My 6 year old wants to know if infinity is an odd or even number. His 38 year old father is keen to know too.

35"Odd" and "even" are typically only applied to natural numbers, of which infinity is not. "1.4" is neither odd nor even as well. – The Chaz 2.0 Jul 02 '11 at 15:34

22Neither or both because $\infty=2\cdot \infty=2\cdot \infty + 1$ – Listing Jul 02 '11 at 15:37

20@The Chaz. Not true. In fact 1.4 = 7/5 is odd when considered as an elt of the subring of rationals expressible with odd denominator. To say that the concept of parity in $\mathbb Z$ doesn't apply to extensions is to miss the point. The question is whether the notion of integer parity can be *extended* in a meaningful way to certain extended "number" systems. – Bill Dubuque Jul 02 '11 at 21:15

1@Bill: I didn't have time to typeout the decimal expansion of an irrational! (But seriously, thanks for the comment) – The Chaz 2.0 Jul 03 '11 at 00:16

6@The Chaz. Ditto for irrationals, e.g. $\:\sqrt{3}\:$ is odd in $\:\mathbb Z[\sqrt{3}]\:.$ See [here](http://math.stackexchange.com/questions/26837/dooddimaginarynumbersexist/26843#26843) for more. – Bill Dubuque Jul 03 '11 at 00:37

1I would answer yes. ;) – Michael Brown Jul 03 '11 at 00:53

Alas! $$$$$$$$. – The Chaz 2.0 Jul 03 '11 at 11:43

Related MathOverflow question: http://mathoverflow.net/questions/69461/onthedifferencebetweentwoconceptsofevencardinalitiesisthereamodelof – JDH Jul 04 '11 at 12:38

4Infinity is not a number to start with :) – Cano64 Aug 19 '15 at 16:29

@Cano64 what is a number? – CcVHKakalLLOOPPOkKkkKk Apr 29 '20 at 21:03
6 Answers
In the context of transfinite ordinals, the usual definition is that an ordinal number $\alpha$ is even if it is a multiple of $2$, specifically: if there is another ordinal $\beta$ such that $2\cdot\beta=\alpha$. In other words, the order type $\alpha$ can be viewed as $\beta$ many pairs in sequence, or in other words, $\alpha$ is leftdivisible by $2$. Otherwise, it is odd.
It is easy to prove from this definition by transfinite recursion that the ordinals come in an alternating even/odd pattern, and that every limit ordinal (and hence every infinite cardinal) is even. Many transfinite constructions proceed by doing something different on the even as opposed to the odd stages, just as with finite constructions.
The smallest infinite ordinal is $\omega$, which is even on this definition, since having $\omega$ many pairs in sequence is orderisomorphic to $\omega$, and so $2\cdot\omega=\omega$. Meanwhile, the next infinite ordinal is $\omega+1$, which is odd. The ordinal $\omega+2$ is even, since it is equal to $2\cdot(\omega+1)$, even though it is not $\beta+\beta$ for any $\beta$.
(Please note that $\alpha=2\cdot\beta$ is not at all the same as saying $\alpha=\beta+\beta$, since $\beta$ copies of $2$ is not the same order type as $2$ copies of $\beta$, a phenomenon at the heart of the noncommutativity of ordinal multiplication. )
To explain the idea to a child, I would focus on the principal idea: whether finite or infinite, a number is even when it can be divided into pairs. For finite sets, this is the same as the ability to divide the set into two sets of equal size, since one may consider the first element of each pair and the second element of each pair. In the infinite context, as others have noted, there are numerous concepts of infinity, each with its own concept of even and odd. In my experience with children, one of the easiesttograsp concepts of infinity is provided by the transfinite ordinals, since it can be viewed as a continuation of the usual counting manner of children, but proceeding into the transfinite:
$$1,2,3,\cdots,\omega,\omega+1,\omega+2,\cdots,\omega+\omega=\omega\cdot2,\omega\cdot 2+1,\cdots,\omega\cdot 3,\cdots,\omega^2,\omega^2+1,\cdots,\omega^2+\omega,\cdots\cdots$$
This concept of infinity is attractive to children, because they can learn to count into the infinite this way. Also, this concept of infinity has one of the most successful parity concepts, since one maintains the even/odd pattern into the transfinite. The smallest infinity $\omega$ is even, $\omega+1$ is odd, $\omega+2$ is even and so on. Every limit ordinal is even, and then it repeats even/odd up to the next limit ordinal.
See the Wikipedia entries on transfinite ordinals and ordinal arithmetic for more information about the ordinals.
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78Please record your 6 year old's expression when you explain it this way ;) – Henrik N Jul 02 '11 at 18:03

