To understand how equivalence relation examples look we should look at some common ones. Sp, equivalence relation examples include:

1. Let F be a set of fractions such that

\(\begin{align}\{ \frac{a}{b}:a,b \in Z,\rm{and}\,\,b \ne 0\} \end{align}\)

Then, \(\begin{align}\frac{p}{q}\,R\,\frac{r}{s}\end{align}\), If ps = rq is an equivalence relation.

2. For any set A, the identity relation is an equivalence relation.

3. Equalities is an important example of equivalence relations.

4. Modulus \(\begin{align}( \equiv )\end{align}\) are also a type of equivalence relation.

**Equivalence relation Proof **

Let us look at an example in Equivalence relation to reach the equivalence relation proof.

If R is a relation on the set of ordered pairs of natural numbers such that \(\begin{align}\left\{ {\left( {p,q} \right);\left( {r,s} \right)} \right\} \in R,\end{align}\), only if pq = rs.Let us now prove that R is an equivalence relation.

To do that, we need to prove that R follows all the three properties of equivalence relation, i.e. it is reflexive, symmetric, and transitive.

**Reflexive**

According to this property, if \(\begin{align}\left( {p,p} \right) \in R\end{align}\), for every \(\begin{align}p \in N\end{align}\)

For all ordered pairs of natural numbers:

\(\begin{align}\{ \left( {p,q} \right),\left( {p,q} \right) \in R.\end{align}\)

Here, we can say pq = pq for all natural numbers.

Thus, the reflexive property is proved………….(1)

**Symmetric**

According to this property if \(\begin{align}\left( {p,q} \right) \in R,\end{align}\) then \(\begin{align}\left( {q,p} \right) \in R\end{align}\) should also hold true.

For the set of Natural numbers,

If \(\begin{align}\left\{ {\left( {p,q} \right),\left( {r,s} \right)} \right\} \in R,\rm{then}\,\,\left\{ {\left( {r,s} \right),\left( {p,q} \right)} \right\} \in R.\end{align}\)

Since multiplication is commutative in Natural numbers, we can say ps = rq

Thus, \(\begin{align}\left\{ {\left( {r.s} \right),\left( {p,q} \right)} \right\} \in R\end{align}\)

Thus, the symmetric property is proved……………(2)

**Transitive Property **

According to this property, if \(\begin{align}\left( {p,q} \right),\left( {q,r} \right) \in R,\end{align}\) then \(\begin{align}\left( {p,r} \right) \in R\end{align}\)

For a given set of ordered pairs in Natural numbers,

if \(\begin{align}\{ \left( {p,q} \right),\left( {r,s} \right) \in R\end{align}\) and \(\begin{align}\left\{ {\left( {r,s} \right),\left( {x,y} \right)} \right\} \in R,\end{align}\)

then \(\begin{align}\{ \left( {p,q} \right),\left( {x,y} \right) \in R.\end{align}\)

Let us here assume that {(p,q), (r,s)} R and {(r,s), (x,y)} R.

Then we get that ps= qr and ry = sx.

This further implies that \(\begin{align}\frac{p}{q} = \frac{r}{s}\,\rm{and}\,\frac{r}{s} = \frac{x}{y}\end{align}\)

Thus, \(\begin{align}\frac{p}{q} = \frac{x}{y}\end{align}\)

Thus, py = xq.

Thus,\(\begin{align}\left\{ {\left( {p,q} \right),\left( {x,y} \right)} \right\} \in R.\end{align}\)

Thus, the transitive property is proved……………..(3)

From (1), (2), and (3) we prove equivalence relation in Natural Numbers. This is basically what entails a equivalence relation proof.

**Equivalence Questions**

A relation R is defined on the set of Integers Z by “xRy if x – y leaves a remainder of 0, when divided by 10. ” for x, y ∈ Z. Show that R is an equivalence relation on Z.

**Solution:**

The Hypothesis x-y = 0 is true when x-y is divisible by 10. Let us prove this.

(i) Let x ∈ Z. Then x – x =0 is divisible by 10. Therefore xRx holds true for all x in Z. Hence, R is reflexive.

(ii) Let x, y ∈ Z , and xRy hold true. Then x – y is divisible by 10 and therefore y-x is also divisible by 10.

Thus, xRy ⇒ yRx . Hence, R is symmetric.

(iii) Let x, y, z ∈ Z , and xRy, yRz both hold true. Then x – y and y – z are both divisible by 10.

Therefore x – y = (x – y) + (y – z) is divisible by 10.

Thus, xRy and yRz ⇒ xRz Hence, R is transitive.

