Equivalence Relation On A Set Of Ordered-Pairs

Question

The question is, "Let R be the relation on the set of ordered pairs of positive integers such that $((a, b), (c, d)) ∈ R$ and only if $a+d=b+c$. Show that R is an equivalence relation."

There are two ways to prove this, but I only understand the second one.

The first way to proof: "By algebra, the given conditions is the same as the condition that $f((a,b))=f((c,d))$, where $f((x,y))=x-y$. Therefore, this is an equivalence relation."

I am not remotely sure of what they are doing...

Answer 1

You are given

$$a + d = b + c$$

as being equivalent to $(a, b) R (c, d)$ (i.e. $((a, b), (c, d))\in R$).

Then, you can do

$$a + d - b = c$$ (subtract b from both sides) $$a - b = c - d$$ (subtract d from both sides)

So then define $f((x, y)) = x - y$, and now you have the given statement. Now, it's easy to see the given relation is an equivalence relation, being that it now follows directly from the properties of equality.

Answer 2

They’re using the result of this problem. First they define the function

$$f:\Bbb Z^+\times\Bbb Z^+\to\Bbb Z:\langle m,n\rangle\mapsto m-n\;.$$

From the earlier problem you know that the relation

$$\langle m,n\rangle\sim\langle k,\ell\rangle\quad\text{iff}\quad f(\langle m,n\rangle)=f(\langle k,\ell\rangle)$$

is an equivalence relation on $\Bbb Z^+\times\Bbb Z^+$. Then they observe that

$$\begin{align*} \big\langle\langle m,n\rangle,\langle k,\ell\rangle\big\rangle\in R\quad&\text{iff}\quad a+d=b+c\\ &\text{iff}\quad a-b=c-d&&\text{by ordinary algebra}\\ &\text{iff}\quad f(\langle m,n\rangle)=f(\langle k,\ell\rangle)\\ &\text{iff}\quad\langle m,n\rangle\sim\langle k,\ell\rangle\;. \end{align*}$$

In other words, the relation $R$ is exactly the same as the relation $\sim$ defined using the function $f$, and since $\sim$ is an equivalence relation, so is $R$.

Answer 3

They are cheating. R is in fact a way of extending the natural numbers (or positive integers) to all integers, as shown in Wikipedia.

But even if you know what subtraction means for natural numbers or positive integers, the statement $f((x,y))=x-y$ is not meaningful for natural numbers or positive integers when $x \lt y$ and so should not be used in a proof.