@JDH: Some naive questions: Can you explain to me what you mean when you say $\omega+1$ is odd while $\omega$ is even? What confuses me is that $\omega=\omega+1$, so I don't know how to interpret this statement. Also, why is it that $2\cdot\beta=\beta+\beta$ does not always hold? What exactly is the meaning of $2\cdot \beta$ then? – Eric Naslund Jul 02 '11 at 18:32

4In ordinal arithmetic, $\omega+1$ is the successor to $\omega$, that is, the next ordinal after $\omega$, and so $\omega\lt \omega+1$. Ordinal addition is that $\alpha+\beta$ is the order arising from a copy of $\alpha$ followed by a copy of $\beta$, and $\alpha\cdot\beta$ means $\beta$ copies of $\alpha$. So $2\cdot\omega$ means $\omega$ copies of $2$, which you can see easily is $\omega$, but $\omega\cdot 2$ means two copies of $\omega$, or $\omega+\omega$. Meanwhile, $1+\omega$ is $1$ followed by $\omega$, which is the same as $\omega$, and so $1+\omega=\omega=2\cdot\omega=67\cdot\omega$. – JDH Jul 02 '11 at 18:37

1To continue, the ordinal $\omega+2$ is even because it is equal to $2\cdot(\omega+1)=2\cdot\omega+2=\omega+2$, but it is not $\beta+\beta$ for any $\beta$, since if $\beta$ is finite, then $\beta+\beta$ is also finite, and otherwise $\beta+\beta$ is at least $\omega+\omega$, which is strictly larger than $\omega+2$. So in this context, to get the even/odd/even/odd pattern, one should not use rightdivisibility by $2$, but rather leftdivisibility as I mentioned. – JDH Jul 02 '11 at 18:40

1Note that $\omega+1$ is odd, because if $2\cdot\beta=\omega+1$, then $\beta$ cannot be finite, since then $2\cdot\beta$ would be finite, and $\beta$ cannot be equal to $\omega$, since $2\cdot\omega=\omega\lt\omega+1$, and so $\beta$ must be at least $\omega+1$, but in this case $2\cdot\beta$ would be at least $\omega+2$, which is strictly larger than $\omega+1$. A similar argument establishes the even/odd pattern for all ordinals. – JDH Jul 02 '11 at 18:50

Is there any application of this notion ? Or is it just a definition which extends the one on finite ordinals ? – user10676 Jul 02 '11 at 18:52

1Eric, $\omega + 1$ actually represents the linear order of $\mathbb N \cup \{\infty\}$ if you will. There are countably many numbers that get bigger step by step and one number "$\infty$" that is bigger than all of them. Also $n\in\mathbb N$ (as an ordinal) is a linear order of $n$ numbers. $2\cdot n$ is $n$ linear orders of $2$ numbers concatenated, hence a linear order of $2n$ numbers. $2\cdot \omega$ is $\omega$many (countably many) linear orders of $2$ numbers, just giving a linear order isomorphic to the natural numbers. – Peter Patzt Jul 02 '11 at 18:52

1@user10676, yes, this notion of even/odd for ordinals is just the standard ordinal arithmetic, known since the time of Cantor, and is used frequently in set theory. In transfinite constructions, for example, you can interleave two transfinite processes by doing the first on the even ordinals and the second on the odd ordinals. – JDH Jul 02 '11 at 19:01

Henrik, I think JDH did a good explaination saying, something is even, if one can pair a set with that cardinality. (Or for ordinals a linear order.) E.g. $4$ is even because you can pair $4$ apples, without leaving a single behind. The same can be made to understand for infinite many apples. (OK, here it can also be paired such that a single apple is left behind, but it is all about the definition :D) – Peter Patzt Jul 02 '11 at 19:02

4I'm slowly digesting this answer (and comments) with wikipedia in hand. Once my expression has normalised, I'll explain it to my 6 year old (and record his expression!) – Kevin Jul 02 '11 at 19:18

1Does this actually answer the question? Sorry but my grasp of mathematics isn't good enough to understand it. Also, ∞ is an odd creature in that if you add to it it still remains ∞. Which leaves the question of whether the property odd/even is exclusive in the unlikely case where a value of some sort could be both. – James P. Jul 02 '11 at 19:56

