1. ## Isomorphism correction?

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

We must show that $\displaystyle \mathbb{Q}$ and $\displaystyle \mathbb{Z}$ are not isomorphic. Consider the mapping $\displaystyle \alpha : \mathbb{Q} \rightarrow \mathbb{Z}$. Then we have $\displaystyle \alpha(1/2)=a$. Then we have $\displaystyle \alpha(1/2)^2=a^2$ which can be written as $\displaystyle \alpha((1/2)^2)=a^2$. So, $\displaystyle \alpha(1/4)=a^2$. We can pull this apart to get $\displaystyle \alpha(1/8)+\alpha(1/8)=a^2$. to yield $\displaystyle 1/4=a^2$. But there is nothing in $\displaystyle \mathbb{Z}$ that squares to $\displaystyle 1/4$. Therefore, they are not isomorphic.

2. Originally Posted by sfspitfire23
I'm having trouble showing how $\displaystyle \alpha(1/4)=a^2\Rightarrow 1/4=a^2$. How would I do this?

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

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

Tonio