Q. 8 Show that the relation R in the set A = \{1, 2, 3, 4, 5\} given by
R = \{(a, b) : |a - b| \;is\;even\}, is an equivalence relation. Show that all the
elements of \{1, 3, 5\} are related to each other and all the elements of \{2, 4\}are
related to each other. But no element of \{1, 3, 5\} is related to any element of \{2, 4\}.

 

Answers (1)
S seema garhwal

A = \{1, 2, 3, 4, 5\}

R = \{(a, b) : |a - b| \;is\;even\}

R=\left \{ \left ( 1,1 \right ),\left ( 2,2 \right ),\left ( 3,3 \right ),\left ( 4,4 \right ),\left ( 5,5 \right ),\left ( 1,3 \right ),\left ( 2,4 \right ),\left ( 3,5 \right ),\left ( 3,1 \right ),\left ( 5,1 \right ),\left ( 4,2 \right ),\left ( 5,3 \right )\right \}

Let there be a\in A then (a,a)\in R as \left | a-a \right |=0  which is even number. Hence, it is reflexive

Let  (a,b)\in R where a,b\in A then (b,a)\in R as \left | a-b \right |=\left | b-a \right |

Hence, it is symmetric

Now, let (a,b)\in R \, and\, (b,c)\in R

\left | a-b \right | \, and \, \left | b-c \right |   are even number i.e. (a-b)\, and\,(b-c) are even

then, (a-c)=(a-b)+(b-c) is even                  (sum of even integer is even)

So, (a,c)\in R. Hence, it is transitive.

Thus, it is reflexive, symmetric and transitive i.e. it is an equivalence relation.

The elements of \{1, 3, 5\} are related to each other because the difference of odd numbers gives even number and in this set all numbers are odd.

The elements of \{2, 4\}are related to each other because the difference of even number is even number and in this set, all numbers are even.

The element of \{1, 3, 5\} is not related to  \{2, 4\} because a difference of odd and even number is not even.

 

 

 

 

 

Exams
Articles
Questions