11@James: " Also, ∞ is an **odd** creature in that if you add to it it still remains ∞." perhaps we have solved the mystery. :) – InterestedGuest Jul 02 '11 at 20:06


3@David, on the contrary, Cantor's ordinal numbers are a major part of Cantor's work (http://en.wikipedia.org/wiki/Georg_Cantor), about which Hilbert famously declared, "No one shall expel us from the Paradise that Cantor has created." I would encourage you to learn more about the beautiful mathematical theory of ordinal numbers http://en.wikipedia.org/wiki/Ordinal_number. – JDH Jul 02 '11 at 20:09

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7@David: there is nothing not serious in JDH's answer. Please collect yourself, and try not to say things like *Do any of you have a mathematical education?* because it only contributes noise to the discussion (and yes, JDH and many others here do have an extensive mathematical education...) – Mariano SuárezÁlvarez Jul 02 '11 at 22:26

4It may be easier for the original poster to just dodge the question by raising the kid as an ultrafinitist! – tzs Jul 02 '11 at 23:13

I added an addendum about explaining the concept to children, which in my experience is possible. – JDH Jul 03 '11 at 11:31

4@JDH It's *quite* absurd to think that a six year olds notion of infinity might have something to do with ordinals. It's even more absurd to consider that one might be able to "explain the concept to children". I don't recall ever having encountered a more pedagogically ridiculous proposition. – Bill Dubuque Jul 03 '11 at 14:46

16Bill, despite your emphatic comments, I know for a fact that counting into the ordinals is something that children can easily learn. I have two young children (ages 4 and 9), who are happy to discuss $\aleph_\alpha$ for small ordinals $\alpha$although my daughter's pronunciation sounds more like $\text{Olive}_0$, $\text{Olive}_1$and my son can count up to small countably infinite ordinals. The pattern below $\omega^\omega$ is not difficult to grasp. Below $\omega^2$, it is rather like counting to $100$, since the numbers have the form $\omega\cdot n+k$, essentially two digits. – JDH Jul 03 '11 at 23:56

2@JDH My comment is based upon decades of interacting with hundreds of welleducated *adult* lay folks on topics like Goodstein's theorem. Some of these folks were quite welleducated, e.g. esteemed professors in other fields. Given their difficulty truly grasping ordinals, I am highly skeptical of your claims about (your) children. Not to mention the fact that, except for children of logicians, it is highly unlikely that a 6yearold's informal notion of "infinity" has anything whatsoever to do with ordinals. PS You need to write "@Bill" not "Bill" if you wish me me to be notified. – Bill Dubuque Jul 04 '11 at 15:50

1@JDH Do you *really* think that your children truly understand the concept of a limit ordinal, and why such limit ordinals are considered even, why such parity ordinal arithmetic is consistent, etc? – Bill Dubuque Jul 04 '11 at 15:56

4I don't care for an extended argument with you about it, but no, my children are not undertaking transfinite consistency proofs. My claim about children above, if you read it, was much milder, namely, that children can learn to count in the ordinals and learn that ordinals have the same even/odd pattern as ordinary numbers, and that to be even means to be divisible into pairs. The task simply isn't as difficult as your comments suggest. – JDH Jul 06 '11 at 02:26

10Almost anyone including children can learn to count to $\omega^2$, naming the numbers in turn and describing the general patternand my son can count much higherthe key hurdle is getting to the idea that something can come *after* all the finite numbers, that is, just getting to $\omega$ itself. In my experience (also decades of teaching), I have found that this hurdle can be overcome with a Zenostyle explanation of getting half way to the line, saying $1$, then halfway again, saying $2$, and so on, finally saying $\omega$ when you are at the line. – JDH Jul 06 '11 at 02:27

8Once a person can count to $\omega$, then you just do it again, counting $\omega+1$, $\omega+2$, and so on, and they will guess that the name of the ordinal after this is $\omega+\omega$, which you can explain is also known as $\omega\cdot 2$. Soon you are counting $\omega\cdot 2+1$, $\omega\cdot 2+2$, and so on, and then they will guess $\omega\cdot 3$ and also see how the $\omega\cdot n$ arise, which you can explain are called limit ordinals, and very soon you reach $\omega^2$. And one can carry interested subjects still further along... – JDH Jul 06 '11 at 02:53

3@JDH Thanks for the elaboration. In my experience I think one has to exercise extreme caution lest one attribute inordinate amounts of comprehension when all that may be occuring is simply mimicry. It's not difficult to test true comprehension by asking probing questions. In my opinion, for a sixyear old to truly comprehend the essence of the matter would require a mathematical prodigy. PS Again, you need to say @Bill if you want me to see your replies. – Bill Dubuque Jul 07 '11 at 19:57

