Boolean algebra

Discussion in 'Homework Help' started by lusito92, Feb 4, 2013.

  1. lusito92

    Thread Starter New Member

    Jan 29, 2013
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    Hello,
    Im trying to solve this with identities

    problem.png

    xz+z /z + zxy = z
    z ( x +/x + xy) <-- x +/x = 1 (identity)
    z (xy) = z so now what? Im trying to find identity to get rid xy but I dont know how to do it ..
     
  2. DerStrom8

    Well-Known Member

    Feb 20, 2011
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    Where are you getting xz+z /z + zxy?

    Be careful with your distribution. Once you distribute everything correctly, you'll see a common factor.

    EDIT: Just realized you're using "/" as a "not" symbol. I suggest you use an apostrophe after the variable instead of a / before it.
     
    Last edited: Feb 4, 2013
  3. lusito92

    Thread Starter New Member

    Jan 29, 2013
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    Sorry..
    xz + z(x'+xy) = z <-- x' + xy = x' + y
    xz + z ( x' + y ) = z
    xz + zx' + zy = z < -- distributive
    z ( x + x' + y ) = z <- x + x' = 1
    z ( 1 + y ) = z <-- Im stuck! what can I do ?
     
  4. tshuck

    Well-Known Member

    Oct 18, 2012
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    what is any variable ORed with 1?
     
  5. tshuck

    Well-Known Member

    Oct 18, 2012
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    Code ( (Unknown Language)):
    1. x' + xy = x' + y
    I've never seen this identity, if that was taught as one, great, if not, I think you need to expand this out...
     
  6. lusito92

    Thread Starter New Member

    Jan 29, 2013
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    well in my notes that my teacher gave me says C' +CB = C+B
     
  7. tshuck

    Well-Known Member

    Oct 18, 2012
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    Well, then you're good. I've never seen given like that, but since your teacher gave that, use it!;)
     
  8. lusito92

    Thread Starter New Member

    Jan 29, 2013
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    but Im still stuck :( because you can see I carry out
    xz + z ( x' + y ) = z
    xz + zx' + zy = z < -- I expanded everything, and then the common factor is z and I got..
    z ( x + x' + y ) = z <- since I have another x + x' = 1 according to my teacher A + A' = 1
    z ( 1 + y ) = z <-- now Im not sure, I found another A + 1 = 1 ? is that can work 1 + y what do you think?
     
  9. DerStrom8

    Well-Known Member

    Feb 20, 2011
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    If anything is "OR"ed with a 1, the outcome is 1. So z (1 + y) is the same as z (1). What is Z AND 1?
     
  10. lusito92

    Thread Starter New Member

    Jan 29, 2013
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    z (1 ) = z :) is that the way I should do it ?
     
  11. DerStrom8

    Well-Known Member

    Feb 20, 2011
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    It works, doesn't it? :p:D
     
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  12. tshuck

    Well-Known Member

    Oct 18, 2012
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    ...sounds like a logical progression;)
     
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  13. lusito92

    Thread Starter New Member

    Jan 29, 2013
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    Now I have another that Im stuck sorry
    x'+y' + xyz'= x'+y'+ z'
    No idea how to do it .. the only think I can find x' y' could be xy'' ( xy are not) but there are not common factors or something in this equation that can lead me to do something
     
  14. DerStrom8

    Well-Known Member

    Feb 20, 2011
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    Remember the rule A' + AB' is A' + B'. If you rearrange a bit, you will see that this can help you twice.

    Matt
     
  15. lusito92

    Thread Starter New Member

    Jan 29, 2013
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    Sorry I dont know that rule, but I have this one with only one not .. A + A'B, but iff I have the whole term together because is XYZ' and the rule says A' + AB' ? is that applies?
     
  16. DerStrom8

    Well-Known Member

    Feb 20, 2011
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    Hmm, I was thinking it worked with two NOTs just as it would with only 1. I could be wrong though. I haven't done this in a while :p

    Perhaps I should let someone else jump in here, just in case my memory is failing me....
     
  17. lusito92

    Thread Starter New Member

    Jan 29, 2013
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    No problem :) I would wait lol I really need help :/
     
  18. tshuck

    Well-Known Member

    Oct 18, 2012
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    Let:
    A = x
    B = y'

    EDIT: formatted, per WBahn's advice...
    Code ( (Unknown Language)):
    1.  
    2. x' + xy = x' + y
    3. ^    ^^   ^    ^
    4. |    ||   |    |
    5. A'   A B' A'   B'
    6.  
     
    Last edited: Feb 4, 2013
  19. WBahn

    Moderator

    Mar 31, 2012
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    Try putting it in a CODE block. You probably will still see it as proportional spacing while you are typing, but it will show up as monospaced when posted (or previewed).
     
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  20. tshuck

    Well-Known Member

    Oct 18, 2012
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    This is confusing because it isn't an identity. It is valid, but you only see why when you go through the steps...
    A + A'B : Expand A
    A + AB + A'B : Factor B'
    A + B(A + A') : Using Identity A + A' = 1
    A + B(1) : Using identity A(1) = A
    => A + A'B = A + B

    So, if we were to use another form of B, namely B':
    A + A'B' : Expand A
    A + AB' + A'B' : Factor B'
    A + B'(A + A') : Using Identity A + A' = 1
    A + B'(1) : Using identity A(1) = A
    A + B'
    => A + A'B' = A + B'
     
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