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Question: If A = {a, b, c, d, e} and B = {a, c, e, g} and C = {b, e, f, g}. Verify that (i) \(A\cup B=B\cup...

If A = {a, b, c, d, e} and B = {a, c, e, g} and C = {b, e, f, g}. Verify that
(i) AB=BAA\cup B=B\cup A
(ii) AC=CAA\cup C=C\cup A
(iii) BC=CBB\cup C=C\cup B
(iv) AB=BAA\cap B=B\cap A
(v) BC=CBB\cap C=C\cap B
(vi) AC=CAA\cap C=C\cap A
(vii) (AB)C=A(BC)\left( A\cup B \right)\cup C=A\cup \left( B\cup C \right)
(viii) (AB)C=A(BC)\left( A\cap B \right)\cap C=A\cap \left( B\cap C \right)

Explanation

Solution

Hint: We will substitute the elements of the sets A, B, and C into the operations given to us where the sets A, B and C are given here. Union of sets indicates the elements present in both the sets with repeated elements only taken once and the intersection of sets indicates all the elements which are common in both sets.

Complete step-by-step answer:

We will consider first the operation of union as AB=BA...(i)A\cup B=B\cup A...(i).
Now, we will consider the left hand side ABA\cup B of (i). Since, the elements of A = {a, b, c, d, e} and B = {a, c, e, g}. Therefore, we have
ABA\cup B = {a, b, c, d, e} \cup {a, c, e, g}. Thus, we have that ABA\cup B = {a, b, c, d, e, g}.
Now, we will consider the right side BAB\cup A of equation (i). Since, the elements of A = {a, b, c, d, e} and B = {a, c, e, g}. Therefore, we have
BAB\cup A = {a, c, e, g} \cup {a, b, c, d, e}. Thus, we have that BAB\cup A = {a, b, c, d, e, g}.
Therefore, it is verified that AB=BAA\cup B=B\cup A.
Now, we will consider the expression AC=CA...(ii)A\cup C=C\cup A...(ii).
Now, we will consider the left hand side ACA\cup C of (ii). Since, the elements of A = {a, b, c, d, e} and C = {b, e, f, g}. Therefore, we have
ACA\cup C = {a, b, c, d, e} \cup {b, e, f, g}. Thus, we have that ACA\cup C = {a, b, c, d, e, f, g}.
Now, we will consider the right side CAC\cup A of equation (ii). Since, the elements of A = {a, b, c, d, e} and C = {b, e, f, g}. Therefore, we have
CAC\cup A = {b, e, f, g} \cup {a, b, c, d, e}. Thus, we have that CAC\cup A = {a, b, c, d, e, f, g}.
Therefore, it is verified that AC=CAA\cup C=C\cup A.
Now, we will consider the expression BC=CB...(iii)B\cup C=C\cup B...(iii).
Now, we will consider the left hand side BCB\cup C of (iii). Since, the elements of B = {a, c, e, g} and C = {b, e, f, g}. Therefore, we have
BCB\cup C = {a, c, e, g} \cup {b, e, f, g}. Thus, we have that BCB\cup C = {a, b, c, e, f, g}.
Now, we will consider the right side CBC\cup B of equation (iii). Since the elements of B = {a, c, e, g} and C = {b, e, f, g}. Therefore, we have
CBC\cup B = {b, e, f, g} \cup {a, c, e, g}. Thus, we have that CBC\cup B = {a, b, c, e, f, g}.
Therefore, it is verified that BC=CBB\cup C=C\cup B.
Now, we will consider the expression AB=BA...(iv)A\cap B=B\cap A...(iv).
Now, we will consider the left hand side ABA\cap B of (iv). Since, the elements of A = {a, b, c, d, e} and B = {a, c, e, g}. Therefore, we have
ABA\cap B = {a, b, c, d, e} \cap {a, c, e, g}. Thus, we have that ABA\cap B = {a, c, e}.
Now, we will consider the right side BAB\cap A of equation (iv). Since, the elements of B = {a, c, e, g} and A = {a, b, c, d, e}. Therefore, we have
BAB\cap A = { a, c, e, g} \cap {a, b, c, d, e}. Thus, we have that BAB\cap A = {a, c, e}.
Therefore, it is verified that AB=BAA\cap B=B\cap A.
Now, we will consider the expression BC=CB...(v)B\cap C=C\cap B...(v).
