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Show that A nnB = A nnBimplies A = B...

Show that `A nnB = A nnB`implies `A = B`

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let a be an element
`a in A`
then `a in (A uu B)`
`=> a in (A nn B)` (given)
so, `a in B`
if `b in B`
then `b in (A uu B)`
`=> b in (A nn B) => b in A`
...
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Knowledge Check

  • a.b =0 implies only

    A
    `a = 0`
    B
    ` b = 0`
    C
    `theta = 90^(@)`
    D
    either a = 0 or b = 0 or `theta = 90^(@)`

  • Let A={(xy) : X+y=p}, Be ={(x,y) : x^3+y^3 two sets sch that A nnB=phi for any real number p, than the exhastive intervals of p is

    A
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    B
    `{0}uu[2,oo)`
    C
    `(-oo,-2]uu{0}`
    D
    `{x!x in N ^^ x le 3}`

  • Consider the following relations: 1. A - B = A - (A nnB) 2. A = (A nn B) uu (A -B) 3. A - (B uu C) = (A - B) uu (A -C) Which of these is/are correct ?

    A
    1 and 3
    B
    2 only
    C
    2 and 3
    D
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