Equivalence classes when x &#x2212;<!-- āˆ’ --> y = multiple of 3? š‘„ R š‘¦ exactly whe

Richardtb

Richardtb

Answered question

2022-05-18

Equivalence classes when x āˆ’ y = multiple of 3?
š‘„ R š‘¦ exactly when š‘„ āˆ’ š‘¦ equals a multiple of 3.
There exists equivalence classes [0], [1] and [2]. Prove that [ 1 ] + [ 2 ] is equal to [0].
Not sure about the ambigious part, all I could find from my research was that if it's intepreted different ways the meaning will still be the same e.g. [ 1 ] + [ 2 ] = [ 0 ] is the same as [ 2 ] + [ 1 ] = [ 0 ] or [ 0 ] = [ 2 ] + [ 1 ]. But I am unsure about that.
How do I create these equivalence classes and prove that [ 1 ] + [ 2 ] is equal to class [0] ?

Answer & Explanation

redclick53

redclick53

Beginner2022-05-19Added 12 answers

Step 1
Here, [ 1 ] = { ā€¦ , āˆ’ 5 , āˆ’ 2 , 1 , 4 , 7 , ā€¦ } = { 3 n + 1 āˆ£ n āˆˆ Z } and [ 2 ] = { ā€¦ , āˆ’ 4 , āˆ’ 1 , 2 , 5 , 8 , ā€¦ } = { 3 n + 2 āˆ£ n āˆˆ Z } .
Step 2
And when you add any element of [1] to any element of [2], you get an element of [ 0 ] = { ā€¦ , āˆ’ 6 , āˆ’ 3 , 0 , 3 , 6 , ā€¦ } = { 3 n āˆ£ n āˆˆ Z } .
So, if m āˆˆ [ 1 ] and n āˆˆ [ 2 ], the class of m + n is independent of the choice of m and n, and therefore [ 1 ] + [ 2 ] is not ambiguous (and it turns out that it is equal to [0]).

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