Congruence Class Proofs: Tips and Examples

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Discussion Overview

The discussion revolves around proofs related to congruence classes, specifically within the context of modular arithmetic in $\mathbb{Z}_{3}$. Participants explore various approaches to proving properties of sums of squares in this system, examining cases and providing examples.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Homework-related

Main Points Raised

  • One participant expresses uncertainty about their solution and seeks assistance.
  • Another participant suggests examining all cases and notes that some cases may have been omitted, specifically mentioning $[a]=[b]=[2]$.
  • A participant proposes a method of expressing numbers in the form $[a]=3n+a$ and $[b]=3m+b$, leading to a derived equation involving $a^2 + b^2$.
  • Another participant reiterates the same method and derives the same equation, questioning its interpretative value and the implications of $a$ and $b$ being less than 3.
  • A later reply clarifies that the term $3(3n^3+3m^3+2an+2bn)$ is a multiple of 3 and can be disregarded in modulo 3 calculations, reinforcing the conclusion about $a$ and $b$ being positive integers less than 3.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the completeness of the cases examined or the implications of the derived expressions. There are multiple approaches and interpretations presented without resolution.

Contextual Notes

Some assumptions about the properties of congruence classes and the implications of modular arithmetic are not fully explored, leaving room for further clarification and exploration.

Joe20
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Hi, I have tried the question as attached. I am not sure if I am correct. You help is greatly appreciated. Thanks in advance!
 

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Hi Alexis87,

I like the idea to examine each of the cases. A few were left out - e.g., $[a]==[2]$ - and I don't know if that was intentional on your part or not. It may be a little neater to note that the result of squaring in $\mathbb{Z}_{3}$ can only be $[0]$ or $[1]$, in which case there are only three possibilities to check. Your counterexample looks good.
 
I would consider it simpler to write the first number as [a]= 3n+ a and the second as = 3m+ b where a and b are one of 0, 1, or 2. Then $[a]^2+ ^2= (3n+a)^2+ (3m+ b)^2= 9n^2+ 6an+ a^2+ 9m^2+ 6bm+ b^2= 3(3n^3+3m^3+ 2an+ 2bm)+ a^2+ b^2= 0$.

So we must have a and b less than 3 and $a^2+ b^2= 0$. From that, a= 0, b= 0 so that [a]= = 0.
 
HallsofIvy said:
I would consider it simpler to write the first number as [a]= 3n+ a and the second as = 3m+ b where a and b are one of 0, 1, or 2. Then $[a]^2+ ^2= (3n+a)^2+ (3m+ b)^2= 9n^2+ 6an+ a^2+ 9m^2+ 6bm+ b^2= 3(3n^3+3m^3+ 2an+ 2bm)+ a^2+ b^2= 0$.

So we must have a and b less than 3 and $a^2+ b^2= 0$. From that, a= 0, b= 0 so that [a]= = 0.


Hi HallsofIvy,

may I ask what's the purpose of showing this expression:
=3(3n3+3m3+2an+2bm)+a2+b2=0 ?

Is there any thing we can interpret from this equation?

So for "So we must have a and b less than 3" does it mean that because it is Z subscript 3?
 
The purpose is to realize that $3(3n^3+ 3m^3+ 2an+ 2bn)$ is a multiple of 3 and so can be ignored "modulo 3". That is why I can say "We must have a and b positive integers less than 3 such that $a^2+ b^2= 0$.
 

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