Now, we will consider the left hand side BCB\cap C of (v). Since, the elements of B = {a, c, e, g} and C = {b, e, f, g}. Therefore, we have
BCB\cap C = {a, c, e, g} \cap {b, e, f, g}. Thus, we have that BCB\cap C = {e, g}.
Now, we will consider the right side CBC\cap B of equation (v). Since, the elements of B = {a, c, e, g} and C = {b, e, f, g}. Therefore, we have
CBC\cap B = {b, e, f, g} \cap {a, c, e, g}. Thus, we have that CBC\cap B = {e, g}.
Therefore, it is verified that BC=CBB\cap C=C\cap B.
Now, we will consider the expression AC=CA...(vi)A\cap C=C\cap A...(vi).
Now, we will consider the left hand side ACA\cap C of (vi). Since, the elements of A = {a, b, c, d, e} and C = {b, e, f, g}. Therefore, we have
ACA\cap C = {a, b, c, d, e} \cap {b, e, f, g}. Thus, we have that ACA\cap C = {b, e}.
Now, we will consider the right side CAC\cap A of equation (vi). Since, the elements of A = {a, b, c, d, e} and C = {b, e, f, g}. Therefore, we have
CAC\cap A = {b, e, f, g} \cap {a, b, c, d, e}. Thus, we have that CAC\cap A = {b, e}.
Therefore, it is verified that AC=CAA\cap C=C\cap A.
Now, we will consider the expression (AB)C=A(BC)...(vii)\left( A\cup B \right)\cup C=A\cup \left( B\cup C \right)...(vii).
Now, we will consider the left hand side (AB)C\left( A\cup B \right)\cup C of (vii). Since, the elements of A = {a, b, c, d, e}, B = {a, c, e, g} and C = {b, e, f, g}. Therefore, we have
(AB)\left( A\cup B \right) = {a, b, c, d, e} \cup {a, c, e, g} = {a, b, c, d, e, g}
(AB)C\Rightarrow \left( A\cup B \right)\cup C = {a, b, c, d, e, g} \cup {b, e, f, g}. Thus, we have that (AB)C\left( A\cup B \right)\cup C = {a, b, c, d, e, f, g}.
Now, we will consider the right side A(BC)A\cup \left( B\cup C \right) of equation (vii). Since, the elements of A = {a, b, c, d, e}, B = {a, c, e, g} and C = {b, e, f, g}. Therefore, we have
(BC)\left( B\cup C \right) = {a, c, e, g} \cup {b, e, f, g} = {a, b, c, e, f, g}.
A(BC)\Rightarrow A\cup \left( B\cup C \right) = {a, b, c, d, e} \cup {a, b, c, e, f, g}. Thus, we have that A(BC)A\cup \left( B\cup C \right) = {a, b, c, d, e, f, g}.
Therefore, it is verified that (AB)C=A(BC)...(vii)\left( A\cup B \right)\cup C=A\cup \left( B\cup C \right)...(vii).
Now, we will consider the expression (AB)C=A(BC)...(viii)\left( A\cap B \right)\cap C=A\cap \left( B\cap C \right)...(viii).
Now, we will consider the left hand side (AB)C\left( A\cap B \right)\cap C of (viii). Since, the elements of A = {a, b, c, d, e}, B = {a, c, e, g} and C = {b, e, f, g}. Therefore, we have
(AB)\left( A\cap B \right) = {a, b, c, d, e} \cap {a, c, e, g} = {a, c, e}
(AB)C\Rightarrow \left( A\cap B \right)\cap C = {a, c, e} \cap {b, e, f, g}. Thus, we have that (AB)C\left( A\cap B \right)\cap C = {e}.
Now, we will consider the right side A(BC)A\cap \left( B\cap C \right) of equation (viii). Since, the elements of A = {a, b, c, d, e}, B = {a, c, e, g} and C = {b, e, f, g}. Therefore, we have
(BC)\left( B\cap C \right) = {a, c, e, g} \cap {b, e, f, g} = {e, g}
A(BC)A\cap \left( B\cap C \right) = {a, b, c, d, e} \cap {e, g}. Thus, we have that A(BC)A\cap \left( B\cap C \right) = {e}.
Therefore, it is verified that (AB)C=A(BC)\left( A\cap B \right)\cap C=A\cap \left( B\cap C \right).

Note: We are using intersection as taking common points between the sets. But when we are using union then we are selecting all elements which are included (repeated taken only once) under union of the sets. One should not get confused while using intersection and union. This confusion is the most common reason for mistakes committed during the exams.