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In a recent post I inquired about arithmetic between integers and congruence classes. I learned a lot from the comments and I want to do right by giving a more adequate context to my predicament.

I have an equation $(a_{1}+a_{2})(N_{1} +N_{2})=a_{2}N_{2}+β$ ($β=N_{1}(a_{1}+a_{2})+a_{1}N_{2}$). $a,k∈ℤ$ and $N∈\left\{[0],A,B,C,D\right\}$. I wanted to have $[k]=N∈ℤ/5ℤ$ and $A=[1]$, $B=[2]$, $C=[3]$, and $D=[4]$. Thus, I could have an equation like such:

$(3-3)(D +A)=-3A+β$.

However, I was skeptical about the ability to perform arithmetic with integers and congruences classes. Not only to see if it is adequate to call such an object an equation, but also to know if $-3A=-A-A-A$. Now I know that it isn't feasible. You cannot perform arithmetic with integers and congruence classes.

I'm hoping there is some other way I can get the equation I want.

What I'm trying now is having each letter represent $f(5m+n)$. So $A=f(5m+1)$, $B=f(5m+2)$, all the way to $D=f(5m+4)$. $f(5m+0)=z$. $f(k)=N∈\left\{z,A,B,C,D\right\}.$ This way, it seems valid that $-3A=-A-A-A$. Is this a feasible alternative? Thank you for the input. I'm always open to information that will help me succeed.

EDIT: $5m+n=k$. While $k,m,n∈ℤ$, $0≤n<5$. $N∈\left\{ z, A, B, C, D \right\}$. $f:ℤ→\left\{ z, A, B, C, D \right\}$. 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 $A=f(5m+1)$.

Eugene Mason IV
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  • Comments are not for extended discussion; this conversation has been moved to chat. – Xander Henderson Dec 25 '22 at 14:51
  • @XanderHenderson thank you for the clarification. By chance, can you verify how many comments count as extended discussion and how I can utilize chat voluntarily? I was not aware this site had a chat area. Thank you. – Eugene Mason IV Dec 26 '22 at 01:49
  • The site automatically raises flags after 20 or so comments. If a back-and-forth goes on for several comments, new commenters will be given the option to move the conversation into chat. Otherwise, non-moderator users cannot manually move comments to chat. – Xander Henderson Dec 26 '22 at 03:31
  • Ah, okay. Thank you you for this information. – Eugene Mason IV Dec 26 '22 at 08:59
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    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 – Bill Dubuque Jan 01 '23 at 17:31
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    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}\ \ $ – Bill Dubuque Jan 01 '23 at 17:32
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    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\ \ $ – Bill Dubuque Jan 01 '23 at 17:32

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