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Let `A={x inN:x^(2)-5x+6=0},B={x inW:0lexlt2}`and `C={x inN:xlt3}.` Verify that (i) `Axx(BuuC)=(AxxB)uu(AxxC)` (ii) `Axx(BnnC)=(AxxB)nn(AxxC)` |
Answer» We have `A={x inN:x^(2)-5x6=0}={x inN:(x-2)(x-3)=0}={1,2).` `:." "A={2,3},B={0,1}" and "C={1,2}.` (i) `(BuuC)={0,1}uu{1,2}={0,1,2}.` `:." "Axx(BuuC)={2,3}xx{0,1,2}` `={(2,0),(2,1),(2,2),(3,0),(3,1),(3,2)}.` Now, `(AxxB)={2,3}xx{0,1}` `={(2,0),(2,1),(3,0),(3,1)}` and `(AxxC)={2,3}xx{1,2}` `={(2,1),(2,2),(3,1),(3,2)}.` `:." "(AxxB)uu(AxxC)={(2,0),(2,1),(2,2),(3,0),(3,1),(3,2)}.` Hence, `Axx(BuuC)=(AxxB)uu(AxxC).` (ii) `(BuuC)={0,1}nn{1,2}={1}.` `:." "Axx(BnnC)={2,3}xx{1}={(2,1),(3,1)}.` And, `(AxxB)nn(AxxC)={(2,0),(2,1),(3,0),(3,1)}nn{(2,1),(2,2),(3,1),(3,2)}` `={(2,1),(3,1)}.` Hence, `Axx(BnnC)=(AxxB)nn(AxxC).` |
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