Partitioning the Positive Integers

Every positive integer has a unique representation $2^k m$ , where $k \in \mathbb{N} \cup \{0\}$ and $m$ is odd. If $A_1 = \{ 2^k m \; | \; \mbox{k is even and m is odd} \}$ and $A_2 = \{ 2^k m \; | \; \mbox{k is odd and m is odd} \}$ , then $A_1 \cup A_2 = \mathbb{N}$ and $A_1 \cap A_2 = \emptyset$ . Furthermore, since multiplying by $2$ changes the parity of $k$ , $x \in A_i \Rightarrow 2x \not\in A_i$

It only requires $2$ subsets:

Go through the numbers in order. It doesn't matter which subset the odd numbers are put into. Simply place each even number $n$ into whichever subset doesn't contain $\frac{n}{2}$ .

You can prove n=2 is possible inductively. Once you've placed the numbers 1,2,...,2m you can place 2m+1 in either set and 2m+2 in whichever set m+1 is not in.

Let's assume there will be 2 subsets required !

1. In first let's take all odd nos.

   We can't take doubles so let's take.
   quadruples(4th multiples) of all odds.
   

   And similarly, 16th , 64th, 128th.......
   multiples of odds.
   (because we can't take 8th multiples as it'll.
   be double of 4th multiples and so on! )
   
   
   

2. Our 2nd set is already created as it is the set of
   remaining nos. It'll contain all doubles of odds, 8th multiples.
   of Odds and so on. Hence both set Fulfil the mentioned condition.

Using the fundamental theorem of arithmetic, each positive integer can be uniquely represented in the form $(2^k)m$ where m is odd (being the product of all the prime factors that are not 2, or 1 in the case of powers of 2) and k = 0,1,2,... .

Define two partitions with k = 0,1,2,... and m is an odd positive integer: $P_1$ = { $(2^k)m$ | k is even} and $P_2$ = { $(2^k)m$ | k is odd}

For each n = $(2^k)m$ , 2n = $(2^{k+1})m$ and by definition in the opposite partition from n, since k and k+1 have opposite odd/even parity.

This is my method which is not nearly as simple and easy as Binky Mh's solution, and is quite unnecessarily long, but I decided to write down my own solution anyways(for practice. Also do not that creating this solution was much easier than communicating it (in fact, it took a very little amount of time)). Note that, for this solution, 0∈N.We will also be calling the 2 sets G and W.The way you can do this is to first partition the set {x:x∈N} into an infinite series of mutually exclusive sets {x*(2^n):n∈N} by first creating such a set for x=1, than repeating the cycle of creating a new of such set for the lowest x not in any previous set, forever. This works as every x will always have a unique prime factorization which will never include 2 (as otherwise x/2 would have already been in a previous set), and each set is just a list of numbers who's prime factorization is that of x with additions (this term is very confusing, but I am too lazy to think up a better term) of 2, so will be different as their prime factorization will be different as their prime factorization excluding 2 will be different, it is also quite obvious how the lowest possible value not in any previous set will incorporate all possible values over the infinite cycles. For each set {x(2^n):n∈N}, we take the smallest value and put it into G, and than repeat the cycle where the next smallest value (which will be double the one we previously put in, hence making it necessary to put into the other set (but as it is the other set, and this is the only value double the previous value, we will validate the previous value)) into the set (either G or W) which is not in where you just put the previous smallest value (which is now gone as you put it into the other set) forever, the infinite nature of this means every value will be included in W or (exclusive) G, but the double of this value will be in the other set which is not your set, and we will be doing this for all the series of nth_term=x(2^n)/set {x(2^n):n∈N} so hence partitioning all positive integers.

Partition the set of all positive integers into many geometric series with ratio 2. Consider 2 sets: A and B. For each geometric series, starting from its initial element, we put each element alternately into set A and B. Then we are done.

Basically separate the odd and even numbers.

We need two 2 subsets. Put any odd number x in either set, at will. The constraint only demands that 2x be in the other subset, but we could put 4x together with x, and any mx where m is a power of 4. Put 2x,8x,32x...(2mx) into the other.