Thread: Show that this is an equivalence relation

1. Yes, this shows the relation is transitive by your definition.

2. Thanks mate!

Any idea on the following:Function is defined as followed:
⊕:
by

Show that,
~ and ~
then
~

3. Suppose (n,m)~(n,m). Then, by definition, n+m=m+n ⇒ 0=0. This clearly true; thus, ~ is refexive.

Suppose (n,m)~(k,l), where (n,m) and (k,l) are arbitrary elements of N^2. Then n+l=m+k ⇔ m+k=n+l. Therefore, (k,l)~(n,m). Consequently, ~ is symmetric.

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