Question

he set E of even integers and O of odd integers are called equivalence classes containing respectively
a)one and zero
b)zero and one
c)zero
d)one

Answer 1

Hint: In this question we are given two equivalence classes which are the set E of even integers and O of odd integers. Therefore, we should first see the definition of odd and even equivalence classes and see to which equivalence class zero and one belong and that would be the required answer to this question.

Complete step-by-step answer:
We know that a relation $\sim$ is called an equivalence relation on a set A if it satisfies the following criteria
i) It is reflexive, i.e. for any element a in A, a should be related to a, i.e. $a\sim a$ …………………..(1.1)
ii) It is symmetric, i.e. for any two elements a and b in A, if a is related to b, then b should be related to a, i.e. $a\sim b\Rightarrow b\sim a$ ……………………….(1.2)
iii) It is transitive, i.e. for any three elements a, b and c in A, if a is related to b, and b is related to c then a should be related to c, i.e. $a\sim b\mathrm{\&}b\sim c\Rightarrow a\sim c$ ……………….(1.3)
Now, we should try to find out the equivalence relation for which odd and even integers are equivalence classes. The equivalence relation is
 $R=\{a\sim b\text{ if }b=a+2n|a,b,n\in Z\}$ ………………………..(1.4)
i.e. two elements a and b of Z will be related to each other if $a=b+2n$ where n is an integer. We see that
i) For any integer a, a can be written as $a=a+2\times 0$ , which is of the form (1.4) with n=0. Therefore, $a\sim a$ and thus the relation is reflexive.
ii) For any integers a and b, if $a\sim b$ , i.e. if $b=a+2r$ for some integer r , then we can write the equation as $a=b-2r$ which is of the form (1.4) with n=-r. Therefore, $a\sim b\Rightarrow b\sim a$ and thus the relation is symmetric.
 iii) For any integers a,b and c, if $a\sim b$ , i.e. if $b=a+2r_1$ for some integer $r_1$ and if $b\sim c$ , i.e. if $c=b+2r_2$ for some integer $r_2$ , then we can write the equation as $c=b+2r_2=a+2r_1+2r_2=a+2(r_1+r_2)$ which is of the form (1.4) with $n=r_1+r_2$ . Therefore, $a\sim b,b\sim c\Rightarrow a\sim c$ and thus the relation is transitive.
Thus, as the given relation satisfies the reflexive, symmetric and transitive, it is verified to be an equivalence relation.
Thus, the equivalence class containing zero will contain all elements which are related to 0 by equation (1.4). Thus, the equivalence class of 0 will be given by
Equivalence class of 0 = $\{m|m=0+2n=2n,n\in Z\}=\{...-4,-2,0,2,4,...\}$
which is the set of all even integers E. Therefore, the set of even integers contains 0.....................(1.5)
And, the equivalence class containing one will contain all elements which are related to 1 by equation (1.4). Thus, the equivalence class of 0 will be given by
Equivalence class of 1 = $\{m|m=1+2n=2n,n\in Z\}=\{...-3,-1,1,2,3,...\}$
which is the set of all odd integers O. Therefore, the set of odd integers contains 1.....................(1.6)
Thus, from (1.5) and (1.6) we find that E contains 0 and O contains 1. Thus, option (b) is the correct answer to this question.

Note: We should note that in this question, we should first form the equivalence classes of zero and one and check whether they are the set of even or odd numbers. For this we should explicitly verify the reflexive, symmetric and transitive relations as equivalence classes can be formed only by equivalence relations which satisfy these properties.