5@BillDubuque: I'm always amazed at what children are capable of, or anyone with an open mind. They give it 100% compared to adults, who typically have reservations or fear of failure. Children are young and thus approach everything with no knowledge of ever failing at anything they do. – surfasb Dec 18 '11 at 16:23

3I decided a couple of years ago that ordinals were the way to answer my own sixyearold's questions about "infinity", and that they led most naturally to an understandable explanation of how infinite quantities work. For example, there is a simple and correct answer to the first question a sixyearold asks, "What is infinity?", namely, (understanding "infinity" as $\omega$) it is the smallest number that you cannot reach by counting from 0. My experience has borne out that this was a good way to proceed. Cardinal infinities, on the other hand, are quite peculiar and unintuitive. – MJD Mar 18 '12 at 02:22

2As enlightening as this answer is, I'm not sure it answers the question. Yes, limit ordinals, including $\omega$, the "smallest infinity", are even. But what does this say about whether "infinity" is odd or even? Are you assuming that the questioner's "infinity" means $\omega$? If so, why? Isn't $\omega+1$ also infinite? – Ari Brodsky Jan 09 '13 at 09:13

Regarding my idea above that an honest and natural way to answer kids’ questions about ‘infinity’ is to understand ‘infinity’ as meaning $\omega$. Ten years on I still think this was a good approach. [I wrote up a blog post, “how to explain infinity to kids”](https://blog.plover.com/math/infinityforkids.html) with more details about this. My experiences closely match those of JDH. – MJD Feb 27 '22 at 02:30
I suggest that you read the discussion at Is infinity a number? first (since of course you need to answer that question to answer this question). There are some senses in which infinity is a number, and there are some senses in which infinity is not a number, and it all depends on what exactly you mean by "number," which in turn depends on what applications you have in mind.
On the other hand, there is a useful sense in which infinity is even. To explain this we have to replace "numbers" with cardinalities of sets.
Definition: A set $S$ has even cardinality if it can be written as the disjoint union of two subsets $A, B$ which have the same cardinality.
In other words, we need to be able to divide $S$ into pairs. This definition reduces to the ordinary definition for finite sets, but an infinite set always has even cardinality. For example, the cardinality of the natural numbers $\mathbb{N}$ is even because we can pair up even numbers with odd numbers.
This definition of "even" came up in my answer to this question, where precisely the above property turned out to be relevant.
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1An interesting thing about this definition of "even" is that there is a model of ZF where the axiom of choice does not hold but for all infinite cardinals $a+a=a$. – Asaf Karagila Jul 02 '11 at 17:47

1@Asaf, do you know anything about the equivalence or (likely) inequivalence over ZF of the assertion that every infinite set can be partitioned into pairs (subsets of size 2) with the assertion that every infinite set can be partitioned into two equinumerous sets? The second implies the first, by considering the bijection, and for linearly orderable sets, the first implies the second, by considering the two projections of the pairs. These two concepts seem to get at the heart of "evenness" for infinite sets, but it is not clear that they are equivalent without the axiom of choice. – JDH Jul 04 '11 at 00:42

@JDH: yes, I was worried about that, too, but decided it wasn't a technicality worth mentioning at this level. Might be worth asking on MO...? – Qiaochu Yuan Jul 04 '11 at 01:02

I don't know the full answer offhand, and I think it would be a good MO question. The issue appears to be a uniformity issue in the pairs, that is, whether one can uniformly pick an ordering on the pairs. So I think it is likely equivalent to AC for families of pairs, since this is what it would take to turn a partion into pairs into a partion into firsts and seconds, which would then be equinumerous. – JDH Jul 04 '11 at 01:34

@JDH: Take the model in which we add a countable set of pairs without a choice function, the union of this family of pairs is infinite, can be split into pairs, but it is Dedekind finite, since splitting will require choosing from infinite many pairs. – Asaf Karagila Jul 04 '11 at 05:20

@Asaf, perhaps that model will work, but are you suggesting that infinite Dedekind finite sets cannot be split into two sets of equal size? This is isn't true in general, since if $A$ is infinite Dedekind finite, then so is the set obtained by having two disjoint copies of $A$, which is clearly splittable. You seem to be addressing $a=a+a$ rather than $a=b+b$. – JDH Jul 04 '11 at 10:21

