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Theorem 2 (For any three set `A;B;C`; prove that `Axx(B-C)=(AxxB)-(AxxC)`

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Theorem 2 (For any three set `A;B;C`; prove that `Axx(B-C)=(AxxB)-(AxxC)`

This is the Solution of Question From RD SHARMA book of CLASS 11 CHAPTER RELATIONS AND FUNCTIONS This Question is also available in R S AGGARWAL book of CLASS 11 You can Find Solution of All Question From RD SHARMA and R S AGGARWAL book from class 6-12 on our app DOUBTNUT for Free #rdsharmasolution #RELATIONSANDFUNCTIONS #rdsharmaclass11 # ...

6.5K viewsAug 13, 2018

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Theorem 1(ii) (For any three set `A;B;C` ; prove that `Axx(BnnC)=(AxxB)nn(AxxC)`

Theorem 1(ii) (For any three set `A;B;C` ; prove that `Axx(BnnC)=(AxxB)nn(AxxC)`

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If `A={1,2,3},B={3,4}" and "C={4,5,6}` then find: (i) `Axx(BnnC)` (ii) `(AxxB)nn(AxxC)`

If `A={1,2,3},B={3,4}" and "C={4,5,6}` then find: (i) `Axx(BnnC)` (ii) `(AxxB)nn(AxxC)`

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Let `A` be a non-empty set such that `AxxB=AxxC` . Show that `B=C` .

Let `A` be a non-empty set such that `AxxB=AxxC` . Show that `B=C` .

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Theorem 1(ii) (For any three set `A;B;C` ; prove that `Axx(BnnC)=(AxxB)nn(AxxC)`

Theorem 1(ii) (For any three set `A;B;C` ; prove that `Axx(BnnC)=(…

1.7K viewsDec 20, 2019

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If `A={1,2,3},B={3,4}" and "C={4,5,6}` then find: (i) `Axx(BnnC)` (ii) `(AxxB)nn(AxxC)`

If `A={1,2,3},B={3,4}" and "C={4,5,6}` then find: (i) `Axx(BnnC)` (ii) `(Axx…

8K viewsApr 19, 2020

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Let `A` be a non-empty set such that `AxxB=AxxC` . Show that `B=C` .

Let `A` be a non-empty set such that `AxxB=AxxC` . Show that `B=C` .

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If `A={1,2,3},B={4},C={5},` then verify that: `Axx(BuuC)=(AxxB)uu(AxxC)` `Axx(BnnC)=(AxxB)nn(Axx...

If `A={1,2,3},B={4},C={5},` then verify that: `Axx(BuuC)=(AxxB)uu(AxxC)…

2.1K viewsSep 19, 2017

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Theorem 1(i) (For any three set `A;B;C`; prove that `Axx(BuuC)=(AxxB)uu(AxxC))`

Theorem 1(i) (For any three set `A;B;C`; prove that `Axx(BuuC)=(A…

46 viewsDec 20, 2019

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$For any three sets `A ,B ,C` prove that: `Axx(B^(prime)uuC^(prime))^\\'=(AxxB)nn(AxxC)` `Axx(B^(p...$

For any three sets `A ,B ,C` prove that: `Axx(B^(prime)uuC^(prime))^…

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`(AuuB)xxC=(AxxC)uu(BxxC)`

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