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Q4(iii):
Show that the following four conditions are equivalent : (iii) $A \cup B = B$
Solution :
Theoretical Foundation & Logical Equivalence
To demonstrate that a condition $A \cup B = B$ is equivalent to $A \subseteq B$, we must prove the bidirectional implication (biconditional statement): $A \subseteq B \iff A \cup B = B$. This relies on the Axioms of Set Theory, specifically the definitions of Union and Subset relations.
Step 1: Forward Implication ($\implies$) - Proving $A \subseteq B \implies A \cup B = B$
Assume $A \subseteq B$. We aim to show that $A \cup B = B$ by demonstrating mutual inclusion ($A \cup B \subseteq B$ and $B \subseteq A \cup B$).
- Part A: Show $A \cup B \subseteq B$
Let $x \in A \cup B$. By the definition of union, $x \in A$ or $x \in B$ [Definition of Set Union].
Case 1: If $x \in A$, then since $A \subseteq B$, it follows that $x \in B$ [Definition of Subset].
Case 2: If $x \in B$, the condition is trivially satisfied.
Thus, $x \in B$ in all cases, confirming $A \cup B \subseteq B$. - Part B: Show $B \subseteq A \cup B$
For any set $B$, it is a fundamental property that any element belonging to $B$ also belongs to the union of $B$ with any set $A$ [Identity/Monotonicity of Union].
Formally, if $y \in B$, then $y \in B$ or $y \in A$, which implies $y \in A \cup B$.
Thus, $B \subseteq A \cup B$.
By the Axiom of Extensionality, since $A \cup B \subseteq B$ and $B \subseteq A \cup B$, we conclude $A \cup B = B$.
Step 2: Reverse Implication ($\impliedby$) - Proving $A \cup B = B \implies A \subseteq B$
Assume $A \cup B = B$. We aim to show $A \subseteq B$.
- Let $x$ be an arbitrary element such that $x \in A$.
- By the property of union, $x \in A \cup B$ [Definition of Union].
- Since we are given that $A \cup B = B$, it follows by substitution that $x \in B$.
- Since $x \in A$ implies $x \in B$, it follows by definition that $A \subseteq B$ [Definition of Subset].
Step 3: Visual Representation (Venn Diagram)
Summary of Logic
| Logical Direction | Process | Conclusion |
|---|---|---|
| $A \subseteq B \implies A \cup B = B$ | Subset Inclusion | Verified via Element-wise Analysis |
| $A \cup B = B \implies A \subseteq B$ | Identity Substitution | Verified via Definition of Union |
Final Solution: Since the forward and reverse implications hold true, it is rigorously established that $A \cup B = B$ is logically equivalent to $A \subseteq B$.
More Questions from Class 11 Mathematics Sets Miscellaneous Exercise on Chapter 1
- Q1: Decide, among the following sets, which sets are subsets of one and another: $A = \{ x : x \in R$ and $x$ satisfy $x^2 – 8x + 12 = 0 \}$, $B = \{ 2, 4, 6 \}$, $C = \{ 2, 4, 6, 8, . . . \}$, $D = \{ 6 \}$.
- Q2(i): In each of the following, determine whether the statement is true or false. If it is true, prove it. If it is false, give an example. (i) If $x \in A$ and $A \in B$, then $x \in B$
- Q2(ii): In each of the following, determine whether the statement is true or false. If it is true, prove it. If it is false, give an example. (ii) If $A \subset B$ and $B \in C$, then $A \in C$
- Q2(iii): In each of the following, determine whether the statement is true or false. If it is true, prove it. If it is false, give an example. (iii) If $A \subset B$ and $B \subset C$, then $A \subset C$
- Q2(iv): In each of the following, determine whether the statement is true or false. If it is true, prove it. If it is false, give an example. (iv) If $A \not\subset B$ and $B \not\subset C$, then $A \not\subset C$
- Q2(v): In each of the following, determine whether the statement is true or false. If it is true, prove it. If it is false, give an example. (v) If $x \in A$ and $A \not\subset B$, then $x \in B$
- Q2(vi): In each of the following, determine whether the statement is true or false. If it is true, prove it. If it is false, give an example. (vi) If $A \subset B$ and $x \notin B$, then $x \notin A$
- Q3: Let $A, B$, and $C$ be the sets such that $A \cup B = A \cup C$ and $A \cap B = A \cap C$. Show that $B = C$.
- Q4(i): Show that the following four conditions are equivalent : (i) $A \subset B$
- Q4(ii): Show that the following four conditions are equivalent : (ii) $A – B = \phi$
- Q4(iv): Show that the following four conditions are equivalent : (iv) $A \cap B = A$
- Q5: Show that if $A \subset B$, then $C – B \subset C – A$.
- Q6: Show that for any sets $A$ and $B$, $A = ( A \cap B ) \cup ( A – B )$ and $A \cup ( B – A ) = ( A \cup B )$
- Q7(i): Using properties of sets, show that (i) $A \cup ( A \cap B ) = A$
- Q7(ii): Using properties of sets, show that (ii) $A \cap ( A \cup B ) = A$.
- Q8: Show that $A \cap B = A \cap C$ need not imply $B = C$.
- Q9: Let $A$ and $B$ be sets. If $A \cap X = B \cap X = \phi$ and $A \cup X = B \cup X$ for some set $X$, show that $A = B$. (Hints $A = A \cap ( A \cup X ) , B = B \cap ( B \cup X )$ and use Distributive law )
- Q10: Find sets $A, B$ and $C$ such that $A \cap B, B \cap C$ and $A \cap C$ are non-empty sets and $A \cap B \cap C = \phi$.
CBSE Solutions for Class 11 Mathematics Sets
Chapters in CBSE - Class 11 Mathematics
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