@JDH: Oh, you are correct. I was thinking about amorphous sets. I will look into it a bit later today. Also the model I gave does not work, since you can obviously take all the pairs which are of even/odd coordinate (there are $\omega$ many pairs). – Asaf Karagila Jul 04 '11 at 10:28

@JDH: I think that I have a proper answer, take a model with countably many Cohen real adjoined to it, and permute the forcing such that not only there is no choice function on the pairs, but the set of pairs is amorphous. The union of the pairs is infinite, and if it can be split into two equinumerous sets that means that either you have split the pairs into two equal parts  which you cannot do; or that you chose one from infinitely many pairs  which you also cannot do. The construction is probably very technical but rather straightforward, I do not mind writing it up sometime if you want. – Asaf Karagila Jul 04 '11 at 11:34

@Asaf, I posted a question about this at MathOverflow (http://mathoverflow.net/questions/69461/onthedifferencebetweentwoconceptsofevencardinalitiesisthereamodelof). Please post a fuller account of your remark there. Your argument seems to separate the two notions of evenness, but it doesn't actually seem to answer my question above, which is about whether there is a model where *every* infinite set can be divided into pairs, but some cannot be cut in half. – JDH Jul 04 '11 at 12:25

Be aware there are many different notions of infinity in mathematics, so the answer to your query will depend on the particular notion of infinity that you have in mind, and how it interacts with the operations and relations of the extended "number" system. For example, if your notion of $\infty$ satisfies $\:1 +\infty = \infty\:$ then this may yield an obstruction to extending parity arithmetic.
Here is a simple example that has some hope of being comprehensible to a 6yearold. I will explain it in a language that is hopefully comprehensible to his 38yearold father. Consider the ring of polynomial functions with integer coefficients, i.e $\rm\:\mathbb Z[x] = \{\: a_0 + a_1\ x\ +\:\cdots\: + a_n\: x^n\ :\ a_i \in \mathbb Z\:\}\:.\:$ If we consider these functions in a neighborhood of $\rm\:+\infty\:$ we obtain an ordered ring. Namely, define $\rm\ f(x) > g(x)\:$ if this holds true on some neighborhood $\rm\:(x_0,\:+\infty)\:$ of $\rm\:+\infty\:,\:$ i.e. if there is some $\rm\:x_0\:$ such that it holds true for all $\rm\:x > x_0\:,\:$ i.e. if it is "eventually" true. One easily checks that this is welldefined. Indeed, since polynomials have only a finite number of roots, they eventually have constant sign. Thus if $\rm\:f\ne g$ then eventually $\rm\:fg > 0\:$ or $<0$ so eventually $\rm\:f>g\:$ or $\rm\:g>f\:$. In fact one easily deduces that this is equivalent to defining the sign of a polynomial to be the sign of its leading coefficient (the leading term eventually dominates lowerdegree terms). This makes it clear that every polynomial is either positive, negative or zero, and the positive polynomials are closed under addition and multiplication (these are precisely the properties required in general to define a total order on a ring, compatible with the ring operations).
This ring $\rm\:\mathbb Z]x]\:$ has "infinite" elements, e.g. $\rm\:x > n\:$ for all integers $\rm\:n\:$ since $\rm\:x  n\:$ is eventually $> 0\:.\:$ Can we extend parity arithmetic from $\:\mathbb Z\:$ to such infinite elements? In fact we can, in two different ways. First, we can define $\rm\:x\:$ to be even. Since, by the Factor Theorem, $\rm\:f(x) = f(0) + x\ g(x)\:$ for some $\rm\:g(x)\in \mathbb Z[x]\:,\:$ this amounts to defining the parity of $\rm\:f(x)\:$ to be the parity of its constant coefficient $\rm\:f(0)\:.\:$ Alternatively we can define $\rm\:x\:$ to be odd. Again, by the Factor Theorem, we have $\rm\:f(x) = f(1) + (x1)\ g(x)\:$ for some $\rm\:g(x)\in \mathbb Z[x]\:.\:$ Since $\rm\:x1\:$ is even, this amounts to defining the parity of $\rm\:f(x)\:$ to be that of $\rm\:f(1)\:,\:$ i.e. the sum of its coefficients. Both definitions lead to a consistent parity arithmetic in the extension ring $\rm\:\mathbb Z[x]\:.\:$ But in general there is no compelling reason to decide which parity we should assign to the infinite element $\rm\:x\:.\:$
In contrast, there are "number" systems extending the integers where parity arithmetic has a unique extension. For example, the rational numbers (fractions) expressible with odd denominator have parity arithmetic given by defining the parity of $\rm\: m/(2\:n+1)\:$ to be the parity of $\rm\:m\:.\:$ Also the Gaussian integers $\ m + n\ i\:,\ m,n\in \mathbb Z\:,\ i = \sqrt{1}\:,\: $ have parity arithmetic given by defining $\:i\:$ to be odd. On the other hand, there are also such number rings with no extension of parity, or with numerous possible extensions. For further discussion see my post here.
Also, as JDH mentioned, parity arithmetic extends in some sense to more exotic structures such as ordinals, which may or may not satisfy your definition of a "number". Based on a few decades teaching such concepts, I suspect that you'll have much more luck teaching a 6yearold a concept of parity of polynomials vs. ordinals. Indeed, my experience is that many adult educated layfolks have difficulty comprehending ordinals (I've had hundreds of interactions with such adults based on my popular posts about Goodstein's Theorem, e.g. see my sci.math post of Dec 11 1995; update: now migrated here).
A nice introduction to the many different notions of infinity in mathematics is Rudy Rucker's book: Infinity and the Mind. Unlike many other popularizations, the author has expertise in the field, having completed a Ph.D. on a related topic. Moreover, Rucker has gone to great lengths to make the presentation faithful to the mathematics but still accessible to an educated layperson.
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3The concept of an "educated layperson" seems to me contradictory in nature. – dberm22 Jun 30 '15 at 19:02

