# Find feild where 1 = 0 and show that it is the only one

1. Jan 13, 2009

### rosh300

1. The problem statement, all variables and given/known data
Define a paddock to be a set in which A1 - A4, M1 - M4 and D holds but instead of 1$$\neq$$ 0, we have 1 = 0. Find an example of a paddock, and show that your example is the only one

2. Relevant equations
A1 - A4, M1 - M4 and D are all axioms

A1: a + b = b + a
A2: (a + b) + c = a + (b + c)
A3: there is a 0 s.t a + 0 = 0 + a = a
A4 for every a there exsist -a s.t a + (-a) = 0

multiplication axioms:
M1: a.b = b.a
M2: (a.b).c = a.(b.c)
M3: there is a 1 s.t a.1 = 1.a = a
M4: for every a there exsist a-1 s.t a.a-1 = 1

D (a+b)c = ab + bc
a, b, c, 1, 0, inverses all belong in the set

3. The attempt at a solution
the set {1, 0} closed under +1 and x1 (modulo multiplication of 1 and modulo addition of 1).
1 = 0 because both 0 and 1 are the neutral element,
if it is true how how would i go about showing its the only one. im guessing it requires proof by contradiction.

2. Jan 14, 2009

### HallsofIvy

Staff Emeritus
M3 tells you that a*1= a for all a.

By D, a*(b+ 0)= a*b+ a*0 but by A3, b+ 0 = b so a*(b+ 0)= a*b. That is, a*(b+0) is equal to both a*b and a*b+ a*0 so they are equal: a*b= a*b+ a*0 and, adding the additive inverse of a*b to both sides, a*0= 0 for all a.

Now you have both a*1= a and a*0= 0 as well as 0= 1. Put them together.

3. Jan 18, 2010

### vintwc

May I ask, how does putting 'a*1= a and a*0= 0 as well as 0= 1' together explicitly shows that this example of a paddock is the only one?

4. Jan 19, 2010

### HallsofIvy

Staff Emeritus
a*1= a and a*0= 0 are the same thing because 0= 1. Therefore a= 0 for all a. The "paddock" contains only a single element.