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An equivalence relation $R$ is defined on $\mathbb Z^2$ by $(a,b) R (c,d)$ iff $2a − b=2c-d$. Determine the distinct equivalence classes.

Hello all, I can show the distinct equivalence classes $$\begin{align} [0,0]&=\{(a,b) \in \mathbb Ζ \times \mathbb Ζ \colon (0,0)R(a,b)\}= \{(a,2a),a \in \mathbb Ζ\}\\ [0,k]&=\{(a,b)\in \mathbb Ζ \times \mathbb Ζ \colon (0,k)R(a,b)\} = \{(a,2a+k) \colon a \in \mathbb Ζ\}\\ [k,0]&=\{(a,b) \in \mathbb Ζ \times \mathbb Ζ \colon (k,0)R(a,b)\} = \{(a,2(a-k)) \colon a \in Ζ\} \end{align}$$ now I think that I must show now that random $(x,y) \in \mathbb Z \times \mathbb Z$ with i) $x>y$ and ii) $y>x$ belongs to $[0,k]$ and $[k,0]$ classes I've shown before but I don't know how to do it ...

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    There is something that doesn't sound right: it seems like you are saying that $(a,b)$ and $(c,d)$ are in the relation $R$ if $2a - b= 2c-d$ and $2a-b$ is even. If that's the case then $R$ is not an equivalence relation, hence it doesn't make sense to talk about equivalence classes: to see that is not an equivalence relation consider the pair $(0,3)$, $(0,3)R(0,3)$ doesn't hold, because $2*0-3=3$ which is not even. – 2017-02-18
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    (0,3)~(0,3) iff 2*0-3=2*0-3 <=> -3=-3...it is equivelance. – 2017-02-18
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    Yeah but you said *relation $R$ is defined on $\mathbb Z\times\mathbb Z$ by $(a,b)R(c,d)$ if $2∗a−b=2∗c−d2∗a−b=2∗c−d$ is even*: which I read as you have an equality that has to hold and you need also that $2*c-d$ is even. Did I misunderstood you? What does that "is even" mean? – 2017-02-18
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    is even.. means that 2*a-b=2*c-d must be even..i ll rewrite..(a,b)~(c,d)<=>2*a-b=2c-d. thats all. – 2017-02-18

2 Answers 2

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$\newcommand{\Set}[1]{\left\{ #1 \right\}}$$\newcommand{\Z}{\mathbb{Z}}$Given two non-empty sets $X, Z$, a function $f : X \to Z$ induces the equivalence relation $$\text{$x R y$ iff $f(x) = f(y)$.}$$ If moreover $f$ is surjective, then the equivalence classes are given by $$ f^{-1} (z) = \Set{ x \in X : f(x) = z } $$ as $z$ ranges in $Z$.

In your case $X = \Z \times \Z$, $Z = \Z$, and $f(a, b) = 2 a - b$. Note that $f$ is clearly surjective, as $f(0, -z) = z$ for $z \in \Z$.

It follows that the equivalence classes are of the form $$ C(z) = \Set{ (a, b) : a, b \in \Z, f(a, b) = 2 a - b = z } $$ as $z$ ranges in $\Z$. In other words, rewriting $b = 2 a - z$, $$ C(z) = \Set { (a, 2 a - z) : a \in \Z }. $$

Perhaps one could add, if one is already familiar with these topics, that $f$ is a group morphism, so that by the first isomorphism theorem the classes are really the cosets with respect to the kernel $\ker(f) = \Set{ (a, 2 a) : a \in \Z }$.

