Is there any set of numbers that, when multiplied with anything from $\mathbb{N}$, results in an uneven number? I noticed the (very easy to see) fact that, whenever you multiply any natural number $n$ with anything having a prime factor of $2$, you get an even number.
For example:
$ 3 \times 2 = 6$, even
$5 \times 6 = 30 $, even
and so on. It's easy to see that this is true because everytime you multiply, you add prime factors to the set of the resulting set of prime factors. E.g. 
$ 3 \times (2 \times 3) = 3 \times 6 = 18$, prime factors are $2, 3, 3$. And whenever you have a prime factor of 2, the number is even.
Now I ask myself: is it possible to construct a set $\mathbb{G}$ of numbers in a way that $\forall x \in \mathbb{G}: \forall n \in  \mathbb{N}: \lfloor x \times n\rfloor = $ uneven? Maybe even without flooring them? $\mathbb{G}$ may be in $\mathbb{R}$, but it would be even more interesting without rounding and with $\mathbb{G} \subset \mathbb{N}$, but is this even possible?
It seems way easier to create even numbers than to create odd numbers from any set of natural numbers. Is this true? 

Any real number $r$ is equal to the limit of a series of this form:
$$
n + \frac{a_1}{2} + \frac{a_2}{2^2} + \frac{a_3}{2^3} +
\cdots + \frac{a_i}{2^i} + \cdots, \tag1
$$
where $n$ is an integer and each $a_i$ is either $0$ or $1.$
Moreover, if $r$ is any number in your desired set $\mathbb G,$ we can rule out the case where the limit is an integer (because then if we multiplied by $2$ the result would be even).
So we must have at least one $0$ term in the sequence, since the series
$$\frac12 + \frac1{2^2} + \frac1{2^3} + \cdots + \frac1{2^i} + \cdots$$
has limit $1$ and we would then have $r = n+1,$ an integer.
Now let $a_k = 0,$ that is, $k$ is a positive integer and the $k$th term is a zero term. Let $r$ be the limit of the series in $(1),$ and consider $2^k r,$ which is the limit of the series
$$
2^k n + 2^{k-1} a_1 + 2^{k-2} a_2 + \cdots + 2^2 a_{k-2} + 2a_{k-1} + a_k + \frac{a_{k+1}}2 + \frac{a_{k+2}}{2^2} + \frac{a_{k+3}}{2^3} + \cdots.
$$
Thus $2^k r$ is the sum of an integer
$$
2^k n + 2^{k-1} a_1 + 2^{k-2} a_2 + \cdots + 2^2 a_{k-2} + 2a_{k-1} + a_k
$$
and the limit of the series
$$
\frac{a_{k+1}}2 + \frac{a_{k+2}}{2^2} + \frac{a_{k+3}}{2^3} + \cdots.
$$
Now there are two cases. 
In the first case, the limit of the series 
$\frac{a_{k+1}}2 + \frac{a_{k+2}}{2^2} + \frac{a_{k+3}}{2^3} + \cdots$ 
is $1,$ and therefore $2^k r$ is an integer,
$2^{k+1}r$ is an even integer, and $r \not\in \mathbb G.$
In the second case, the limit of the series 
$\frac{a_{k+1}}2 + \frac{a_{k+2}}{2^2} + \frac{a_{k+3}}{2^3} + \cdots$ 
is less than $1,$ and therefore 
$$
\lfloor 2^k r \rfloor = 2^k n + 2^{k-1} a_1 + 2^{k-2} a_2 + \cdots + 2^2 a_{k-2} + 2a_{k-1} + a_k.
$$
But since $a_k = 0,$ and all the other terms have at least one factor of $2,$
it follows that $\lfloor 2^k r \rfloor$ is even and that $r \not\in \mathbb G.$
So in either case $r \not\in \mathbb G.$
There is then no way for any real number to be in your desired set $\mathbb G.$
