up any more of your time. I very much appreciate the input you've offered.and thank you for bearing with me.

It seems you are attempting to apply modular arithmetic and/or congruences without having a firm grasp on the underlying theory, i,e. you are putting the cart before the horse. Forget modular arithmetic and take a step backwards and tell us what problem you are really trying to solve (using mod arithmetic). Then there may be some hope of figuring out what you intend.

Unfortunately I can't really grasp what you are trying to do from your question, that's one of the main issues. Other than that, $+,×$ are all well defined binary operations on congruence classes $[a]_n+[b]_n=[a+b]_n$, but I don't know if that's what you are looking for. Good luck.

@BillDubuque BertrandWittgenstein'sGhost. Sure thing. Here's what I'm dealing with. I have a set of elements $M=\left\{ A, B, C, D \right\}$. $m∈M$. On there own, these elements are valueless in a mathematical context, mere objects like an apple or banana. However, they are to be paired with an integer $k$ as if to be multiplied, i.e. $km$. Despite it having no intrinsic value, I would like to be able to perform arithmetic between $m$ and $k$. For example, I would like to say that $2m=m+m$. I would like $m_{1}+m_{2}=m_{2}+m_{1}$ to be valid. I would like for $m-m=0$.

Of course, this becomes a problem when you are working with a set of valueless elements. After all, how can you even have $m-m=0$ when $0$ isn't even within the set? Even if you include the identity $0$ $(M∪\left\{0 \right\}=\left\{ 0, A, B, C, D \right\}$) it still doesn't seem to make sense. $A-A=0$? How would that even make sense when $A$ has no value?

As time went on, I also realized that I wanted the addition of any element from $M∪\left\{0 \right\}$ to return another of the same set. For example $A+A=B$ (I realize I'm hinting at a group here). That's when I looked at modular arithmetic. Though, even if it meant that the once valueless elements would represent a set of values, it didn't matter as long as I could perform arithmetic with those elements. Then again, the issue then becomes how to justify the multiplication of an integer $k$ to an element of $M$ returning another element of $M$.

@BertrandWittgenstein'sGhost thank you for your help. Hopefully this explanation shines some more light on the situation.

If you are working with all formal sums $ k_1 A + k_2 B + k_3 C + k_4 D\,$ for integers $\,k_1\,$ then this is the free abelian group (or $\Bbb Z$-module) generated by $A,B,C,D$. Here free means they are free of any nontrivial equalities, i.e. two sums are equal iff they have the same coefs $\,k_i\,$ (or, equivalently, a sum $= 0\iff $ all its coefs $\,k_i = 0).\,$ If instead there are some equalities between sums then you are working in a quotient of the free group, i.e. modulo the subgroup generated by the equalities imposed. Lookup group presentation by generators & relations.

Here the only operation involved is addition. There is no (ring) multiplication, rather $\,nX\,$ denotes repeated addition $\,X +\cdots + X\, (n\,$ times) to get the $n$'th multiple of $X$, e.g. $\,2X = X + X\,$ is doubling, $\,3X = X+ X+ X\,$ is tripling. When the group operation is written multiplicatively then then repeating it yelds $n$'th powers of $X$, e.g. squares $X^2 = X * X,\,$ and cubes $\,X^3 = X * X * X,\,$ etc. In this multiplicative notation your sums would be notated as products $\,A^{k_1} B^{k_2} C^{k_3} D^{k_4}\ \ $

For negative multipliers $-n < 0\,$ we have $\,-nX = n(-X)$ is the $\, n$'th multiple of $\,{-}X = $ additive inverse of $X.\ $ Said multiplicatively $\,X^{-n} = (X^{-1})^{n} = n$'th power of inverse of $X\ \ $

Thank you so much @BillDubuque . This has been so very helpful. I appreciate your time and effort.