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Let $A=\{1,2,3,4,...,9\}$ and $R$ be a relation in $A \times A$ defined by $(a,b)R(c,d)$. If $a+d=b+c$ for $(a,b),(c,d)$ in $A \times A$. Prove that $R$ is an equivalence relation and also obtain the equivalent class $(2,5)$.

I have proved it to be equivalence relation but how to find the obtain the equivalent class $(2,5)$. Could someone help me with this?

user383014
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    Reworded, $(a,b)R(c,d)$ iff $a-b=c-d$ iff $a$ is the same distance above $b$ as $c$ is above $d$. $2$ is three under $5$. What other pairs are there with the first number three less than the second number? That is all that is being asked when being asked "find the equivalence class of $(2,5)$" it is just asking "what other pairs have the first number the same distance below the second number as $(2,5)$" in this specific problem. – JMoravitz Mar 18 '17 at 04:21
  • @JMoravitz Thanks for such a detailed explanation. – user383014 Mar 18 '17 at 04:42

1 Answers1

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You know the theorem that if $A$ is a set and $\sim$ is an equivalence relation on $A$, then $A$ can be split into disjoint classes under this equivalence.

The key to this theorem is the following observation:

Two elements in the same equivalence class are related under $\sim$, while elements of different classes are not related under $\sim$.

That is, given an element, it's equivalence class is precisely the set of all elements it is related to.

Having said that, you just need to figure out all pairs of numbers $(a,b) \subset A \times A$ such that $a - b = 2-5$, with the help of the comment given above. I am sure you can do it from here.

Sarvesh Ravichandran Iyer
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