Discrete Maths: find equivalence classes of a relation with 2 variables

[Mampac]

Hi there,

I have an equivalence relation R defined on the set of integers Z such that

aRb if and only if 7|(a + 6b)​

OK, now I have a really hard time finding what are the equivalence classes of this relation...

By the definition of an equivalence class, an equivalence class [x] means every integer relates to x such that (aRx).

At first, i tried to play with some integers but soon got that it didn't bring me to anywhere. for example, I considered existence of [0], means for any a statement 7|(a + 0) is true. But then I noticed I can find an a for which it's not true (e.g. when a = 1, 7 doesn't divide 1). I went like this through a lot of integers and realized it doesn't make sense at all: whatever x I select, I can find an a such that the statement is not true. Thus, I decided to use some variables instead:

If I replace b with x I get 7|(a + 6x). Now, as far I understood, I need to solve for x.
Well, I can say a + 6x = 7k for some integer k.
x is going to be equal to (7k - a)/6 for some integer k.
I wonder if this makes any sense? This already includes 2 variables and I'm confident this is incorrect...

I can clearly point out what are the equivalence relations if you give me a completely defined relation with all the pairs. But nothing comes to my head in this case... It seems like the answer is either gonna be "there are no equivalence classes" or "there's Z equivalence classes."

Could someone break this down for me?

 

[pka]

I have an equivalence relation R defined on the set of integers Z such that
aRb if and only if 7|(a + 6b) Could someone break this down for me?

Consider \([1]=\{\cdots -6,1,8,15,\cdots\}\) that is the equivalence of one.
\([2]=\{\cdots -5,2,9,16,\cdots\}\) that is the equivalence of two.
What are \([3],~[4],~[5],~[6],~\&~[7],~?\)

 

[Mampac]

Consider \([1]=\{\cdots -6,1,8,15,\cdots\}\) that is the equivalence of one.
\([2]=\{\cdots -5,2,9,16,\cdots\}\) that is the equivalence of two.
What are \([3],~[4],~[5],~[6],~\&~[7],~?\)

OMG my whole way of thinking was so flawed from the very beginning... now i understand equivalence class is a part of the relation
thank youuuu
so
\([1]=\{\cdots -6,1,8,15,\cdots\}\)
\([2]=\{\cdots -5,2,9,16,\cdots\}\)
\([3]=\{\cdots -4,3,10,17,\cdots\}\)
\([4]=\{\cdots -3,4,11,18,\cdots\}\)
\([5]=\{\cdots -2,5,12,19,\cdots\}\)
\([6]=\{\cdots -1,6,13,20,\cdots\}\)
\([7]=\{\cdots 0,7,14,21,\cdots\}\)
and the union of all of these is Z. so neat

also, as far as I noticed, when I'm given Z as the set of definition I should start guessing the equivalence class of [1] and then proceed to the others.
and if it's not Z but a limited set then I should put the elements instead of the integers and try to get something

 

[Mampac]

also i noticed there's Z equivalence classes and they're denoted by [7t + 1], [7t + 2], [7t + 3], [7t + 4], [7t + 5], [7t + 6], [7t + 7] for some integer t because [1] = [8] = [15] and so on! so that's my precise answer