Is powers of rationals dense in $\mathbb R$ Consider $\mathbb {\tilde {Q}} = \{ x^n : x \in \mathbb Q \} $ $n$ is fixed odd integer.
I have two questions here.


*

*Is this set dense in $\mathbb R$ and

*Is there any bijection exists between $\mathbb Q$ and $\mathbb {\tilde Q}$


For the first question, I think the set is dense. Consider $ a,b \in \mathbb R$ WLOG assume $ a,b \ge 0$ and $ b \ge a$. We can find $c \in \mathbb Q$ such that $ a^{1/n} \le c \le b^{1/n}$. And now $ c^n \in \mathbb {\tilde Q}$ and $ a \le c^n \le b$.
So the set is dense. 
Is there a way to define bijection between $\mathbb Q$ and $\mathbb {\tilde Q}$

The set of powers of the rationals, to any power, including even, is infinitely dense.  What happens of the even numbers is that the range of the dense region is not less than zero.
One can demonstrate this by noting that any irrational number must be bracketed by two rational numbers, for example, numbers written in base 10.  So we know, for example, that $\sqrt[3]{2}$ lies variously between $1$ and $2$, between $1.2$ and $1.3$, and $1.25$ and $1.26$, and so forth for each decimal of the cube root of $2$.
For this reason, we would then note that there are always rational powers of a given value, between any previous rational value, and a number not of that particular power.  That is, there must exist a cube between $2.197$ and $2$, such as $2.000376$, and even closer ones.
Thus, the powers are infinitely dense on the real number line.
Since every cube must has exactly one real cube root and likewise true for all odd powers, then there is a bijection between the two sets.  For even numbers, one must restrict the tests to ${R}^+$, the positive reals.  
