For relation defined on $\mathbb Z$ by $aRb$ if $6 \mid (a^2 - b^2)$
I know that the equivalence classes are {0,1,2,3,4,5} but I'm unsure if my method of proof is valid. For the first part of the question, I proved it was an equivalence relation on $\mathbb{Z}$, and now I need to find and prove the equivalence classes.
My thoughts so far:
- Let there be $a,r \in \mathbb{Z}$, so that $6 \mid (a^2-r^2)$.
- I know by Euclidean Division that $6 \mid (a^2-r^2) \equiv a^2=6q+r^2$ , $q \in \mathbb{Z}$
- Then I also know that $0 \leq r^2 <6$
- If $aRr$, the I know $[a]=[r]$ and also $a\in[r]$***
- Then by transitivity, since I proved (in my homework) $bRa \land aRr \Longrightarrow bRr$ and therefore, $[b]=[a]=[r]$ and $a,b \in [r]$
***Can I state this? Or am I missing something key concept before I can say this?
Any feedback on my train of thoughts is appreciate. Pls let me know if you see any holes of logic that I need to fill.
you can use $\pmod 2$ and $\pmod 3$ to describe things, but the result does not fit that well with $\pmod 6$
We always have $x^2 \equiv x \pmod 2,$ so $a^2 \equiv b^2 \pmod 2$ if and only if $a \equiv b \pmod 2.$ For this calculation, we are left with classses even and odd.
Next, squares are $0 \pmod 3$ or $1 \pmod 3.$ As with 2, what we can tell is whether a number is divisible by $3$ or not divisible by $3$.
The equivalence class could be summarized as YY, YN, NY, NN, where the first yes/no is divisbility by $2,$ the second letter is divisibility by $3$
two numbers are equivalent if and only if they have the same pair of letters