Meaning of {(a, b) | x= ± b} ⊆ Z×Z and {(a, b) | x ≡ b (mod7)} ⊆ Z×Z

Question

What elements would be in the given relations?

If Z = all integers

{(a, b) | a = ± b} ⊆ Z×Z

• for this relation I am thinking that the elements would be

  $$\cdots(-2,-2),(-1,-1),(0,0),(1,1),(2,2), (3,-3), (4,-4)\cdots$$

Just any number $(a,b)$ that is equal to one another and b can be either positive or negative.

{(a, b) | a ≡ b (mod7)} ⊆ Z×Z

• For this relation I and stumped. Any clarification or input would be great.

Answer 1

For $\{(a,b)\mid a\cong b\pmod7\}$ we get $\{(a,7\cdot k+a)\mid a,k\in\mathbb Z\}$.

Answer 2

First, it is helpful to remember just what it means for something to be equivalent to something else modulo some number.

For integers $a,b$ and positive integer $n$, the following are equivalent:

• $a\equiv b\pmod{n}$ (read aloud as $a$ is equivalent to $b$ modulo $n$)

• $a-b\equiv 0\pmod{n}$

• $n\mid (a-b)$ (read aloud as $n$ divides $(a-b)$)

• $(a-b)$ is a multiple of $n$

• $\frac{a-b}{n}\in\Bbb Z$

• $\exists k\in \Bbb Z$ such that $(a-b)=n\cdot k$

• $\vdots$

In terms of writing proofs, the final one that I mentioned is often very powerful in proving properties relating to elementary number theory.

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Your second relation is the more standard so I will begin there. You ask if $\{(a,b)~:~a\equiv b \pmod 7\}$ is an equivalence relation over $\Bbb Z$. It is indeed. To see this, we check the properties of what it means to be an equivalence relation: reflexivity, symmetry, and transitivity.

• Is it reflexive? That is to say, if $x\in\Bbb Z$ does that imply that $x$ is related to itself?

Since $x-x=7\cdot 0$ we have that $x\equiv x\pmod{7}$ is true, therefore the relation is reflexive.

• Is it symmetric? That is to say, if $x,y\in\Bbb Z$ (x,y not necessarily distinct) such that $x$ is related to $y$, does that imply that $y$ is related to $x$?

Suppose $x\equiv y\pmod{7}$. That implies that there is some integer $k$ for which $x-y=7\cdot k$. In that case, $y-x = 7\cdot (-k)$, and noting $-k$ is an integer so long as $k$ is we have that indeed $y\equiv x\pmod{7}$ and so the relation is indeed symmetric.

• Is it transitive? That is to say, if $x,y,z\in\Bbb Z$ (x,y,z not necessarily distinct) such that $x$ is related to $y$ and $y$ is related to $z$, does that imply that $x$ is related to $z$?

Suppose $x\equiv y\pmod{7}$ and $y\equiv z\pmod{7}$. This implies that there is some $k$ and some $\ell$ such that $x-y=7\cdot k$ and $y-z=7\cdot \ell$. Then $x-z = (x-y) + (y-z) = 7\cdot k + 7\cdot \ell = 7\cdot (k+\ell)$ and noting $k+\ell$ is an integer we have as a result $x\equiv z\pmod{7}$

All of these together imply that this really is an equivalence relation. This is such a common equivalence relation that you should keep it in your bag of tools to use for various problems as it will most certainly come up again. As for what the equivalence classes look like, they will be:

$7\Bbb Z=\{\dots,-14,-7,0,7,14,21,\dots\}$

$7\Bbb Z + 1 = \{\dots,-13,-6,1,8,15,22,\dots\}$

$7\Bbb Z + 2 = \{\dots, -12,-5,2,9,16,23,\dots\}$

$\vdots$

$7\Bbb Z + 6 = \{\dots,-8,-1,6,13,20,27,\dots\}$

That is to say, any element in one of those equivalence classes is related to any other element of that equivalence class.

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Your first example that you ask about is also an equivalence relation.

It is reflexive since $x=x$ for all $x$.

It is symmetric since if $x=\pm y$ then that implies $y=\pm x$.

It is transitive since if $x=\pm y$ and $y=\pm z$ then $x=\pm z$.

The equivalence classes will be $\{0\}, \{1,-1\}, \{2,-2\},\dots, \{n,-n\},\dots$