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mathmath333

  • one year ago

Set Theory

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  1. mathmath333
    • one year ago
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    \(\large \color{black}{\begin{align} \normalsize \text{A ∩ B = A ∩ C need not imply B = C.} \quad \normalsize \text{ Explain through an example.}\hspace{.33em}\\~\\ \end{align}}\)

  2. Kainui
    • one year ago
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    A = {1} B = {1, 2} C = {1, 3}

  3. ikram002p
    • one year ago
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    huh kai was so fast :P

  4. Kainui
    • one year ago
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    Hahaha there could have been a simpler example: A = {} B = {1} C = {2}

  5. mathmath333
    • one year ago
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    thnx

  6. Kainui
    • one year ago
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    I think that last example I gave sorta shows that it's almost like this is a lot like multiplying by zero. a*b=a*c This doesn't mean b=c, since a=0 is possible.

  7. mathmath333
    • one year ago
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    Let A and B be sets. If A ∩ X = B ∩ X = φ and A ∪ X = B ∪ X for some set X, show that A = B.

  8. ikram002p
    • one year ago
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    in general any case with \[ A\cap B \neq \varnothing \] does not imply A=C

  9. ikram002p
    • one year ago
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    or \(C\cap B \neq \varnothing \)

  10. mathmath333
    • one year ago
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    i need an example like the previous one

  11. Kainui
    • one year ago
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    ikram needs more owl bucks sry

  12. ikram002p
    • one year ago
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    A={1,2,3} B={1,2,3} C={1,2,3,4}

  13. ikram002p
    • one year ago
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    :O @Kainui i dont lol

  14. Kainui
    • one year ago
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    Hahaha jk :P

  15. mathmath333
    • one year ago
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    is that the example for this que Let A and B be sets. If A ∩ X = B ∩ X = φ and A ∪ X = B ∪ X for some set X, show that A = B.

  16. ikram002p
    • one year ago
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    you wanna prove this ?

  17. mathmath333
    • one year ago
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    yes with example

  18. ikram002p
    • one year ago
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    ok example A={1,2,3} B={1.4} C={1,2.3}

  19. mathmath333
    • one year ago
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    but C is not there in the question

  20. ikram002p
    • one year ago
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    sorry i ment to say x :P

  21. ikram002p
    • one year ago
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    now to prove they are two side of the prove 1- show A is a subset of B 2- show B is a subset of A

  22. ikram002p
    • one year ago
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    let \(c\in A \cup X\) so \( c\in A ~or~c\in X \) if c in A then \( c\in B\cup X \) and \( c\in B\) (since c is not in x) thus \(A\subset B\) the other way is alike :P sorry im lazy to type

  23. thomas5267
    • one year ago
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    Is it possible to do something like this? \[ A\cup X=B\cup X\land A\cap X=B\cap X=\emptyset\implies A\cup X\setminus X=B\cup X\setminus X\implies A=B \]

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