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So I was given this problem by my friend, and apparently I got the answer wrong.

Problem:

Consider the set $S=\big\{1,3,2,5,7,4,9,11,6,13,17,8,19,21,10,\ldots\big\}$. Now, form a ratio $\#O:\#E$ such that $O$ is the set of all odd numbers and $E$ is the set of all even numbers.

What is the value of the ratio?

Notation: $\#X = $ the cardinality (cardinal number) of a set $X$.

My Answer:

Obviously, we can see a pattern. Let $o_n$ be the $n^\text{th}$ odd number and $e_n$ be the $n^\text{th}$ even number. Then, $$S=\big\{o_1,o_2,e_1,o_3,o_4,e_2,\ldots\big\}=\bigcup_{n\in O}^\infty\big\{\{o_n,o_{n+1}\}\cup \{e_{(n+1)/2}\}\big\}.$$ So now, we have three such elements to consider: $o_n,o_{n+1}$ and $e_{(n+1)/2}$. Two of those elements belong to the set $O$ and only one element belongs to the set $E$. Thus, $$\#O:\#E=2:1$$ which means that $2/3$ of the set is odd and $1/3$ of the set is even.

However, my friend said the answer was wrong! Actually, $\#O:\#E=1:1$. But I don't understand where I might have gone wrong.

Possible reason I was wrong:

If a set $X=\bigcup_{i=k}^n\big\{x_i\big\}$ then $n=\#X$. Thus, $O$ and $E$ have the same carnality (infinite), so $$\#O:\#E=\infty:\infty=1:1.$$ But are we allowed to reduce ratios like that? I mean, $\infty$ is not a number! My friend said that my reasoning for why the answer is wrong is incorrect, and I understand that... so how else is my answer incorrect? He gave me the following hint:

Hint:

$X=\{a,b,c\}=\{b,a,c\}$.

But how is this useful?

Thank you in advance.

Community
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Mr Pie
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  • The ratio is undefined as $#O=#E = \infty$. – copper.hat May 25 '18 at 05:25
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    Your friend does not appear to know what they are talking about. – Eric Wofsey May 25 '18 at 05:26
  • @EricWofsey yes I see... very strange. Given the hint, I rearranged the set $S$ and saw that $S=\mathbb{Z}^+ =$ the set of all positive integers. I guess he thought that there are as many even numbers as odd numbers, and so $1:1$ but that's absurd, especially given what copper.hat mentioned in their comment, so I wanted to know what the real answer was. – Mr Pie May 25 '18 at 05:29
  • By the way, this post is related $\longrightarrow$ https://math.stackexchange.com/questions/696210/is-infty-infty-1?rq=1 – Mr Pie May 25 '18 at 06:49

1 Answers1

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The ratio of the cardinalities of the two sets is undefined as they are both infinite and "infinity divided by infinity" isn't allowed. You can however talk about the sequence and the relative density of the odds and evens within that sequence.

Let $O$ represent the set of odd natural numbers and let $E$ represent the set of even natural numbers.

Let $A_1=\{1\},A_2=\{1,3\},A_3=\{1,3,2\},A_4=\{1,3,2,5\},A_5=\{1,3,2,5,7\},\dots$

$\dots,A_{3n}=\{1,3,2,5,7,4,\dots,4n-3,4n-1,2n\},\dots$

We can then talk about $\lim\limits_{n\to\infty}\dfrac{|O\cap A_n|}{|A_n|}$ and $\lim\limits_{n\to\infty}\dfrac{|E\cap A_n|}{|A_n|}$ which will be $\dfrac{2}{3}$ and $\dfrac{1}{3}$ respectively.

In that sense, the "ratio" that you might have been thinking of is $2:1$ when treating this in terms of the relative density in which the odds versus evens appear.

Compare this to $[n]=\{1,2,3,\dots,n\}$ and how $\lim\limits_{n\to\infty}\dfrac{|O\cap [n]|}{|[n]|}=\dfrac{1}{2}$ and $\lim\limits_{n\to\infty}\dfrac{|E\cap [n]|}{|[n]|}=\dfrac{1}{2}$

Notice that $\lim\limits_{n\to\infty}A_n=\lim\limits_{n\to\infty}[n]=\Bbb N^+$.

This goes to show that depending on the order in which things are written, certain subsets can seem to "appear more frequently." Once we accurately define the ratios we are talking about we can see exactly what that means and why it happens.

JMoravitz
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  • This is a wonderful answer! It was well-written; I didn't need to double check anything; and I also like your notation, $[n]$ (and I am aware of what $\mathbb{N}^+$ is). If anyone doesn't beat it, I'm gonna accept this! $(+1)$ P.S. You just got my last upvote before I reached my daily voting limit :) – Mr Pie May 25 '18 at 05:44
  • @user477343 as an aside, the notation $[n]$ is in common use in combinatorics but some authors will use it to mean ${1,2,3,\dots,n}$ while others will use it to mean ${0,1,2,3,\dots,n-1}$. In both cases it is meant to generally represent "the simplest $n$-element set" so it almost never matters which it strictly means and the meanings can be used interchangeably in most proofs without affecting anything. Still, it is worth letting readers know which meaning you intend if it could possibly matter. I am guilty of using it to mean whichever of the two is convenient for me at the time. – JMoravitz May 25 '18 at 05:47
  • hahaha yes sometimes I use $[n]$ to mean rounding $n$ to the nearest integer since we have $\lfloor n\rfloor$ for lowest and $\lceil n\rceil$ for highest. But that doesn't matter; as long as I let the reader know what I mean by this certain notation, as you said. And given your answer, you seem to know exactly what you are talking about :) – Mr Pie May 25 '18 at 05:50
  • @user477343 As an additional tangential comment, the reordering of terms in infinite sequences causing changes in the properties we are analyzing like this also comes up in the subject of convergent series. Read about the Riemann series theorem which briefly states that if $\sum a_n$ is a conditionally convergent series, then you can reorder the terms in the sum and change what the series converges to, even being able to pick how to reorder the terms to get it to converge to something specific you choose. – JMoravitz May 25 '18 at 06:03
  • Thank you very much for your comment. Also, congratulations on your answer! $$\color{green}{\text{Accepted.}} \ \ \color{green}{\checkmark}$$ – Mr Pie May 25 '18 at 06:04