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Lets assume that for every $x,y,z$ that belong to an $A$ , $(x+2y)$ and $(y+2z)$ can be divided by $3$.If we want to prove that $(x+2z)$ can also be divided by $3$, is it ok to do the next steps ?

$(x+2y),(y+2z)$ can be divided by $3$, so lets take the sum of them:$(x+2y)+(y+2z)$ = $3y+(x+2z)$ And here we come into conclusion that the sum of them is obviously divided by $3$, $ 3y$ can be divided by $3$ obviously, can we say the same about $(x+2z)$ (that can be divided by $3$ according to the above sum)


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Yes, this is fine. – Brian M. Scott Jan 11 '13 at 15:10
Fine and ok. Now try to re-write your question using LaTeX. Look at the FAQ section for directions. – DonAntonio Jan 11 '13 at 15:12
Ah, too bad Amzoti already did that in your place...I've erased Amzoti's editting so that you can learn and do it yourself. BTW, +1 for showing your work. – DonAntonio Jan 11 '13 at 15:17
yea sorry about that ^_^ ,i will be using latex in my next questions, just had a discrete math exam and i was worried about failing :p , but now its ok – p0ffer Jan 11 '13 at 15:20
the original exercise was to show that $(x+2y)$ is an equivalence relation :p – p0ffer Jan 11 '13 at 15:24

To be more explicit: If $x+2y=3a$ and $y+2z=3b$ then $x+2z=3(a+b-y)$ (by the equation you have shown and assuming "an $A$" means something like "an abelian group $A$").

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