Since R is reflexive symmetric transitive. Hence, R is an equivalence relation on Z.

Let P be the set of all lines in three-dimensional space. A relation R is defined on P by “aRb if and only if a lies on the plane of b” for a, b ∈ P.

Check if R is an equivalence relation.

**Solution:**

(i) Reflexive: Let a ∈ P. Then a is coplanar with itself.

Therefore, aRa holds for all a in P.

Hence, R is reflexive

(ii) Symmetric: Let a, b ∈ P and aRb holds true. Then a lies on the plane of b.

Therefore, b lies on the plane of a. Thus, aRb ⇒ bRa. Hence, R is symmetric.

(iii) Transitive: Let a, b, c ∈ P and aRb, bRc both hold true. Then a lies in the plane of b and b lies in the plane of c. This however does not always imply that a lies in the plane of c.

That is, aRb and bRc do not necessarily imply aRc.

Hence, R is not transitive.

Since R is reflexive and symmetric but not transitive so, R is not an equivalence relation on set Z.

Let us now have a look at some conditions where relations do not follow all the three equivalence properties.

**Reflexive and Symmetric but not Transitive**

Let us look a a case when the set is Reflexive and Symmetric but not Transitive.

Let X = {1,2,3}

A relation R is defined on X as

R = {(1,1),(2,2),(3,3),(1,2),(2,1),(2,3),(3,2)}

If a ∈ X,(a,a) ∈ R, that implies {(1,1),(2,2),(3,3)} ∈ R. Hence, it is reflexive.

If a,b ∈ X, (x,y) ∈ R and (y,x) ∈ R,that implies {(1,2),(2,1),(2,3),(3,2)} ∈ R . Hence, it is symmetric.

If a,b ∈ R and b,c ∈ R, that implies (1,2) ∈ R and (2,3) ∈ R

But \(\begin{align}\left( {1,3} \right) \notin R.\end{align}\) Hence, it is not transitive.

Thus, the relation is reflexive and symmetric but not transitive. Also, not an equivalence relation.

**Symmetric and Transitive but not Reflexive**

Let us look at a case when the set is Symmetric and Transitive but not Reflexive.

Let X = {−3, −4}.

A relation R is defined on X as:

R = {(−3, −4), (−4, −3), (−3, −3)}

Relation R is not reflexive as (−4, −4) ∉ R.

Relation R is symmetric as (−3, −4) ∈ R and (−4, −3)∈ R.

It is seen that (−3, −4), (−4, −3) ∈ R. Further, (−3, −3) also ∈ R.

Hence, the relation R is transitive.

Thus, relation R is symmetric and transitive but not reflexive. And also, not an equivalence relation.

**Reflexive and Transitive but not Symmetric**

Let us have a look at when a set is Reflexive and Transitive but not Symmetric.

A relation R is defined as

R ={(a,b) : a^{3} b^{3}

Clearly (a, a) ∈ R since a = a^{3}

Hence, R is reflexive.

Moving on,

(2, 1) ∈ R (since 2^{3} ≥ 1^{3})

But,

(1, 2) ∉ R (as 1^{3} < 2^{3})

Hence,R is not symmetric.

Again,

Let (a, b), (b, c) ∈ R.

⇒ a^{3} ≥ b^{3} and b^{3} ≥ c^{3}

⇒ a^{3} ≥ c^{3}

⇒ (a, c) ∈ R

Hence, R is transitive.

Thus we can conclude that the relation R is reflexive and transitive but not symmetric. And thus, not an equivalence relation.

**SUMMARY**

- Equivalence relations are a special type of relation. They are derived from the term equivalent meaning to be equal in value, function, or meaning.
- They are Reflexive - x R x, for all x ∈ A
- They are Symmetric - x R y implies y R x, for all x,y ∈ A
- They are Transitive - x R y and y R z imply x R z, for all x,y,z ∈ A
- Only when a relation satisfies all the three properties, it can be classified as an equivalence relation. This is the equivalence relation proof.
- We even looked at cases when sets are reflexive symmetric transitive, and when sets are reflexive and symmetric but not transitive and also when sets are reflexive and transitive but not symmetric and lastly when a set is symmetric and transitive but not reflexive.

**Written by Aarti Agarwal**

**FAQS**

**What is an equivalence relation?**

An equivalence relation on a set A is a relation that is reflexive, symmetric, transitive.

**What is the use of equivalence relations?**

It allows us to partition a set in such a way that, the components of a given part for all our purposes are equal.

**Equivalence relation proof?**

To check for equivalence relation in a given set or subset one needs to check for all its properties. If the set is reflexive symmetric transitive, it is an equivalence relation.