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On $\mathbb{N}\times\mathbb{N}$, Let $\left( a,b\right) \equiv \left( c,d\right) \Leftrightarrow a+d=c+b$.

Show that it is equivalence relation. Find equivalence of class of this.

My answer.

For all $a,b\in\mathbb{N}$, we have $a+b=a+b$. Clear. So, The relation is reflexivity.

For all $a,b,c,d\in\mathbb{N}$, we assume $a+b=c+d$. Hence, also we know $c+d=a+b$. Thus, the relation is symetric.

For all $a,b,c,d,e,f\in\mathbb{N}$, we assume $a+d=c+b$ and $c+f=e+d$. We will show that $a+f=e+b$. Sum of both of sides $a+b=c+d$ and $c+f=e+d$, then we obtain $(a+f)+(c+d)=(c+d)+(e+b)$. Hence, by the cancelation, we obtain $(a+f)=(e+b)$. Thus, The relation $R$ is transitivity.

Therefore, the relation $R$ is equivalence relation. Can you check my answer,so, what is the equivalence classes of this?

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    Your proof that is a equivalence relation is correct – 2017-02-02
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    In your argument for symmetry you're writing $a+b=c+d$ instead of $a+d=c+b$. – 2017-02-02
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    So think about a graph of this on $\mathbb{N} \times \mathbb{N}$ to determine the equivalence class – 2017-02-02
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    I just would prefer to remember that what you are doing it a relation in the $\mathbb{N}^2$, so, e.g, after your ''hence'' put like $(a,b)$ relates to itself in $\mathbb{N}^2$ – 2017-02-02
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    @HenningMakholm You are right. – 2017-02-02
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    **Hint-ish trivia:** This equivalence relation was considered by Hamilton as part of a construction of negative numbers. – 2017-02-02
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    You're not being asked to find "the equivalence class" of your relation, but to find "the equivalence **classes**". There are many of them. – 2017-02-02

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Note that $a+d=c+b$ is true iff $a-b=c-d$. Let us interpret a pair such as $(a,b)$ as: $a$ is the amount of income this month, $b$ as the expenditure this month for a single person. We say two persons are equivalent if the money they saved this month (that is income minus the expenses) is the same for these two persons.

Now have fun solving your problem without using plus and minus signs or brackets.