Integer power of fraction never is two For any 
$$\forall a,b,n \in\mathbb{N}\wedge n\ge 2\Rightarrow\Big(\frac{a}{b}\Big)^{n}\ne 2$$ 
Here is my proof of it :
There are $3$ possibilities:


*

*$a = b$

*$a > b$

*$b < a$

*$ a = b $ if two numbers are equal, their ratio is $1$ and $1^n$ is always $1$, and $$1 \neq 2$$ 

*$ a < b $ A smaller number divided by a bigger number is less than one, and a number less than one when raised to integer power can only get smaller, so it cannot be $2$.

*$ a > b $ By taking the $n$-th root on both sides:
$$\frac{a}{b} = \sqrt[n]{2}$$  
As the $n$-th sqrt of two can never be expressed as a ratio (Pythagoras proved this I think), this possibility is False too.

Is my proof sensible? Is it rigorous? Is it even fully correct? Is the last item of my last too big / complex to be used in something so simple?
I seek any suggestion, improvement or correction.
 A: Note that your proof of the $a>b$ case is all you need as your reasoning applies to the other two cases, so you can certainly shorten the proof.
The fact that you quote is actually a step in the standard proof that $\sqrt[n]2$ is irrational. As such, you should probably prove it from first principles to avoid any circular logic.
Have you seen the proof that $\sqrt 2$ is irrational? Can you adapt this proof to this more general case?
A: Your  proof  for  the  first  two  cases  are  perfect.
For  the  case  $a>b$  ;  you  could  do  this,in  case  you  do  not  have  the  proof  that  :
$\sqrt[n]{2}$ is irrational for $n\ge 2$ (as Mathmo123  says  , is  required)
$$(a/b)^n ={a^n}/{b^n}=2$$
$$or, a^n=2.b^n$$
As  $2$  is  a  prime number ,  this  above  equation  tells  that $2|a$. 
Let  $$a=2k$$ 
 Then  $$2^n.k^n=2.b^n$$
$$2^{n-1}.k^n=b^n$$
Again  this  means $$2|b$$  as  well  so  let  $$b=2j$$
Then  we  get  $$2^{n-1}.k^n=2^n.j^n$$
$$or\ \ \ , k^n=2.j^n$$
The  same  procedure  applied again  shows  that  $$a=2^s\ \ \ and\ \ \ b=2^l\ \ \ for\ \ \ some\ \ \ s,l\in \mathbb N$$
Then  $${(a/b)}^n={({2^s}/{2^l})}^n = (2^{s-l})^n=2^{ns-n}$$
So  this  can  be $2$  iff  $$n(s-l)=2$$ 
Now  $2$  being  a  prime ,  we  have  $2$  possibilities :


*

*$n=2$,$s-l=1$

*$n=1$,$s-l=2$
Both  of  them  gives  $$(a/b)^n = 4\neq 2$$  So  we  have  it .Proved .
A: Let $$\left(\frac ab\right)^n=2$$ or
$$a^n=2b^n,$$ where $a,b$ have no common factor (otherwise they can be simplified).
The last equation shows that $a^n$ is even. So is $a$, as odd numbers given only odd powers. Then $a^n$ is a multiple of $2^n$ and $a^n/2$ is a multiple of $2^{n-1}$, i.e. is even ($n>1$). Then $b^n$ is even and so is $b$.
$a$ and $b$ are both even, a contradiction.

Second proof:
Let $a^n=2b^n$, and denote $\alpha$ and $\beta$ the exponents of $2$ in the prime decompositions of $a$ and $b$.
Then as the decomposition is always unique,
$$(2^\alpha\cdot p)^n=2\cdot(2^{\beta}\cdot q)^n$$implies
$$\alpha n=\beta n+1$$ which is impossible.
A: If $(a/b)^n=2$, then $a^n=2b^n$. We claim that the only integer solution to this equation is when $a=b=0$.
We'll prove this by contradiction. Suppose that there was a non-zero solution $(a,b)$. Choose a solution with minimal value of $|a|+|b|$. Since $2b^n$ is even, it follows that $a^n$ must be even, and therefore $a$ is also even. Now we may write $a=2c$ to obtain that $(2c)^n=2b^n$, so canceling a $2$ from both sides yields $b^n=2^{n-1}c^n$. Therefore $b$ is even, so $b=2d$ for some integer $d$, yielding
$$
2^nd^n=2^{n-1}c^n\implies c^n=2d^n.
$$
Thus we have obtained a solution $(c,d)$ to the original equation which is strictly smaller, in the sense that
$$|c|+|d|=\frac{|a|+|b|}{2}<|a|+|b|.$$
Therefore there is no non-zero integer solution to the equation.
