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Consider the set $S = \{-5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5\}$

And define a relation $R$ on $S$ by $(x, y) \in R$ if and only if $x + > 2y \cong 0 (mod 3)$

Show that the relation $R$ is an equivalence relation and list the equivalence class containing the element 1.

I know how to prove that the set is symmetric and reflexive but I'm stuck at transitive.

There is a solution but I don't understand it at all:

The relation is transitive, for if $(x, y)\in R$ and $(y, z)\in R$ then $x+2y \cong 0$(mod 3) and $y+2z \cong $0(mod 3)

so $x+2z = x+2y - 2y + 2z \cong x+2y + y+2z \cong (x+2y)+(y+2z) \cong > 0 + 0 \cong 0$(mod 3) and thus $(x, z) \in R$

This is the part I understand the least:

$x+2y - 2y + 2z \cong x+2y + y+2z \cong (x+2y)+(y+2z)$

1 Answers 1

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In this step 2y is added. And if we added 2y then also we have to subtract 2y.

So we have,

x+2z = x+2y-2y+2z

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    Also, don't forget that $-2y\equiv y$, since it's modulo $3$. – 2017-02-25
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    Thank you Arthur! That thing that really made me confused. I need to repeat some congruence exercises – 2017-02-25