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# 代數的證明isomorphism

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- 約翰教練Lv 49 years agoFavorite Answer
a*b=a+b+ab=(a+1)(b+1)-1(a)(1) Closure:Givena,b∈G ,claim: (a*b)∈GSupposenot, thenSincea,b∈G⊂ℝ ,a*b=(a+1)(b+1)-1∈ℝ, by any two elements in ℝ is closed under additionand multiplicationSincea*b∈ℝ and a*b∉G,a*b=-1⇒ (a+1)(b+1)-1=-1=0-1 ⇒ (a+1)=0or (b+1)=0 ⇒ a=-1 or b=-1That’sa contradiction on a,b∈G ⇒ (a*b)∈G(2) Associativity:Givena,b,c∊G⊂ℝ, then(a*b)*c=(a+b+ab)*c=(a+b+ab)+c+(a+b+ab)c=a+(b+c+bc)+a(b+c+bc)=a*(b+c+bc)=a*(b*c),by addition and multiplication is associative and commutative in ℝ(3) Identityelement:Lete=0∈G, then for any a∈Ge*a=e+a+ea=0+a+0=a,by 0 is the identity element under addition in ℝwehave that e is the left hand side identity in G(4) InverseelementGivena∈G⊂ℝ, then set a-1= -a/(1+a)Sincea∈G ⇒ a≠-1 ⇒ a-1∈ℝAssumea-1=-1 ⇒ -a=-1-a ⇒ a=a+1, that’s a contradiction on a=a+1 for any a is real ⇒ a-1∈(ℝ\{-1})=GAnda-1 *a=a-1+a+a-1a=(-a+a+a2-a2)/(1+a)=0=eWehave that a-1 is the left hand side inverse of a in GBy theorem, both of identity and inverseexist for left hand side with (1),(2),(3),(4) , we can conclude that is a group (b)Onto: Give a’∈ℝx, ∃a=a’-1∊G suchthat Φ(a)=a’One to one: Since Φ is onto, for any a’,b’∊ℝx, ∃a,b∈G such thatΦ(a)=a’, Φ(b)=b’. For a’=b’,we have that a+1=Φ(a)=Φ(b)=b+1 ⇒ a=bHomomorphism: Given a,b∈G, then Φ(a*b)=(a*b)+1=a+b+ab+1=(a+1)(b+1)=Φ(a)Φ(b)With above three properties, we canconclude that <G,*> and <ℝx,∙> are isomorphic and Φ is isomorphism

2011-11-21 00:16:55 補充：

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