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    so are all classes are C(z)={(a,2a−z):a∈Z} i am ok with that ..that means that for every z ε Ζ i get one class? how many classes we have? – 2017-02-18
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    Exactly, there are as many classes as there are integers. Each integer $z$ uniquely determines a class $C(z)$, and all classes are of this form. – 2017-02-18
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    and for every a ε Ζ gives me a diferent representative of the class? – 2017-02-18
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    Yes, representatives are for instance $(0, -z)$. Easy to see they are pairwise not in relation. – 2017-02-18
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    how could i use the same way the f^-1(z) which give me the classes with the following? :R is defined on N^2 , (x,y)R(z,w) <=> x+w=y+z – 2017-02-18
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    @kotsossgouras, that's the relation that it's used to construct the integers starting from the natural numbers. So the function you want is $\newcommand{\N}{\mathbb{N}}$ $\newcommand{\Z}{\mathbb{Z}}$$g : \N \times \N \to \Z$ such that $g(a, b) = a - b$. – 2017-02-18
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    Representatives are trickier here, because there is $(0, 0)$, and then $(n, 0)$ and $(0, n)$, for $n > 0$. – 2017-02-18
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    1/2 ty for your help i really appreciate it. so the classes here are C(z)={(a,a-z) ,a∊Ν,z∊Z}, right? how will I show that there are only three? i had another way to find classes, with this way i tried to find the classes for the first problem i posted ..but i liked more the way you show me, i just try to make the connection..i ll write the 3 classes that the second R has with a way i knew: 1) (0,0)R(x,y)⇔0+y=0+x⇔y=x ,so [(o,o)]_R={(x,x):x∊N} 2) (n,0)R(x,y)⇔ n+y=x+0⇔ x=n+y,so [(k,0)]_R = {(n+y ,y): y,n ∊N} 3) (0,n) R (x,y)⇔0+y=n+x⇔y=n+x,so [(0,n)]_R = {(x,n+x):y,n∊ N} – 2017-02-19
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    2/2 3) (0,n) R (x,y)⇔0+y=n+x⇔y=n+x,so [(0,n)]_R = {(x,n+x):y,n∊ N} now for random (l , m) ∊NxN we have i) if l>m then l=m+k ,k ε N => (l ,m)=(m+K,m) so (m+k,m)R(x,y)⇔m+k+y=m+x⇔x=k+y.∊[(n,0)_R] ii) the same way m>l∊[(0,n)]_R – 2017-02-19
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    Thx for the appreciation. I have a hard time reading your maths, if you could try and typeset proper LaTeX, that would help. Anyway, if in $(a, b)$ one has $a = b$, then this is in the class of $(0, 0)$; if $a > b$, then it is in the class of $(a - b, 0)$, if $a < b$, then it is in the class of $(0, b - a)$. – 2017-02-19
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    1/2 ..Mr. caranti.sry for my typing i wrote at word and copy paste i thought it would work but it didn't, and i dont know to write at latex :( ... i have one question...F^-1 = { (a,b) ε N x N : f ((a,b)) = a-b =z } => C(z)= {(a,a-z), z ε Ζ } this gives me the classes right? – 2017-02-19
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    2/2 how we show from C(z) that we have only 3 classes?the ones you mentioned.. and why we have not as many classes as there are integers?how we can show that? ty – 2017-02-19
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    But there are indeed as many classes as the integers, as this is precisely the construction of the integers from the naturals! – 2017-02-19
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    ok we have infinite classes i get it now. before we use f we must show that is well defined> like this? f : Z x Z -> Z , with f ((a,b)) =2a-b.. if (a,b) =(b,c) then (a,b)R(c,d)<=>2a-b=2d-c => f(a,b)=f(b,c) so f well defined. is this correct? – 2017-02-19
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    if we have an f :Z/ R -> Z f ( [(a,b)] )= 2a-b ( i mean that f takes classes instead for (a,b) ε ΖxΖ) in that case we should show that f is 1-1 and onto right? is this the same like f^-1 before? – 2017-02-19
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    Please *please* ***please*** use LaTeX, I cannot follow! Anyway, there is absolutely no problem of well-definition here. – 2017-02-19
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    Mr andrea will you help me pls with an isomorphick i cant show?i cant find proper type for f,i will writhe a new question.. – 2017-02-24
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For start observe that the classes that you have described are not distinct: for instance you have that $[a,0]=[0,-2a]$ for every $a \in \mathbb Z$.

I assume you are looking for a set of representatives for the different equivalence classes. Nevertheless in the case at hand you can notice that for every pair $(a,b) \in \mathbb Z \times \mathbb Z$ there is always a (unique) $d \in \mathbb Z$ such that $$(a,b)R(0,d)$$ it is sufficient to take $$d=-(2a-b)=b-2a\ .$$

That $(a,b)R(0,d)$ follows from the definition of $d$. From this it follows that (using your notation) $[a,b]=[0,d]$ for every pair $a,b \in \mathbb Z$ and $d$ given by the equation above. This proves that each equivalence class is of the form $[0,d]$ for some $d$.

It remains to be proven that the $d$ above is unique, but that's easy: by definition $[0,d]=[0,d']$ if and only if $(0,d)R(0,d')$ which holds only when $$2*0-d=2*0-d'$$ that is only when $d=d'$.

With this we conclude that all the equivalence classes for $R$are of the form $[0,d]$, where the $d$ is uniquely determined by the class.

Hope this helps.

p.s. Observe that what we have proven is that the pairs $(0,d)$ form a set of good representatives for the equivalence classes.