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Math Help - Isomorphism correction?

  1. #1
    Senior Member sfspitfire23's Avatar
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    Isomorphism correction?

    I'm having trouble showing how \alpha(1/4)=a^2\Rightarrow 1/4=a^2. How would I do this?

    We must show that \mathbb{Q} and \mathbb{Z} are not isomorphic. Consider the mapping \alpha : \mathbb{Q}  \rightarrow \mathbb{Z}. Then we have \alpha(1/2)=a. Then we have \alpha(1/2)^2=a^2 which can be written as \alpha((1/2)^2)=a^2. So, \alpha(1/4)=a^2. We can pull this apart to get \alpha(1/8)+\alpha(1/8)=a^2. to yield 1/4=a^2. But there is nothing in \mathbb{Z} that squares to 1/4. Therefore, they are not isomorphic.
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  2. #2
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    Quote Originally Posted by sfspitfire23 View Post
    I'm having trouble showing how \alpha(1/4)=a^2\Rightarrow 1/4=a^2. How would I do this?

    We must show that \mathbb{Q} and \mathbb{Z} are not isomorphic. Consider the mapping \alpha : \mathbb{Q} \rightarrow \mathbb{Z}. Then we have \alpha(1/2)=a. Then we have \alpha(1/2)^2=a^2 which can be written as \alpha((1/2)^2)=a^2. So, \alpha(1/4)=a^2. We can pull this apart to get \alpha(1/8)+\alpha(1/8)=a^2. to yield 1/4=a^2. But there is nothing in \mathbb{Z} that squares to 1/4. Therefore, they are not isomorphic.
    You write your questions in a very sloppy way...this can't be good for you: I suppose you mean to prove \mathbb{Q}\,,\,\mathbb{Z} aren't isomorphic...as rings, right?
    Well, I can't understand why you think, or want it to be true, that if \alpha(1\slash 8)+\alpha(1/8)=a^2 , then 1\slash 4=a^2...

    Following your idea, I'd simply say \alpha(1/2)=a\Longrightarrow a^2=\alpha(1/2)^2=\alpha(1/4)\Longrightarrow a=\alpha(1/2)=\alpha(2\cdot 1/4)=2\alpha(1/4)=2a^2 , and get from here your contradiction.

    Tonio
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