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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.

user3602479
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2 Answers2

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Note that $\left( a,b\right) R \left( c,d\right)$ if and only if $a-b = c-d$.

pepa.dvorak
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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? – hardmath Feb 03 '17 at 14:04
  • 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. – pepa.dvorak Feb 03 '17 at 14:58
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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!

Eduard6421
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  • If you need a step by step demonstration i can help you with it! – Eduard6421 Feb 03 '17 at 13:22
  • That would be really nice :) – user3602479 Feb 03 '17 at 13:28
  • 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.

     – Eduard6421 Feb 03 '17 at 13:33
  • 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).

    – Eduard6421 Feb 03 '17 at 13:36
  • Thank you, I got it.This was so helpful :D – user3602479 Feb 04 '17 at 12:22
  • The first step is showing the reflexivity, right? not simetry. The second step is for simetry. – user3602479 Feb 04 '17 at 12:27
  • Yes,that s right! My bad :( – Eduard6421 Feb 04 '17 at 14:01