prove that $a \equiv b \mod m$ is an equivalence relation on the integers

Question

prove that $a \equiv b \mod m$ is an equivalence relation on the integers

I believe there are 3 properties that it must meet to prove and equivalence relationship. Any reference material would be greatly appreciated

Answer 1

We simply use

• the definition of congruence modulo $n$: for any two integers $a, b$, $a \equiv b \pmod m \iff a - b = km,$ for some $k \in \mathbb{Z}$.
• the definitions of the three properties any equivalence relation must, by definition, satisfy: reflexivity, symmetry, transitivity.

Reflexivity: for all $a \in \mathbb{Z}, a\sim a.$

• Is there a $k\in\mathbb Z$ such that for integer $a$, $a-a=mk$?

Symmetry: for all $a, b \in \mathbb{Z},$ if $a \sim b $, then $ b\sim a$.

• Is there a $k_2\in\mathbb Z$ such that for integers $a,b$, if $k_1$ is a integer such that $a\equiv b \pmod m$ so that $a-b=mk_1$ for some integer $k_1$, then $b-a=mk_2$, and hence $b \equiv a \pmod m$?

Transitivity: for all $a, b \in \mathbb{Z}$, if $a \sim b,$ and $ b\sim c$, then $a\sim c$

• Can it be shown that: $$a\equiv b\pmod{m}\longrightarrow \exists k_1\in\mathbb Z,~~~ a-b=mk_1$$ and $$b\equiv c \pmod m\longrightarrow \exists k_2\in\mathbb Z,~~~ b-c=mk_2,$$ that there must exist some $k_3 \in\mathbb Z$ such that $a-c=mk_3$?

You need only show that indeed answer the above questions, and show why your answer holds, to prove that equivalence modulo $m$ is an equivalence relation.

Answer 2

$a\equiv b$ mod $m$ iff $m\mid (a-b)$ to see that this is an equivelance relation we need to check its reflexive, symmetric and transitive.

1) reflexive, means $a\equiv a$ this is true since $m\mid (a-a)=0$.

2)symmetric, means if $a\equiv b$ then $b\equiv a$ now if we have $a\equiv b$ then $m\mid (a-b)$ that is $$a-b=mk$$ for some integer k but then $$b-a=m(-k)$$ which implies that $m\mid (b-a)$ i.e. $b\equiv a$.

3)transitive, means if $a\equiv b$ and $b\equiv c$ then $a\equiv c$ assume that $a\equiv b$ and $b\equiv c$ that is $m\mid(a-b)$ and $m\mid (b-c)$ which implies that $$a-b=mk_1$$ and $$b-c=mk_2$$ now by adding these two equations we get$$a-c=m(k_1+k_2)$$ that is $m\mid(a-c)$ i.e $a\equiv c$ as required.

Answer 3

According to the definition, $a\equiv b~~~(\text{mod}~m)$ iff $\exists k\in\mathbb Z,~~~ a-b=mk$ wherein $a$ and $b$ are integers. Now check that:

• Does exist $k\in\mathbb Z$ such that for integer $a$, $a-a=mk$?

• Does exist $k'\in\mathbb Z$ such that for integers $a,b$, if $a-b=mk$ then $b-a=mk'$? where $k$ is an integer?

• Can we conclude from: $$a\equiv b~~~(\text{mod}~m)\longrightarrow \exists k\in\mathbb Z,~~~ a-b=mk$$ and $$b\equiv c~~~(\text{mod}~m)\longrightarrow \exists k'\in\mathbb Z,~~~ b-c=mk'$$ that there exists $k''\in\mathbb Z$ such that $a-c=mk''$?

Answer 4

For $m=0$, we are talking equality.

For $m \ne 0$, prove first that $a \equiv b \pmod m$ if and only if $a$ and $b$ divided by $m$ yield the same remainder.

Then it becomes obvious that this is an equivalence relation.

Answer 5

Hint $\rm\,\ a\equiv b\:\ (mod\ m)\iff f(a) = f(b),\ $ for $\rm\ f(n) = n\ mod\ m =\, $ remainder left when dividing $\rm\,n\,$ by $\rm\,m.\:$ But it's easy to prove that all relations of this type are equivalence relations.

More generally, suppose $\rm\ u\sim v\ \smash[t]{\overset{\ def}{\iff}}\, f(u) \approx f(v)\ $ for a function $\rm\,f\,$ and equivalence relation $\,\approx.\, \ $ Then the equivalence relation properties of $\,\approx\,$ transport (pullback) to $\,\sim\,$ along $\rm\,f\,$ as follows:

• reflexive $\rm\quad\ f(v) \approx f(v)\:\Rightarrow\:v\sim v$

• symmetric $\rm\,\ u\sim v\:\Rightarrow\ f(u) \approx f(v)\:\Rightarrow\:f(v)\approx f(u)\:\Rightarrow\:v\sim u$

• transitive $\rm\ \ \ u\sim v,\, v\sim w\:\Rightarrow\: f(u)\approx f(v),\,f(v)\approx f(w)\:\Rightarrow\:f(u)\approx f(w)\:\Rightarrow u\sim w$

Such relations are called (equivalence) kernels. One calls $\, \sim\,$ the $\,(\approx)\,$ kernel of $\rm\,f.$