A relation R on a set A is said to be an equivalence relation if and only if the relation R is reflexive, symmetric and transitive.
Reflexive: A relation is said to be reflexive, if (a, a) âˆˆ R, for every aâˆˆA
Symmetric: A relation is said to be symmetric, if (a, b) âˆˆ R, then (b, a) âˆˆ R
Transitive: A relation is said to be transitive if (a, b) âˆˆ R and (b, c) âˆˆ R, then (a, c) âˆˆ R
Equivalence Relation Proof
Here is an equivalence relation example to prove the properties.
Let us assume that R be a relation on the set of ordered pairs of positive integers such that ((a,b), (c,d))âˆˆ R if and only if ad=bc. Is R an equivalence relation?
In order to prove that R is an equivalence relation, we must show that R is reflexive, symmetric and transitive.
Proof
The Proof for the given condition is given below :
Reflexive
According to the reflexive property, if (a, a) âˆˆ R, for every aâˆˆA
For all pairs of positive integers,
((a,b),(a,b))âˆˆ R.
Clearly, we can say
ab = ab for all positive integers.
Hence reflexive property is proved.
Symmetric
From the symmetric property,
if (a, b) âˆˆ R, then we can say (b, a) âˆˆ R
For the given condition,
if ((a,b),(c,d)) âˆˆ R, then ((c,d),(a,b)) âˆˆ R.
If ((a,b),(c,d))âˆˆ R, then ad = bc and cb = da
since multiplication is commutative.
Therefore ((c,d),(a,b)) âˆˆ R
Hence symmetric property is proved.
Transitive
From the transitive property,
if (a, b) âˆˆ R and (b, c) âˆˆ R, then (a, c) also belongs to R
For the given set of ordered pairs of positive integers,
((a,b), (c,d))âˆˆ R and ((c,d), (e,f))âˆˆ R,
then ((a,b),(e,f) âˆˆ R.
Now, assume that ((a,b), (c,d))âˆˆ R and ((c,d), (e,f)) âˆˆ R.
Then we get, ad = cb and cf = de.
The above relation implies that a/b = c/d and that c/d = e/f,
so a/b = e/f we get af = be.
Therefore ((a,b),(e,f))âˆˆ R.
Hence transitive property is proved.
Sample Problem
Go through the equivalence relation examples and solutions provided here
Question 1 :
Let assume that R be a relation on the set \(\mathbb{R}\) real numbers defined by xRy if and only if xy is an integer. Prove that R is an equivalence relation on \(\mathbb{R}\).
Solution :
Reflexive: Consider x belongs to \(\mathbb{R}\),then x – x = 0 which is an integer. Therefore xRx.
Symmetric: Consider x and y belongs to \(\mathbb{R}\) and xRy. Then x – y is an integer. Thus, y – x = – ( x – y), y – x is also an integer. Therefore yRx.
Transitive: Consider x and y belongs to \(\mathbb{R}\), xRy and yRz. Therefore xy and yz are integers. According to the transitive property, ( x – y ) + ( y – z ) = x – z is also an integer. So that xRz.
Thus, R is an equivalence relation on \(\mathbb{R}\).
Question 2:
Show that the relation R is an equivalence relation in the set A = { 1, 2, 3, 4, 5 } given by the relation R = { (a, b):ab is even }.
Solution :
R = { (a, b):ab is even }. Where a, b belongs to A
Reflexive Property :
From the given relation,
a – a =  0 =0
And 0 is always even.
Thus, aa is even
Therefore, (a, a) belongs to R
Hence R is Reflexive
Symmetric Property :
From the given relation,
a – b = b – a
We know that a – b = (b – a)= b – a
Hence a – b is even,
Then b – a is also even.
Therefore, if (a, b) âˆˆ R, then (b, a) belongs to R
Hence R is symmetric
Transitive Property :
If ab is even, then (ab) is even.
Similarly, if bc is even, then (bc) is also even.
Sum of even number is also even
So, we can write it as ab+ bc is even
Then, a – c is also even
So,
a – b and b – c is even , then ac is even.
Therefore, if (a, b) âˆˆ R and (b, c) âˆˆ R, then (a, c) also belongs to R
Hence R is transitive.
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