代數學的證明 In integral set Z

a*b = a+b if a is even, and = a-b if a is odd.

First, a*b in Z for all a, b in Z.
So, "*" is a binary operation on Z.

0 is the unit element for the operation "*":
a*0 = a+0 or a-0 depends whether a is even or odd.
Both cases lead to the result a*0 = a.
Since 0 is even, 0*a = 0+a = a.

Let a in Z, we have to show that a has an inverse.
When a even, a*(-a) = a+(-a) = 0;
when a odd, a*a = a-a = 0.
This proved that each a in Z has an inverse under the opeartion "*".

For a, b, c in Z.
Case 1: both a, b are even, then
(a*b)*c = (a+b)*c = a+b+c, a*(b*c) = a+(b*c) = a+b+c.
Case 2: both a, b are odd, then
(a*b)*c = (a-b)*c = (a-b)+c, a*(b*c) = a-(b-c) = (a-b)+c.
Case 3: a even, b odd, then
(a*b)*c = (a+b)*c = (a+b)-c, a*(b*c) = a+(b-c) = (a+b)-c.
Case 4: a odd. b even, then
(a*b)*c = (a-b)*c = (a-b)-c, a*(b*c) = a-(b+c) = (a-b)-c.
Since all cases lead to (a*b)*c = a*(b*c), the associative property
is proved.


Thus, (Z,*) is a group.

Note that, a*b≠b*a. For example,
suppose that a odd and b even, nonzero, then
a*b = a-b while b*a = b+a = a+b.
Since b≠0, a+b≠a-b, that is, a*b≠b*a.
Therefore, (Z,*) is non-abelian.



2011-11-02 02:11:56 補充：
Let a in Z, we have to show that a has an inverse.
When a even, a*(-a) = a+(-a) = 0,
　　　　　 and ; (-a)*a = (-a)+a = 0.
when a odd, a*a = a-a = 0.
This proved that each a in Z has an inverse under the opeartion "*".