Question

3.If  A={x:0≤x≤5}; B={x:2≤x≤4}; C= {x:1≤ x ≤ 5} and D ={x:5≤x ≤7} then find a) A Δ B b) B U C c) A – B d) C –D.​

Answer 1

Answer:

(i)

A={x∈Z:0≤x≤12}={0,1,2,3,4,5,6,7,8,9,10,11,12}

R={(a,b):∣a−b∣is a multiple of 4}

For any element a∈A, we have (a,a)∈R as ∣a−a∣=0 is a multiple of 4.

∴R is reflexive.

Now, let (a,b)∈R⇒∣a−b∣ is a multiple of 4.

⇒∣−(a−b)∣=∣b−a∣ is a multiple of 4.

⇒(b,a)∈R

∴ is symmetric.

Now, let (a,b),(b,c)∈R

⇒∣a−b∣ is a multiple of 4 and ∣b−c∣ is a multiple of 4.

⇒(a−b) is a multiple of 4 and (b−c) is a multiple of 4.

⇒(a−c)=(a−b)+(b−c) is a multiple of 4.

⇒∣a−c∣ is a multiple of 4.

⇒(a,c)∈R.

∴R is transitive.

Hence, R is an equivalence relation.

The set of elements related to 1 is {1,5,9} since

∣1−1∣=0 is a multiple of 4,

∣5−1∣=4 is a multiple of 4, and

∣9−1∣=8 is a multiple of 4

Hence, R is an equivalence relation.

(ii)

A={x∈Z:0≤x≤12}={0,1,2,3,4,......,11,12}

R={(a,b):a=b}={(0,),(1,1),(2,2),........,(11,11),(12,12)}

For any element a∈A, we have (a,a)∈R, since a=a.

∴R is reflexive.

Now, let (a,b)∈R.

⇒a=b

⇒b=a

⇒(b,a)∈R

∴R is symmetric.

Now, let (a,b)∈R and (b,c)∈R.

⇒a=b and b=c

⇒a=c

⇒(a,c)∈R

∴R is transitive.

Hence, R is an equivalence relation.

The elements in R that are related to 1 will be those elements from set A which are equal to 1.

Hence, the set of elements related to 1 is {1}.