I have a set $A_n$ = $ \{2^i(2n-1): i \in \mathbb{N} \cup {0} \}$
It is said that the set $P = $ $\{A1, A2, A3,...\}$ partitions the natural numbers.
I am attempting to solve this by the definition of a partition, I can see that no $An = \emptyset$, but I can't see how the union of all the $An$'s would give $\mathbb{N}$ can anyone help?
On
Consider $A_1$. This is the set $\{2^i\}$, which contains the elements $1,2,4,8,16,\cdots$. Now consider $A_2$, which is $2^i\times3$ which contains the elements $3,6,12,24,\cdots$. Do something similar for $A_4$ and so on. You should see that the least element of $A_{k}$ is always the $k$th odd number, which is $2k-1$ , and so you can generate every odd element of $\mathbb{N}$ in such a way. By the definition of the set, we can multiply every element by $2$, which makes every even number. Now you need only formalise this proof.
A good way to get some intuition is to take some numbers and see which set they belong to. For example, $12=2^2\cdot 3 \in A_3$ You divide out all the factors of $2$ and are left with an odd number $N$. You let $N=2n-1, n=\frac 12(N+1)$ and that is the set $A_n$ it belongs to.