Question

LET A ,B, AND C BE THE SETS SUCH THAT A UNION B = A UNION C AND A INTERSECTION B =A INTERSECTIO C.SHOW THAT B=C.
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Answer 1

Given,

$\longrightarrow\sf{A\cup B=A\cup C}$

Taking intersection with $\sf{C,}$

$\longrightarrow\sf{(A\cup B)\cap C=(A\cup C)\cap C}$

$\longrightarrow\sf{(A\cap C)\cup(B\cap C)=(A\cap C)\cup (C\cap C)}$

$\longrightarrow\sf{(A\cap C)\cup(B\cap C)=(A\cap C)\cup C}$

Cancelling $\sf{A\cap C}$ from both sides,

$\longrightarrow\sf{B\cap C=C}$

$\Longrightarrow\sf{C\subseteq B\quad\quad\dots(1)}$

Similarly,

$\longrightarrow\sf{A\cup B=A\cup C}$

Taking intersection with $\sf{B,}$

$\longrightarrow\sf{(A\cup B)\cap B=(A\cup C)\cap B}$

$\longrightarrow\sf{(A\cap B)\cup(B\cap B)=(A\cap B)\cup (C\cap B)}$

$\longrightarrow\sf{(A\cap B)\cup B=(A\cap B)\cup (B\cap C)}$

Cancelling $\sf{A\cap B}$ from both sides,

$\longrightarrow\sf{B=B\cap C}$

$\Longrightarrow\sf{B\subseteq C\quad\quad\dots(2)}$

Combining (1) and (2), we get,

$\longrightarrow\underline{\underline{\sf{B=C}}}$

Hence Proved!

Properties Used:-

$\sf{1.\quad A\cap B=B\cap A\quad (Commutative\ Law)}$

$\sf{2.\quad A\cap A=A\quad(Idempotent\ Law)}$

$\sf{3.\quad (A\cup B)\cap C=(A\cap C)\cup(B\cap C)\quad(Distributive\ Law)}$

$\sf{4.\quad A\cap B=B\quad\iff\quad B\subseteq A}$

$\sf{5.\quad A=B\quad\iff\quad A\subseteq B\quad\!and\quad\!B\subseteq A}$

Answer 2

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

FORMULA TO BE IMPLEMENTED

$1. \: \sf{ \: A \cup \: B = \: B \cup \:A }$

$2. \: \sf{ \: A \cap \: B = \: B \cap \:A }$

$3. \: \: \sf{ A \cap \: (A \cup \: B )\: } = A$

$4. \: \: \sf{ (A\cap\: B)\cup (A \cap \: C ) \: = A\cap\:( B \cup \: C ) \: }$

GIVEN

$\sf{ \: A \cup \: B = A \cup \: C \: \: and \: A \cap \: B = A \cap \: C \: \: }$

TO PROVE

$\sf{B = C}$

PROOF

$\sf{B}$

$\sf{ = B \cap \: (B \cup \: A )\: } \: \: (using \: \: formula \: 3)$

$\sf{ = B \cap \: (A \cup \: B )\: }$

$\sf{ = B \cap \: (A \cup \: C )\: } \: \:( \: given)$

$\sf{ =( B \cap \: A )\cup (B \cap\: C )\: }(using \: \: formula \: 4)$

$\sf{ =(B \cap\: C )\cup ( B \cap \: A ) \: \: }(using \: \: formula \: 1)$

$\sf{ =( C\cap\: B)\cup ( A \cap \: C ) \: \: }(given)$

$\sf{ =( C\cap\: B)\cup ( C \cap \: A ) \: \: }(using \: \: formula \: 2)$

$\sf{ = C \: \cap\: ( \: B\cup \: A ) \: \: }(using \: \: formula \: 4)$

$\sf{ = C \: \cap\: ( \: C\cup \: A ) \: \: }(given)$

$\sf{ = C} \: \: \: \: (using \: \: formula \: 3)$

Hence proved