Prove that the multiplication $*:M \times M \to M$ defined by this table:
* | 0 1
--------
0 | 0 0
1 | 0 1
together with the commutative group (M,+), is a field (M,+,*).
Group axioms:
1) Closure:
$0*0=0 \in M$
$0*1=0 \in M $
$ 1*0=0 \in M$
$ 1*1=1 \in M $
2) Inverse element:
$1*1=1 $
$ 1*1=1 $
$\forall i,a,e \in M : a*i=i*a=e$
here the inverse element is i=e=1.
3) Identity element:
$ 0*1=0 $
$1*0=0$
$1*1=1$
$\forall e,a\in M: e*a=a*e=a $ with $e=1$
4) Associativity:
4.1) $0*(0*0)=0 \leftrightarrow (0*0)*0=0$
4.2) $0*(0*1)=0 \leftrightarrow (0*0)*1=0$
4.3) $0*(1*0)=0 \leftrightarrow (0*1)*0=0$
4.4) $0*(1*1)=0 \leftrightarrow (0*1)*1=0$
4.5) $1*(0*0)=0 \leftrightarrow (1*0)*0=0$
4.6) $1*(0*1)=0 \leftrightarrow (1*0)*1=0$
4.7) $1*(1*0)=0 \leftrightarrow (1*1)*0=0$
4.8) $1*(1*1)=1 \leftrightarrow (1*1)*1=1$
$\forall a,b,c\in M : a*(b*c)=(a*b)*c$
the addition $+:M \times M \to M$ defined by:
$+ | 0$ $1$
------------
$0 | 0$ $1$
$1 | 1$ $0$
Field condition:
1F) Commutativity:
$0*0=0=0*0$
$ 1*0=0=0*1 $
$ 1*1=1=1*1 $
2F) Distributivity:
2.1F) $0*(0+0)=0 \leftrightarrow (0*0)+(0*0)=0$
2.2F) $0*(0+1)=0 \leftrightarrow (0*0)+(0*1)=0$
2.3F) $0*(1+0)=0 \leftrightarrow (0*1)*(0*0)=0$
2.4F) $0*(1+1)=0 \leftrightarrow (0*1)+(0*1)=0$
2.5F) $1*(0+0)=0 \leftrightarrow (1*0)+(1*0)=0$
2.6F) $1*(0+1)=1 \leftrightarrow (1*0)+(1*1)=1$
2.7F) $1*(1+0)=1 \leftrightarrow (1*1)+(1*0)=1$
2.8F) $1*(1+1)=0 \leftrightarrow (1*1)+(1*1)=0$
$\forall a,b,c\in M : a*(b+c)=(a*b)+(a*c)$
Could you please tell me if I have made a mistake?
And what about 2) the inverse element is that correct?(because in the "usual multiplication" the inverse element i is $i=a^{-1}$ defined?!)