Question

Prove or disprove: For any sets A and B, P(A x B) = P(A) x P(B).​

Answer 1

Answer:

$\huge \bf{answer:-} \\ \rm \: For \: any \: set  \: AA, we \: let \:  P(A) \: P(A) \:  denotes \: \\ \rm \: the \: power \: set \: of  \: AA  \\ \rm i.e. \: the \: set \: of \: all \: subsets \: of  \: AA. \\ \rm \: We \: will \: show \: that: \\ \\ { \bigstar} \rm \: P(A)∪P(B)⊆P(A∪B)P(A)∪P(B)⊆P(A∪B) \\ </p><p> \rm \: Since \:  A⊂A∪B \: so \: every \: subset \: of  \: AA \\ \:   \rm \: is \: also \: a \: subset \: of  \: A∪BA∪B, \\ \rm \: therefore \:  P(A)⊆P(A∪B)P(A)⊆P(A∪B). \\ \rm \: Like \: \rm \: wise, P(B)⊆P(A∪B)P(B)⊆P(A∪B). \\ \rm \: Therefore, P(A)∪P(B)⊆P(A∪B)P(A)∪P(B)⊆P(A∪B). \\ P(A∪B)⊈P(A)∪P(B)P(A∪B)⊈P(A)∪P(B) \\ </p><p></p><p>Suppose A={1}A={1}, B={2}B={2} . So, A∪B={1,2}A∪B={1,2}.</p><p></p><p>P(A)={∅,{1}},P(A)={∅,{1}}, P(B)={∅,{2}}P(B)={∅,{2}}</p><p></p><p>P(A∪B)={∅,{1},{2},{1,2}}⊈{∅,{1},{2}}=P(A)∪P(B)</p><p>$

Answer 2

my sister's right answer friend ⬆⬆⬆⬆⬆

the is Right in the poem