Answer
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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 $A\cup B=B\cup A...(i)$.
Now, we will consider the left hand side $A\cup B$ of (i). Since, the elements of A = {a, b, c, d, e} and B = {a, c, e, g}. Therefore, we have
$A\cup B$ = {a, b, c, d, e} $\cup $ {a, c, e, g}. Thus, we have that $A\cup B$ = {a, b, c, d, e, g}.
Now, we will consider the right side $B\cup A$ of equation (i). Since, the elements of A = {a, b, c, d, e} and B = {a, c, e, g}. Therefore, we have
$B\cup A$ = {a, c, e, g} $\cup $ {a, b, c, d, e}. Thus, we have that $B\cup A$ = {a, b, c, d, e, g}.
Therefore, it is verified that $A\cup B=B\cup A$.
Now, we will consider the expression $A\cup C=C\cup A...(ii)$.
Now, we will consider the left hand side $A\cup C$ of (ii). Since, the elements of A = {a, b, c, d, e} and C = {b, e, f, g}. Therefore, we have
$A\cup C$ = {a, b, c, d, e} $\cup $ {b, e, f, g}. Thus, we have that $A\cup C$ = {a, b, c, d, e, f, g}.
Now, we will consider the right side $C\cup A$ of equation (ii). Since, the elements of A = {a, b, c, d, e} and C = {b, e, f, g}. Therefore, we have
$C\cup A$ = {b, e, f, g} $\cup $ {a, b, c, d, e}. Thus, we have that $C\cup A$ = {a, b, c, d, e, f, g}.
Therefore, it is verified that $A\cup C=C\cup A$.
Now, we will consider the expression $B\cup C=C\cup B...(iii)$.
Now, we will consider the left hand side $B\cup C$ of (iii). Since, the elements of B = {a, c, e, g} and C = {b, e, f, g}. Therefore, we have
$B\cup C$ = {a, c, e, g} $\cup $ {b, e, f, g}. Thus, we have that $B\cup C$ = {a, b, c, e, f, g}.
Now, we will consider the right side $C\cup B$ of equation (iii). Since the elements of B = {a, c, e, g} and C = {b, e, f, g}. Therefore, we have
$C\cup B$ = {b, e, f, g} $\cup $ {a, c, e, g}. Thus, we have that $C\cup B$ = {a, b, c, e, f, g}.
Therefore, it is verified that $B\cup C=C\cup B$.
Now, we will consider the expression $A\cap B=B\cap A...(iv)$.
Now, we will consider the left hand side $A\cap B$ of (iv). Since, the elements of A = {a, b, c, d, e} and B = {a, c, e, g}. Therefore, we have
$A\cap B$ = {a, b, c, d, e} $\cap $ {a, c, e, g}. Thus, we have that $A\cap B$ = {a, c, e}.
Now, we will consider the right side $B\cap A$ of equation (iv). Since, the elements of B = {a, c, e, g} and A = {a, b, c, d, e}. Therefore, we have
$B\cap A$ = { a, c, e, g} $\cap $ {a, b, c, d, e}. Thus, we have that $B\cap A$ = {a, c, e}.
Therefore, it is verified that $A\cap B=B\cap A$.
Now, we will consider the expression $B\cap C=C\cap B...(v)$.
Now, we will consider the left hand side $B\cap C$ of (v). Since, the elements of B = {a, c, e, g} and C = {b, e, f, g}. Therefore, we have
$B\cap C$ = {a, c, e, g} $\cap $ {b, e, f, g}. Thus, we have that $B\cap C$ = {e, g}.
Now, we will consider the right side $C\cap B$ of equation (v). Since, the elements of B = {a, c, e, g} and C = {b, e, f, g}. Therefore, we have
$C\cap B$ = {b, e, f, g} $\cap $ {a, c, e, g}. Thus, we have that $C\cap B$ = {e, g}.
Therefore, it is verified that $B\cap C=C\cap B$.
Now, we will consider the expression $A\cap C=C\cap A...(vi)$.
Now, we will consider the left hand side $A\cap C$ of (vi). Since, the elements of A = {a, b, c, d, e} and C = {b, e, f, g}. Therefore, we have
$A\cap C$ = {a, b, c, d, e} $\cap $ {b, e, f, g}. Thus, we have that $A\cap C$ = {b, e}.
Now, we will consider the right side $C\cap A$ of equation (vi). Since, the elements of A = {a, b, c, d, e} and C = {b, e, f, g}. Therefore, we have
$C\cap A$ = {b, e, f, g} $\cap $ {a, b, c, d, e}. Thus, we have that $C\cap A$ = {b, e}.
