Bijective map on $(\Bbb N \times\Bbb N)/R$

Question

I'm not sure how to tackle this problem.

Consider the equivalence relation $R$ on $\Bbb N \times\Bbb N$ given by : $$(a, b)R(c, d) \iff a + d = b + c$$

(i) Show that $R$ is an equivalence relation.

(ii) Construct a natural bijective map $f : (\Bbb N \times\Bbb N)/R\to\Bbb Z$

Now I know how to prove that $R$ is an equivalence relation. What I'm having trouble in is figuring out what $(\Bbb N\times\Bbb N)/R$ means. I understand that in the space $\Bbb N\times\Bbb N$ there are pairs of elements such as $(a,b)$ and $(c,d)$ but I can't seem to grasp how the space $(\Bbb N\times\Bbb N)/R$ will look like. ($R$ here is the equivalence relation not the set of real numbers).

Answer 1

An equivalence relation tells you when two things are "the same", in some sense. So to answer this question you need to figure out what is "the same" about two pairs $(a,b)$ and $(c,d)$ if they are related. If you think about it and use your imagination, (please do before reading on)...

$(a,b)\mathbin{R}(c,d)$ if and only if $a-b=c-d$, in other words the differences of the elements (first minus second) in the ordered pairs are the same.

So what this relation "really means" is that it is

a way of defining (signed) integers as pairs of natural numbers.

For example,

$(3,7)\mathbin{R}(1,5)$ because $3+5=7+1$, or, informally, because both represent the integer $-4$.

Your bijection $f:{\Bbb N}\times{\Bbb N}/R\to{\Bbb Z}$ could be given by

$f([(a,b)])=a-b$.

Answer 2

$Z'=(\Bbb N \times \Bbb N)/R$ is the standard construction of $\Bbb Z$ from the natural numbers: note that you can rewrite your relation as $$(a,b)R(c,d)\iff a-b=c-d$$

(This is not really allowed, as we're trying to define what we mean by $-$(substraction) with this construction)

That is, you're representing each number integer as a substraction of two natural numbers: associating $0$ to $[(a,a)]=[(4,4)]$ etc.

The map is just $f:Z'\to \Bbb Z, [(a,b)]\mapsto a-b$ .

Answer 3

$(\Bbb N \times \Bbb N)/R$ is the set of equivalence classes under the relation $R$. Under this particular relation $R$, each equivalence class consists of all pairs of positive integers which have the same difference (since $(a,b)R(c,d) \Leftrightarrow a+c=b+d \Leftrightarrow a-b=c-d$. If you define a map $f$ from $\Bbb N \times \Bbb N$ to $\Bbb Z$ by mapping each pair $(a,b)$ to $a-b$, this map is surjective, since every integer $z$ can be written as the difference of two positive integers (if $z>0$ then $z=(z+1)-1$; if $z=0$ then $z=1-1$; if $z<0$ then $z=1-(1-z)$). Moreover two pairs $(a,b)$ and $(c,d)$ will map to the same integer $z \Leftrightarrow a-b=z=c-d \Leftrightarrow (a,b)R(c,d)$. (In other words, the kernel of $f$, i.e. the set of pairs mapping to $0$, is the set of pairs of the form $(a,a)$.) Thus $f$ induces an bijective map $\overline f$ from the set of equivalence classes under the relation $R$ to $\Bbb Z$, where $\overline f$ maps each equivalence class to the integer corresponding to the difference of any pair belonging to that class (all pairs in the class have the same difference).