How to prove a set of real number is linearly independent over rational number When I consider the set
$$ \left\{ 2^{1/1},  2^{1/2}, 2^{1/3}, \ldots, 2^{1/n} \right\},$$ 
I don't know how to prove 
$$ λ_1 2^{1/1} +  λ_2 2^{1/2} + λ_3 2^{1/3} + \cdots + λ_n 2^{1/n} =0 $$ 
is a trivial relation where 
$$ λ_1 = λ_2 = \cdots = λ_n = 0. $$ 
Alternatively, how to prove $2^{1/n}$ cannot be expressed as a linear combination of $2^{1/m}$ and $2^{1/p}$ where $n$, $m$, $p$ are not equal.
 A: Let us use induction. 
For $n = 1$, we see that 
$$ \lambda_1 2^1 = 0 $$
implies 
$$ \lambda_1 = 0. $$
So the result holds trivially.
Suppose the result holds for all natural numbers $k$ less than $n$. 
Let $\lambda_1, \ldots, \lambda_{n-1}, \lambda_n$ be any rational numbers such that 
$$
\lambda_1 2^{1/1} + \cdots + \lambda_{n-1} 2^{\frac{1}{n-1}} + \lambda_n 2^{1/n} = 0. \tag{1}
$$
If $\lambda_1 = 0$, then we can use our inductive hypothesis to conclude that 
$$ \lambda_1 = \cdots = \lambda_n = 0. $$
So suppose that $\lambda_1 \neq 0$. Then (1) gives
$$
\begin{align}
0 &\neq 2 = 2^{1/1} \\
&= - \frac{ \lambda_2 2^{1/2} + \cdots + \lambda_{n-1} 2^{\frac{1}{n-1}} + \lambda_n 2^{1/n}  }{ \lambda_1} \\
&= \frac{ - \lambda_2 }{ \lambda_1} 2^{1/2} + \cdots + \frac{ - \lambda_{n-1} }{ \lambda_1 } 2^{1/(n-1)} + \frac{ - \lambda_n }{ \lambda_1 } 2^{1/n}  .
\end{align}
$$
Thus we obtain
$$
2 = \frac{ - \lambda_2 }{ \lambda_1} 2^{1/2} + \cdots + \frac{ - \lambda_{n-1} }{ \lambda_1 } 2^{1/(n-1)} + \frac{ - \lambda_n }{ \lambda_1 } 2^{1/n}. \tag{2}
$$
I'm not sure how to proceed from here.
Hope it might be easier for you to take it from here.
PS: 
Since all the numbers $2^{1/2}, \ldots, 2^{1/n}$ are distinct positive irrational numbers and since the $\lambda$'s are rational, we can conclude that the quantity on the right-hand side of (2) above must also be irrational and so cannot equal $2$. 
Hence (1) cannot hold unless all the $\lambda$'s are $0$.
