性质30a=0,(-1)a=-a,k0=0. 因为a+0a=1a+0a=(1+0)a=1a=a 所以0a=0+0a=(-a+a)+0a a+(a+0a)=0 又因为a+(-1)a=1a+(-1a=[1+(-1)a=0a=0 所以(-1)a=0+(-1a=(-a+a)+(-1)a a+|a+(-1)ax]=-a+0=-a TiO k0=k[a+(1a]=ka+(k)a =[k+(-k)a=0a=0 上页 下页上页 下页 性质３ 0 = 0 , ( − 1 ) = − , k 0 = 0 . 因为  + 0 = 1 + 0 = ( 1 + 0 ) = 1 =  所以 ( 0 ) 0 0 0 0 ( ) 0 = − + + = = + = − + +         又因为  + (− 1) = 1 + (− 1) = [1 + (− 1)] = 0 = 0 所以           = − + + − = − + = − − = + − = − + + − [ ( 1 ) ] 0 ( 1 ) 0 ( 1 ) ( ) ( 1 ) 而 [ ( )] 0 0 0 [ ( 1 ) ] ( ) = + − = = = + − = + −       k k k k k k