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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?!)

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    it's correct because false <-> false.2012-11-29
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    Only $a\cdot(b+c) = a\cdot b + a\cdot c$ is true, $a + (b\cdot c) \neq (a+b)\cdot(a+c)$. Moreover it is a field because $\mathbb{Z}_p$ is a field for any prime $p$.2012-11-29
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    sure?! or are you joking?because if 1=true it will never implicate 0=false.2012-11-29
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    Existence of inverses if fine. Note that you only have to show left inverses (or, if you prefer, right inverses) because you prove (later) that the multiplication is commutative. For 2.6F, 2.7F, note that both sides are equal to 1, not to 0.2012-11-29
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    @dtldarek ok, so it is still field, even when the 2. distributive law is not fullfilled?2012-11-29
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    @phil There is no second distributive law. Real numbers are a field, does $2+(2\cdot2) = (2+2)\cdot(2+2)$?2012-11-29
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    well :), then my source was wrong...I will delete that.2012-11-29
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    @BrettFrankel now I see it, with all these numbers you slightly lose track.About the inverse:that means that 0*0=0 doesn't have inverse or is in this case 0 also an inverse?2012-11-29
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    $0$ never has an inverse since one can prove that $0x=0$ for all $x$ and this would imply $0=1$ if there were an inverse.2012-11-29

0 Answers 0