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Homework Help: Binary field F2

  1. Sep 23, 2010 #1

    Suy

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    1. The problem statement, all variables and given/known data

    Find all solutions to the system of equations with over the binary field F2
    w+x +z=0
    x+y =1
    enter you answer as a list of point (w,x,y,z)
    2. Relevant equations



    3. The attempt at a solution
    This question should be easy, but I just don't understand binary field F2,
    I know 0+0=0 1+1=0 0+1=0
    Also , i tried to put (0,0,1,1)
    which is
    0+0 +1=0
    0+1 =1
    but it said wrong, please help me!
     
  2. jcsd
  3. Sep 24, 2010 #2

    Office_Shredder

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    Staff Emeritus
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    Gold Member

    0+0+1=1, not 0.

    You start with x+y=1. There are two possibilities: x=1 and y=0 or x=0 and y=1 (there are only four choices for x and y, and these are the only two that work). Solve for possible choices of w+x+z=0 for each case using similar logic
     
  4. Sep 24, 2010 #3

    Suy

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    0+0+0=0
    1+1+0=0
    0+1+1=0
    is that mean there are 9 choices?
    if 0+1=1, 0+0+0=0?
    answer is (0,0,1,0)?
    or 1+0=1
    (1,1,0,0)
    (0,1,0,1)
    there is three answer, which one is right???
    btw, what exactly is F2? I tried to google, but there isn't any answer
    ty
     
    Last edited: Sep 24, 2010
  5. Sep 24, 2010 #4

    HallsofIvy

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    The only possible values in the binary field are 0 and 1. There are two possible values for x, two for y, and two for z so there are (2)(2)(2)= 8 possible values for x, y, and z:
    0, 0, 0
    0, 0, 1
    0, 1, 0
    0, 1, 1
    1, 0, 0
    1, 0, 1
    1, 1, 0
    1, 1, 1
    0, 0, 1

    Since your two equations have 4 unknowns, x, y, z, and w, there are 2(2)(2)(2)= 16 possible sets of values but you don't have to look explicitely at all 16.

    Since x occurs in both equations, if x= 0, we have w+ z= 0 and y= 1. So we only have to solve w+ z= 0. There are 4 cases:
    w= 0, z= 0. Then w+ z= 0+ 0= 0
    w= 1, z= 0. Then w+ z= 1+ 0= 1
    w= 0, z= 1. Then w+ z= 0+ 1= 1
    w= 1, z= 1. Then w+ z= 1+ 1= 0.
    Two solutions are x= 0, y= 1, z= 0, w= 0 and x= 0, y= 1, z= 1, w= 1.

    If x= 1, then we have w+ 1+ z= 0 and 1+ y= 0. y= 1 is the same as y= "-1"= 1 and w+ 1+ z= 0 is the same as w+ z= "-1"= 1 ("-1" in this field is 1 because 1+ 1= 0).
    w= 0, z= 1 and w= 1, z= 0 both give w+ z= 1 so we also have x=1, y= 1, z= 1, w= 0 and x= 1, y= 1, z= 0, w= 1 .

    That gives a total of 4 distinct solutions.
     
  6. Sep 24, 2010 #5

    Suy

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    Thx! I probably confused with 1+1=0 those stuff, but like u said there is four answer, does it matter which one I put?
     
  7. Sep 24, 2010 #6

    Suy

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    Oh nvm, I put all four answer and it's correct
     
  8. Sep 25, 2010 #7
    hey i would like to know what exactly you put for your answer
     
  9. Sep 25, 2010 #8

    Suy

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    (x,x,x,x),(x,x,x,x)
    you can click preview to see if the format is correct
     
  10. Sep 25, 2010 #9
    Thanks that worked, are by any chance in math 211 using webwork
     
  11. Sep 25, 2010 #10
    are you by any chance...***
     
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