# Binary field F2

1. Sep 23, 2010

### Suy

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

2. Sep 24, 2010

### Office_Shredder

Staff Emeritus
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

3. Sep 24, 2010

### Suy

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?
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
4. Sep 24, 2010

### HallsofIvy

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.

5. Sep 24, 2010

### Suy

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?

6. Sep 24, 2010

### Suy

Oh nvm, I put all four answer and it's correct

7. Sep 25, 2010

### enggstu

hey i would like to know what exactly you put for your answer

8. Sep 25, 2010

### Suy

(x,x,x,x),(x,x,x,x)
you can click preview to see if the format is correct

9. Sep 25, 2010

### enggstu

Thanks that worked, are by any chance in math 211 using webwork

10. Sep 25, 2010

### enggstu

are you by any chance...***