In 19-31,(1) prove that the relation is an equivalence relation, and (2) describe the distinct equivalence dasses of each relation.

R is the relation defined on Z as follows: For every ( m , n ) ∈ Z .

m R n             ⇔     4 | ( m 2 − n 2 ) .

To prove:

(1) The given relation is an equivalence relation,

(2) Describe the distinct equivalence classes of each relation.

Given information:

R is the relation defined on Z as follows: For all ( m, n ) ∈ Z ,

m R n ⇔ 4 | ( m²− n² ).

Proof:

A=ZR={(m,n)∈A×A|4( m 2 − n 2 )}

(1). To prove: R is an equivalent relation

A relation will be equivalence relation if it is reflexive, symmetric and transitive

Reflexive:

Let x∈A=Z

x2−x2=0=0⋅4

By the definition of divides, 4 divides x2−x2 and thus (x,x)∈R

Since (x,x)∈R for all x∈A, and thus R is indeed reflexive.

Symmetric:

Let us assume that (x,y)∈R. by the definition of R :

4|(x2−y2)

By the definition of divides, there exists an integer k such that:

x2−y2=4k

Multiply each side by − 1:

−(x2−y2)=−(4k)

Use the distributive property:

y2−x2=4(−k)

By the definition of divides, 4 divides y − x as − k is an integer (because k is an integer).

(y,x)∈R

Since (x,y)∈R implies (y,x)∈R, R is symmetric.

Transitive:

Let us assume that (x,y)∈R and (y,z)∈R.

By the definition of R :

4|( x 2− y 2)4|( y 2− z 2)

By the definition of divides, there exist integers k and l such that:

x2−y2=4ky2−z2=4l

Add the previous two equations:

(x2−y2)+(y2−z2)=4k+4l

Combine like terms:

x2−z2=4k+4l

By the definition of divides, 4 divides x2−z2 as k+l is an integer (because k is an integer and l is an integer).

(x,z)∈R

Since (x,y)∈R and (y,z)∈R implies (x,z)∈R, R is transitive.

Equivalence relation:

Since R is reflexive, symmetric and transitive, R is also an equivalence relation.

(2). Let m∈Z