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Math Help - Algebraic Manipulation

  1. #1
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    Cool Algebraic Manipulation

    This is Boolean Algebra. + means OR, the . means AND and the x! means X NOT.

    Use algebraic manipulation to prove that x +yz = (x + y).(x+z)



    x + yz = (x+y) AND (x + z) ...

    x +yz = xx + xz + xy + xy + yz
    = x(x + z + y) + yz
    = x + z + y = 1
    = x AND 1 = x
    x +yz =x + yz

    is that correct?

    Next problem: prove that (x+y)(x+y!) = x
    = xx + xy! + xy + yy!
    = x(x + y! + y) + yy!
    what does x + y! + y evaluate to? is it 1?
    in that case;
    =x AND 1 = x
    But what does yy! evaluate to?
    please help.

    I also need to use algebraic manipulation to prove that xy + yz +x!z = xy + x!z using the consensus property.

    Thank you for your time,
    Sham
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  2. #2
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    Re: Algebraic Manipulation

    Hi Sham,
    Proving Boolean algebra equations is just like proving other equations. You should start with one side, and, by a chain of equations, reach the other side.

    (x+y)(x+z) = xx+xz+xy+yz = x + xz +xy +yz = x(1 + z +y) + yz = x1+yz = x + yz

    (x+y)(x+y!) = xx +xy +xy! +yy! = x + xy +xy! + 0 = x(1+y+y!) = x1 = x
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  3. #3
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    Re: Algebraic Manipulation

    John, thanks so much! But what is (1 +y + y!) = to? I know y*y! = 0 but what about (1+y+y!)?
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  4. #4
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    Re: Algebraic Manipulation

    y OR y! is 1. If y is false then not y is true

    Also 1 OR [anything]=1
    Do you understand the OR opperation?
    X OR Y is asking if either X is true or Y is true. Therefore 1 OR X is asking if either 1 is true or X is true. 1 is always true so regardless of the value of X: 1 OR X is always true
    Last edited by Shakarri; September 12th 2013 at 09:55 AM.
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  5. #5
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    Re: Algebraic Manipulation

    Hi Shamieh,
    In Boolean algebra 1 + anything = 1; i.e. true or anything is true no matter what the truth value of anything.
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  6. #6
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    Re: Algebraic Manipulation

    Hi Shamieh,
    In Boolean algebra 1 + anything = 1; i.e. true or anything is true no matter what the truth value of anything.
    So for example, x(1 + x +y +x +y +x +z +x +y) you are saying that that will evaluate to 1? Getting x AND 1 which evaluates to x for example? And I'm not sure what shakari is talking about because y AND y! definitely evaluates to 0. Unless my book is lying...
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