1. 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.

2. Originally Posted by sfspitfire23
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