Therefore, it is verified that $A\cap C=C\cap A$.
Now, we will consider the expression $\left( A\cup B \right)\cup C=A\cup \left( B\cup C \right)...(vii)$.
Now, we will consider the left hand side $\left( A\cup B \right)\cup C$ 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
$\left( A\cup B \right)$ = {a, b, c, d, e} $\cup $ {a, c, e, g} = {a, b, c, d, e, g}
$\Rightarrow \left( A\cup B \right)\cup C$ = {a, b, c, d, e, g} $\cup $ {b, e, f, g}. Thus, we have that $\left( A\cup B \right)\cup C$ = {a, b, c, d, e, f, g}.
Now, we will consider the right side $A\cup \left( B\cup C \right)$ 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
$\left( B\cup C \right)$ = {a, c, e, g} $\cup $ {b, e, f, g} = {a, b, c, e, f, g}.
$\Rightarrow A\cup \left( B\cup C \right)$ = {a, b, c, d, e} $\cup $ {a, b, c, e, f, g}. Thus, we have that $A\cup \left( B\cup C \right)$ = {a, b, c, d, e, f, g}.
Therefore, it is verified that $\left( A\cup B \right)\cup C=A\cup \left( B\cup C \right)...(vii)$.
Now, we will consider the expression $\left( A\cap B \right)\cap C=A\cap \left( B\cap C \right)...(viii)$.
Now, we will consider the left hand side $\left( A\cap B \right)\cap C$ 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
$\left( A\cap B \right)$ = {a, b, c, d, e} $\cap $ {a, c, e, g} = {a, c, e}
$\Rightarrow \left( A\cap B \right)\cap C$ = {a, c, e} $\cap $ {b, e, f, g}. Thus, we have that $\left( A\cap B \right)\cap C$ = {e}.
Now, we will consider the right side $A\cap \left( B\cap C \right)$ 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
$\left( B\cap C \right)$ = {a, c, e, g} $\cap $ {b, e, f, g} = {e, g}
$A\cap \left( B\cap C \right)$ = {a, b, c, d, e} $\cap $ {e, g}. Thus, we have that $A\cap \left( B\cap C \right)$ = {e}.
Therefore, it is verified that $\left( A\cap B \right)\cap C=A\cap \left( B\cap C \right)$.
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.
Complete step-by-step answer:
We will consider first the operation of union as $A\cup B=B\cup A...(i)$.
Now, we will consider the left hand side $A\cup B$ of (i). Since, the elements of A = {a, b, c, d, e} and B = {a, c, e, g}. Therefore, we have
$A\cup B$ = {a, b, c, d, e} $\cup $ {a, c, e, g}. Thus, we have that $A\cup B$ = {a, b, c, d, e, g}.
Now, we will consider the right side $B\cup A$ of equation (i). Since, the elements of A = {a, b, c, d, e} and B = {a, c, e, g}. Therefore, we have
$B\cup A$ = {a, c, e, g} $\cup $ {a, b, c, d, e}. Thus, we have that $B\cup A$ = {a, b, c, d, e, g}.
Therefore, it is verified that $A\cup B=B\cup A$.
Now, we will consider the expression $A\cup C=C\cup A...(ii)$.
Now, we will consider the left hand side $A\cup C$ of (ii). Since, the elements of A = {a, b, c, d, e} and C = {b, e, f, g}. Therefore, we have
$A\cup C$ = {a, b, c, d, e} $\cup $ {b, e, f, g}. Thus, we have that $A\cup C$ = {a, b, c, d, e, f, g}.
Now, we will consider the right side $C\cup A$ of equation (ii). Since, the elements of A = {a, b, c, d, e} and C = {b, e, f, g}. Therefore, we have
$C\cup A$ = {b, e, f, g} $\cup $ {a, b, c, d, e}. Thus, we have that $C\cup A$ = {a, b, c, d, e, f, g}.
Therefore, it is verified that $A\cup C=C\cup A$.
Now, we will consider the expression $B\cup C=C\cup B...(iii)$.
Now, we will consider the left hand side $B\cup C$ of (iii). Since, the elements of B = {a, c, e, g} and C = {b, e, f, g}. Therefore, we have
$B\cup C$ = {a, c, e, g} $\cup $ {b, e, f, g}. Thus, we have that $B\cup C$ = {a, b, c, e, f, g}.
Now, we will consider the right side $C\cup B$ of equation (iii). Since the elements of B = {a, c, e, g} and C = {b, e, f, g}. Therefore, we have
$C\cup B$ = {b, e, f, g} $\cup $ {a, c, e, g}. Thus, we have that $C\cup B$ = {a, b, c, e, f, g}.
