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Discrete Mathematics With Applicat...

5th Edition
EPP + 1 other
ISBN: 9781337694193

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Discrete Mathematics With Applicat...

5th Edition
EPP + 1 other
ISBN: 9781337694193
Textbook Problem
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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 determine

To prove:

(1) The given relation is an equivalence relation,

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

Explanation

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 xA=Z

x2x2=0=04

By the definition of divides, 4 divides x2x2 and thus (x,x)R

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

Symmetric:

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

4|(x2y2)

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

x2y2=4k

Multiply each side by − 1:

(x2y2)=(4k)

Use the distributive property:

y2x2=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)RR 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:

x2y2=4ky2z2=4l

Add the previous two equations:

(x2y2)+(y2z2)=4k+4l

Combine like terms:

x2z2=4k+4l

By the definition of divides, 4 divides x2z2 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)RR is transitive.

Equivalence relation:

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

(2). Let mZ

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