@dberm22 If you search on this and closely related phrases you will find that they are commonly used, e.g. [this Google search.](http://www.google.com/search?q=educated+lay+person) – Bill Dubuque Jun 30 '15 at 19:10

You are also welcome to comment to my answer to this question that I posted in December. – Anixx Apr 28 '20 at 06:30
JH Conway's Surreal Numbers have a welldefined notion of Omnific Integer which extends the definition of integer from finite numbers. I believe this splits infinite integers between odd and even numbers according to whether they are twice an integer or not, and such that Omnific Integers which differ by 1 always have opposite parities.
I would not recommend the theory to a 6yearold, but Knuth's "Surreal Numbers" would be a good introduction for his father, and might give some interesting ideas on how to explore the idea of numbers with a child who is asking interesting questions.
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+1, esp. for the remark that it is not recommend for 6yearolds. I find it quite bizarre that anyone thinks that a 6yearold has any hope of comprehending ordinals (let alone surreals). This would be a challenge even for most prodigies at age 6. – Bill Dubuque Jul 03 '11 at 20:27
"Infinity" is not a number, but there are numbers that are infinite, including cardinals, ordinals, infinite nonstandard reals, and other things. Some of those can be considered even numbers.
The "infinity" you encounter in calculus would not normally be considered a number.
There are also other notions of "infinity", such as the ones involved in the Dirac delta function and its derivatives. But it's hard to see how to view those as being numbers.
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Look at this my answer. In that approach the "numerocity" ("refined cardinality") of natural numbers is denoted $\omega_$ and the numerocity of nonnegative integers is $\omega_+=\omega_+1$. Similarly, the numerocity of all integers is $\omega_+\omega_+$ and the numerocities of either odd or even numbers are $\frac{\omega_+\omega_+}2$.
Now, we can ask a question, whether we can generalize the concept of even and odd numbers to those infinite numbers, for instance whether $\omega_+$, $\omega_$ or $\frac{\omega_++\omega_}2$ are even or odd?
It seems to me that we can, but we have several choices with no one looking more natural than others.
$\omega_$ has resemblance to $0$ (we cannot divide by it or take logarithm) and $\omega_+$ resembles $1$. Also, regular parts $\operatorname{reg} (\omega_)^n=B_n(0)$ and $\operatorname{reg} (\omega_+)^n=B_n(1)$ So, there is a reason to count $\omega_$ as even and $\omega_+$ as odd.
On the other hand, the regular part of $\omega_$ is $1/2$ and that of $\omega_+$ is $1/2$. So there is a reason to count them both neither even nor odd and count their mean $\frac{\omega_+\omega_+}2$ as even since it has regular part zero.
The logarithm of $\omega_+$ has regular part $\gamma$, if we exponentiate the regular part of the logarithm, we come to a conclusion the absolute valuelike measure of $\omega_+$ is $e^{\gamma}$. So, there is a reason why we better consider $\omega_+e^\gamma$ as an odd number...
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