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? 
 A: There is no real number $x$ such that $\lfloor x\times n\rfloor$ is odd for all positive integers $n$.
Proof by contradiction: Suppose $x$ were such  number. Then $x\pm2$ would also be such numbers, since $\lfloor x\times n\rfloor$ changes by an even amount, $\pm2n$, when we change $x$ by $2$. So by repeatedly adding or subtracting $2$, we can arrange that $0\leq x<2$.  Now let $z=2-x$, so $0<z\leq2$ and 
$$
\lceil nz\rceil=2n-\lfloor nx\rfloor
$$
is odd for all positive integers $n$.
In particular, taking $n=1$, we see that $\lceil z\rceil$, previously known to be $1$ or $2$, is in fact $1$, i.e., $0<z\leq1$. Therefore, in the progression $z,2z,3z,4z,\dots$, each term exceeds the previous one by at most $1$. But none of these terms can be in the interval $(1,2]$, because $\lceil nz\rceil$ is always odd, never $2$. So this progression starts in the interval $(0,1]$, never enters $(1,2]$, and never jumps over $(1,2]$ either (because any two consecutive terms differ by at most $1$). So the progression is forever stuck in $(0,1]$. This means that, for all positive integers $n$, we have $nz\leq1$, i.e., $z\leq1/n$. But that is impossible, since $z>0$.
A: 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 $\mathbb G.$
A: A number is even if it has a factor of $2$, and it is odd if it does not. There is no whole number that you can multiply by that will ‘delete’ a factor of two, so we have to look to rational numbers. Clearly $\frac12$ will ‘delete’ a factor of two when you multiply it by an even number, but we have a problem; the amount of times you have to multiply by $\frac12$ depends on the number of times $2$ appears in the unique factorisation of the even number you’re trying to make odd. If your number has a factor of $2^n$, you have to multiply by $\frac{1}{2^n}$ to make it odd. If you multiply by $\frac12$ too many times you will end up with a fraction, that is neither even nor odd, and not an integer, and clearly not desirable in this context (unless you really want to go the flooring route, but I’m not considering that here).
In general, there is no set of numbers that you can multiply any even number by to make it odd. But given the set of numbers that only have $2$ appearing in their unique prime factorisation $n$ times (that is, $2^n$ is a factor), multiplying by any number of the form $\frac{k}{2^n}$ where $k$ is an odd number will turn this even number into an odd number. This is the closest thing to what you’re after that I think you can achieve. 
Your intuition about it being “easier to create even numbers than odd numbers” is fairly reasonable, depending on how exactly you mean it.
