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Let the collection be denoted by

$\{A_k: k \in\mathbb{N}\cup\{0\}\}$, where each set $A_k$ = $\{2^kn: n \in \mathbb{N}\: and\: n\: is\: odd\}$

To prove that none of the sets in the collection are the empty set (condition (i)), it's clear that $2^k$ is in every set $A_k$ by letting $n=1$.

Now I'm having trouble proving condition (ii) of a partition $\mathcal{P}$, which is that if $X \in\mathcal{P}$ and $Y\in\mathcal{P}$, then $ X = Y$ or $X \cap Y = \emptyset$.

After writing out the first few sets $A_k$ it's hard to determine whether condition (ii) is true at first glance,

$A_0 = \{1,3,5,7,9,11,13,...\}$

$A_1 = \{2,6,10,14,18,22,26,...\}$

$A_2 = \{4,12,20,28,36,44,52,...\}$

$A_3 = \{8,24,40,56,72,88,104,...\}$

and so on

My proof for condition (ii) currently,

Let $X_a, Y_b \in \mathcal{P}$

Suppose $X_a \neq\ Y_b$ and $X_a \cap Y_b \neq \emptyset$

Let $x \in X_a \cap Y_b$

$x = 2^an$ such that n is odd and $x = 2^am$ such that m is odd

$2^an = 2^bm$ if and only if $a = b$ and $n = m$ (where I think my proof is faulty since I'm not sure how to prove this)

In order for this to happen, $X_a$ must equal $Y_b$. This is a contradiction so either $X_a = Y_b$ or $X_a \cap Y_b = \emptyset$

And for proving the last condition (iii) of a partition $\mathcal{P}$, which is that the union of all the sets in the collection equals $\mathbb{N}$, I'm not sure whether to split the proof into evens and odds or to show that some arbitrary natural number, $m \in \mathbb{N}$ belongs to the union of all sets $A_k$. Which approach is more feasible? And am I on the right track to proving condition (ii)?

Peter Taylor
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Chris Gong
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2 Answers2

2

Given a number $n$ if it is odd it belongs to $A_0$. If not divide by 2, getting $n=2n_1$. If $n_1$ is odd then $n\in A_1$. If not repeat with $n_1$, divide it by 2. If $n_1=2n_2$ with $n_2$ odd, then $n=2^2n_2$, belongs to $A_2$, and so on. Keep dividing by 2 until you get an odd number if after $k$ divisions you got an ood number means $n=2^km$ with $m$ odd and belongs to$A_k$. Note that as each division by 2 yields a smaller number given any $n$ this process has to terminate with at most $n$ steps (in fact $\log_2 n$ steps, that is no important).

Another way of viewing this is: write the number in base 2. If the rightmost $k$ digits are $0$ then the number belongs to $A_k$.

P Vanchinathan
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  • I see how this proves condition (iii), but is this also applicable for proving condition (ii)? – Chris Gong Nov 19 '16 at 10:56
  • Keep on dividing by 2 until you get an odd number. A number leading to odd number in 3rd $(A_3)$ round cannot also be a number becoming odd in the 4th stage $(A_4)$. So they are disjoint. – P Vanchinathan Nov 19 '16 at 14:29
  • Ah ok I see, because dividing a number in $A_3$ by 2, 3 times will get you an odd number, but every number in $A_4$ will still be divisble by 2 after dividing by 2, 3 times. Therefore, they are disjoint. Just out of curiosity, is the other answer posted a valid proof since it seems that part of the other answer is applicable here? – Chris Gong Nov 19 '16 at 17:26
  • Yes, the answer by Toan Quang Pham is also correct. Various answers are merely attempt to express the same thing different ways. I also mentioned in the last sentence of my answer about binary expansion of number. It will be instructive if you spend say 15 minutes carrying out this procedure for all numbers from 1 to $50$ and see which $A_k$ each of them gets placed. I strongly urge you do that. – P Vanchinathan Nov 20 '16 at 02:06
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For (ii): I see some typos but in general your solution is correct. And for the proof of $2^an=2^bm$ if and only if $a=b,m=n$, you can say that from Fundamental theorem of arithmetic, we follow $a=b,m=n$. It's pretty much obvious to imply $a=b,m=n$ right away without mentioning the theorem.

For (iii): Your second idea is the way to go: Prove that for any $m \in \mathbb{N}$ then $m \in A_i$ for some $i$. This is also obvious since we can represent $m=2^xy$ with odd $y$, which follow immediately $m \in A_x$.

This is also based on Fundamental theorem of arithmetic since there is only one way to represent $m$ as $2^xy$.

Note on Fundamental theorem of arithmetic: It basically says that all positive integers greater than $1$ factor uniquely into primes.

Tengu
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