Proof of equivalence relation

[UOAMCBURGER]

<Moderator's note: Moved from a technical forum and thus no template.>

Not sure this should be under Linear and Abstract Algebra, but regardless I need help with a question in my mathematical proofs course.
Here it is:
Let ∼ be a relation defined on Z by x ∼ y if and only if 5 | (2x + 3y).
(a) Show that ∼ is an equivalence relation on Z

So I know for ~ to be an equivalence relation the relation needs to have the following properties;
1. Reflexive
2. Symmetric
3. Transitive

My attempt at the problems is this:
Proving the relation is reflexive seems easy enough, since if xRx (x~x) then 5 | (2x + 3x) = 5 | 5x where x is an element of Z, therefore we can clearly see the relation is reflexive since 5 does divide a multiple of 5.

Proving symmetric: Assume that xRy (x~y), that is 5 | (2x+3y) which is equivalent to 5a = 2x+3y right, where a is an element of Z. Now show that yRx (y~x), that is 5 | (2y+3x)... but from here can i say this is also equivalent to 5a = 2y+3x or will i have to use another variable instead of a since 2x+3y does not necessarily equal 2y+3x?
This is where I am up to.

 

[fresh_42]

<Moderator's note: Moved from a technical forum and thus no template.>

Not sure this should be under Linear and Abstract Algebra, but regardless I need help with a question in my mathematical proofs course.
Here it is:
Let ∼ be a relation defined on Z by x ∼ y if and only if 5 | (2x + 3y).
(a) Show that ∼ is an equivalence relation on Z

So I know for ~ to be an equivalence relation the relation needs to have the following properties;
1. Reflexive
2. Symmetric
3. Transitive

My attempt at the problems is this:
Proving the relation is reflexive seems easy enough, since if xRx (x~x) then 5 | (2x + 3x) = 5 | 5x where x is an element of Z, therefore we can clearly see the relation is reflexive since 5 does divide a multiple of 5.

Proving symmetric: Assume that xRy (x~y), that is 5 | (2x+3y) which is equivalent to 5a = 2x+3y right, where a is an element of Z. Now show that yRx (y~x), that is 5 | (2y+3x)... but from here can i say this is also equivalent to 5a = 2y+3x or will i have to use another variable instead of a since 2x+3y does not necessarily equal 2y+3x?
This is where I am up to.

If $5a=2x+3y\,,$ what is $5x - 2x + 5y - 3y\,$ resp. what can you say about the result's divisibilty?

 

[UOAMCBURGER]

If $5a=2x+3y\,,$ what is $5x - 2x + 5y - 3y\,$ resp. what can you say about the result's divisibilty?

I don't understand where 5x−2x+5y−3y came from?

 

[UOAMCBURGER]

I don't understand where 5x−2x+5y−3y came from?

oh right that is what we are tryign to prove right? that 5c = 5x−2x+5y−3y = 3x+2y for some c in Z ?

 

[UOAMCBURGER]

oh right that is what we are tryign to prove right? that 5c = 5x−2x+5y−3y = 3x+2y for some c in Z ?

what does resp mean?

 

[fresh_42]

I don't understand where 5x−2x+5y−3y came from?

Just for fun. Calculate it: $5x−2x+5y−3y= 3x+2y=5(x+y)-5a=5(x+y-a)$ and read it from right to left: $5\,|\,5(x+y-a)=...$

Edit: The trick is to convert $2x$ into $3x$. Since we don't have to bother multiples of $5$ nor signs, adding or subtracting $5x$ does no harm to the result.

 

[fresh_42]

what does resp mean?

respectively

 

[UOAMCBURGER]

Just for fun. Calculate it: $5x−2x+5y−3y= 3x+2y=5(x+y)-5a=5(x+y-a)$ and read it from right to left: $5\,|\,5(x+y-a)=...$

Oh right, 3x+2y=5(x+y)-5a since 5a = 2x+3y. Then it is clear that 5(x+y-a) is divisible by 5 ie 5 | 5(x+y-a).
Hence symmetry is proven.