Therefore, it is verified that $B\cup C=C\cup B$.
Now, we will consider the expression $A\cap B=B\cap A...(iv)$.
Now, we will consider the left hand side $A\cap B$ of (iv). Since, the elements of A = {a, b, c, d, e} and B = {a, c, e, g}. Therefore, we have
$A\cap B$ = {a, b, c, d, e} $\cap $ {a, c, e, g}. Thus, we have that $A\cap B$ = {a, c, e}.
Now, we will consider the right side $B\cap A$ of equation (iv). Since, the elements of B = {a, c, e, g} and A = {a, b, c, d, e}. Therefore, we have
$B\cap A$ = { a, c, e, g} $\cap $ {a, b, c, d, e}. Thus, we have that $B\cap A$ = {a, c, e}.
Therefore, it is verified that $A\cap B=B\cap A$.
Now, we will consider the expression $B\cap C=C\cap B...(v)$.
Now, we will consider the left hand side $B\cap C$ of (v). Since, the elements of B = {a, c, e, g} and C = {b, e, f, g}. Therefore, we have
$B\cap C$ = {a, c, e, g} $\cap $ {b, e, f, g}. Thus, we have that $B\cap C$ = {e, g}.
Now, we will consider the right side $C\cap B$ of equation (v). Since, the elements of B = {a, c, e, g} and C = {b, e, f, g}. Therefore, we have
$C\cap B$ = {b, e, f, g} $\cap $ {a, c, e, g}. Thus, we have that $C\cap B$ = {e, g}.
Therefore, it is verified that $B\cap C=C\cap B$.
Now, we will consider the expression $A\cap C=C\cap A...(vi)$.
Now, we will consider the left hand side $A\cap C$ of (vi). Since, the elements of A = {a, b, c, d, e} and C = {b, e, f, g}. Therefore, we have
$A\cap C$ = {a, b, c, d, e} $\cap $ {b, e, f, g}. Thus, we have that $A\cap C$ = {b, e}.
Now, we will consider the right side $C\cap A$ of equation (vi). Since, the elements of A = {a, b, c, d, e} and C = {b, e, f, g}. Therefore, we have
$C\cap A$ = {b, e, f, g} $\cap $ {a, b, c, d, e}. Thus, we have that $C\cap A$ = {b, e}.
Therefore, it is verified that $A\cap C=C\cap A$.
Now, we will consider the expression $\left( A\cup B \right)\cup C=A\cup \left( B\cup C \right)...(vii)$.
Now, we will consider the left hand side $\left( A\cup B \right)\cup C$ 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
$\left( A\cup B \right)$ = {a, b, c, d, e} $\cup $ {a, c, e, g} = {a, b, c, d, e, g}
$\Rightarrow \left( A\cup B \right)\cup C$ = {a, b, c, d, e, g} $\cup $ {b, e, f, g}. Thus, we have that $\left( A\cup B \right)\cup C$ = {a, b, c, d, e, f, g}.
Now, we will consider the right side $A\cup \left( B\cup C \right)$ 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
$\left( B\cup C \right)$ = {a, c, e, g} $\cup $ {b, e, f, g} = {a, b, c, e, f, g}.
$\Rightarrow A\cup \left( B\cup C \right)$ = {a, b, c, d, e} $\cup $ {a, b, c, e, f, g}. Thus, we have that $A\cup \left( B\cup C \right)$ = {a, b, c, d, e, f, g}.
Therefore, it is verified that $\left( A\cup B \right)\cup C=A\cup \left( B\cup C \right)...(vii)$.
Now, we will consider the expression $\left( A\cap B \right)\cap C=A\cap \left( B\cap C \right)...(viii)$.
Now, we will consider the left hand side $\left( A\cap B \right)\cap C$ 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
$\left( A\cap B \right)$ = {a, b, c, d, e} $\cap $ {a, c, e, g} = {a, c, e}
$\Rightarrow \left( A\cap B \right)\cap C$ = {a, c, e} $\cap $ {b, e, f, g}. Thus, we have that $\left( A\cap B \right)\cap C$ = {e}.
Now, we will consider the right side $A\cap \left( B\cap C \right)$ 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
$\left( B\cap C \right)$ = {a, c, e, g} $\cap $ {b, e, f, g} = {e, g}
$A\cap \left( B\cap C \right)$ = {a, b, c, d, e} $\cap $ {e, g}. Thus, we have that $A\cap \left( B\cap C \right)$ = {e}.
Therefore, it is verified that $\left( A\cap B \right)\cap C=A\cap \left( B\cap C \right)$.
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.
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