Equivalence class of 0 for the relation a ~ b iff 2a+3b is divisible by 5

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SUMMARY

The equivalence class of 0 for the relation defined by a ~ b if and only if 2a + 3b is divisible by 5 is represented as [0] = {5n | n is an integer}. The reasoning provided confirms that both 0 and multiples of 5 satisfy the relation, as 2(0) + 3(5n) results in a multiple of 5. However, the discussion highlights that the relation does not qualify as an equivalence relation since it fails to meet all necessary criteria, particularly the requirement for symmetry. Therefore, the concept of equivalence classes is not applicable in this context.

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jeszo
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Homework Statement



~ is a equivalence relation on integers defined as:
a~b if and only if 2a+3b is divisible by 5

What is the equivalence class of 0

Homework Equations





The Attempt at a Solution



[0] = {0, 5n} n is an integer

My reasoning for choosing 0 is that if a=0 and b=0, the relation is satisfied since 2(0)+3(0) = 0 and 5 divides 0, so 0~0
My reasoning for choosing 5n is that if a=0 and b=5n, or the other way around, then 2(0)+3(5n)=3(5n), which is a multiple of 5, and thus divisible by 5, satisfying the relation, making 0~5n

I got 0 marks for the question, but I don't know if it's because the statement isn't assembled properly, or if it's because I don't know what an equivalence class is. Please help me see my error

 
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jeszo said:

Homework Statement



~ is a equivalence relation on integers defined as:
a~b if and only if 2a+3b is divisible by 5

What is the equivalence class of 0

Homework Equations





The Attempt at a Solution



[0] = {0, 5n} n is an integer

My reasoning for choosing 0 is that if a=0 and b=0, the relation is satisfied since 2(0)+3(0) = 0 and 5 divides 0, so 0~0
My reasoning for choosing 5n is that if a=0 and b=5n, or the other way around, then 2(0)+3(5n)=3(5n), which is a multiple of 5, and thus divisible by 5, satisfying the relation, making 0~5n

I got 0 marks for the question, but I don't know if it's because the statement isn't assembled properly, or if it's because I don't know what an equivalence class is. Please help me see my error

In order to have an equivalence relation, we need both a ~ b and b ~ a, so we need both 2a + 3b and 2b + 3a to be divisible by 5.

RGV
 
In response to Ray Vickson:
With 0 and 5n as the equivalence class for 0, wouldn't it still hold true that 0~0,5n~0 and 0~5n? Since, 2(5n)+3(0)=5(2n) and 2(0)+3(5n)=5(3n)?
 
It's redundant to include the zero in {0, 5n}, since if n=0, then 5n=0.

A better notation would be, [0] = {5n| n is an integer.}

Regarding Ray Vickson's comment, I agree with you.

The equivalence class of 0, is the set of all integers related to 0. I.e. it's the set of all integers, m, such the m~0 .
 
What, exactly, was the question? I suspect it was to determine whether or not this was an equivalence relation and, if so find the equivalence class containing 0.

As Ray Vickson said, this is NOT an equvalence relation and so does NOT have "equivalence classes".
 

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