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The relation $R$ is defined for all positive integers such that $(a,b) R (c,d) \longleftrightarrow a+d=b+c$. Show that $R$ is an equivalence relation.

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    Which are three conditions to satisfy to get an equivalence relation? – 2017-02-03
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    [Also](http://math.stackexchange.com/questions/1106295/equivalence-relation-on-mathbbn-times-mathbbn). – 2017-02-03

2 Answers 2

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For R to be an equivalence relationship it would have to suffice all the following conditions :

1) it has to be a reflexive relationship ! ( (a,b) R (a,b) )

2) it has to be simetric ( (a,b) R (c,d) only if (c,d) R (a,b) )

3) it has to be transitive ( if ( a,b) R (c,d) and (c,d) R (e,f) then (a,b) R (e,f) ) ;

Hope this was helpful!

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    If you need a step by step demonstration i can help you with it! – 2017-02-03
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    That would be really nice :) – 2017-02-03
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    Firstly we have to prove that every pair (a,b) is in relationship with itself. You defined R as :two pairs : (x,y) (z,t) are in the relationship R if (x + t) =(y + z) 1)Applying the definition of this relationship on the pairs (a,b) , (a,b) we get that (a+b) = (b + a) which is true. As suchs (a,b) is in the R relationship with (a,b) => R is simetric. 2)We presume the pairs (a,b) are in relationship with (c,d) ((a + c) = (b + d) We have to show that (c,d) is in a Relationship with (a,b); => (c +a ) = (d + b) .Because + is commutative ( a + c = c + a ) the equality is true. – 2017-02-03
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    3) We presume (a,b) is in a relationship with (c,d) and (c,d) is in a relationship with (e,f); from the first relationship (a+c) = (b+d) ; from the second relationship (c+e) = (d+f); now you have to show that ( a + e ) = ( b + f). – 2017-02-03
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    Thank you, I got it.This was so helpful :D – 2017-02-04
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    The first step is showing the reflexivity, right? not simetry. The second step is for simetry. – 2017-02-04
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    Yes,that s right! My bad :( – 2017-02-04
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Note that $\left( a,b\right) R \left( c,d\right)$ if and only if $a-b = c-d$.

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    How far does this go toward showing $R$ is an equivalence relation? You've rewritten the definition, but made no mention of reflexivity, symmetry, or transitivity. Please review [How do I write a good answer?](http://math.stackexchange.com/help/how-to-answer) – 2017-02-03
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    From this all the necessary properties are evident, because instead of calculating anything, one just considers the property "difference of coordinates" which is evidently equivalence, i.e. from a two-dimensional problem it shifts to a one-dimensional. – 2017-02-03