Suppose $x,y\in R$ and $k\in N$ where $x^k=y$. Then prove ${x^a:a\in Q}={y^a:a\in Q}$ Suppose $x,y\in R$ and $k\in N$ where $x^k=y$. Then prove {$x^a:a\in Q$}={$y^a:a\in Q$}  
All I can intuitively see is that {$x^a:a\in Q$}={$x^{ka}:a\in Q, k\in N$}. This seems false because you can choose some random natural number then ka!=a and $x^a!=x^{ka}$ 
 A: *

*{$x^a:a\in \mathbb Q$} $\subset ${$y^b:b\in \mathbb Q$}


Let $a \in${$x^a:a \in \mathbb Q$}
$a=\frac{c}{d}$ where c,d $\in \mathbb Z$
So $x^a=y^\frac{a}{k}=y^\frac{c}{dk}$ and $dk \in \mathbb Z$.
Then $a \in ${$y^b:b\in \mathbb Q$}.


*

*{$y^b:b\in \mathbb Q$} $\subset ${$x^a:a \in \mathbb Q$}


Let $b \in${$y^b:b\in \mathbb Q$}
$b=\frac{c}{d}$ where $c,d$ $\in \mathbb Z$
So $y^b=x^{kb}=x^\frac{ck}{d}$ and $ck \in \mathbb Z$.
Then $b \in ${$x^a:a\in \mathbb Q$}.
Finally, {$y^b:b\in \mathbb Q$} = {$x^a:a\in \mathbb Q$}
A: Why don't you start with $\{y^a:a\in Q\}=\{(x^k)^a:a\in Q\}=\{x^{ka}:a\in Q\}$? Now let $b=ka\in kQ =\{kq:q\in Q\}$. But if $k$ is a non-zero integer then $kQ=Q$ (if $r\in kQ$ then $r=kq$ for some $q\in Q$ but $kq$ is rational, so $r=kq\in Q$; and if $q\in Q$ then $q=k\frac qk\in kQ$ since $\frac qk\in Q$). 
Thus $\{x^{ka}:a\in Q\}=\{x^{ka}:ka\in kQ\}=\{x^b:b\in kQ\}=\{x^b:b\in Q\}$. Since $b$ is a dummy variable, we may replace it with any other dummy variable, so $\{x^b:b\in Q\}=\{x^a:a\in Q\}$. We had started with $\{y^a:a\in Q\}$ and ended with $\{x^a:a\in Q\}$. 
A: Thought of this, 
Take real numbers m and n and a rational number p/q such that p and q are natural numbers. Now for every number m^(p/q) can be written as n^p, such that m^(1/q) = n.
Using this analogy, the set { x^a : a belongs to Q } is the same as { x^k : k belongs to N }.
The same can be thought for y^a, taking y = x^n where n is a natural number. Then, do the analogy in reverse, and you raise every term to n. This is also reduced to the same expression as before. This proves the statement.
If this is correct, then negative sign can be incorporated in the exponents for the analogy. 
