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Q4(iv):
Show that the following four conditions are equivalent : (iv) $A \cap B = A$
Solution :
Step 1: Theoretical Foundation of Set Equivalence
To demonstrate that the condition $A \cap B = A$ is equivalent to the fundamental inclusion relation of subsets, we must analyze the logical implications of the intersection operation under the axioms of Set Theory [Per Zermelo-Fraenkel Set Theory axioms].
Step 2: Logical Derivation of the Condition
We analyze the statement $A \cap B = A$ using the definition of set intersection: $A \cap B = \{x : x \in A \text{ and } x \in B\}$.
- Forward Implication ($\implies$): Assume $A \cap B = A$. By the definition of intersection, for any element $x \in (A \cap B)$, it must be that $x \in A$ and $x \in B$. Since our assumption states $A \cap B = A$, any element $x \in A$ must also be an element of $A \cap B$. Therefore, if $x \in A$, then $x \in (A \cap B)$, which implies $x \in A$ and $x \in B$. Consequently, $x \in B$. Thus, $A \subseteq B$.
- Backward Implication ($\impliedby$): Assume $A \subseteq B$. To show $A \cap B = A$, we must demonstrate both $A \cap B \subseteq A$ and $A \subseteq A \cap B$.
- By the definition of intersection, $A \cap B \subseteq A$ is always true [By the properties of set operations].
- If $A \subseteq B$, then for any element $x \in A$, it follows that $x \in B$. Since $x \in A$ and $x \in B$, it follows that $x \in (A \cap B)$. Thus, $A \subseteq A \cap B$.
- Since $A \cap B \subseteq A$ and $A \subseteq A \cap B$, by the principle of double inclusion, $A \cap B = A$.
Step 3: Visual Representation of the Equivalence
The following SVG illustrates the scenario where $A \subseteq B$, resulting in $A \cap B = A$.
Step 4: Formal Summary of Equivalent Conditions
The condition $A \cap B = A$ holds if and only if $A$ is a subset of $B$. This relationship is fundamentally defined by the behavior of elements residing in the intersection of two sets where the constraints of the subset relation are satisfied.
Final Solution: The condition $A \cap B = A$ 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(iii): Show that the following four conditions are equivalent : (iii) $A \cup B = B$
- 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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