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.

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    $\begingroup$ "Odd" and "even" are typically only applied to natural numbers, of which infinity is not. "1.4" is neither odd nor even as well. $\endgroup$ Commented Jul 2, 2011 at 15:34
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    $\begingroup$ Neither or both because $\infty=2\cdot \infty=2\cdot \infty + 1$ $\endgroup$
    – Listing
    Commented Jul 2, 2011 at 15:37
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    $\begingroup$ @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. $\endgroup$ Commented Jul 2, 2011 at 21:15
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    $\begingroup$ @The Chaz. Ditto for irrationals, e.g. $\:\sqrt{3}\:$ is odd in $\:\mathbb Z[\sqrt{3}]\:.$ See here for more. $\endgroup$ Commented Jul 3, 2011 at 0:37
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    $\begingroup$ Infinity is not a number to start with :) $\endgroup$
    – Cano64
    Commented Aug 19, 2015 at 16:29

7 Answers 7


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 left-divisible 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 order-isomorphic 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 non-commutativity 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 easiest-to-grasp 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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    $\begingroup$ Please record your 6 year old's expression when you explain it this way ;) $\endgroup$
    – Henrik N
    Commented Jul 2, 2011 at 18:03
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    $\begingroup$ @James: " Also, ∞ is an odd creature in that if you add to it it still remains ∞." perhaps we have solved the mystery. :) $\endgroup$ Commented Jul 2, 2011 at 20:06
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    $\begingroup$ Bill, 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. $\endgroup$
    – JDH
    Commented Jul 3, 2011 at 23:56
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    $\begingroup$ Almost anyone including children can learn to count to $\omega^2$, naming the numbers in turn and describing the general pattern---and my son can count much higher---the 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 Zeno-style explanation of getting half way to the line, saying $1$, then half-way again, saying $2$, and so on, finally saying $\omega$ when you are at the line. $\endgroup$
    – JDH
    Commented Jul 6, 2011 at 2:27
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    $\begingroup$ Once 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... $\endgroup$
    – JDH
    Commented Jul 6, 2011 at 2:53

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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    $\begingroup$ An 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 $a+a=a$. $\endgroup$
    – Asaf Karagila
    Commented Jul 2, 2011 at 17:47
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    $\begingroup$ @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 "even-ness" for infinite sets, but it is not clear that they are equivalent without the axiom of choice. $\endgroup$
    – JDH
    Commented Jul 4, 2011 at 0:42
  • $\begingroup$ @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...? $\endgroup$ Commented Jul 4, 2011 at 1:02
  • $\begingroup$ 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. $\endgroup$
    – JDH
    Commented Jul 4, 2011 at 1:34
  • $\begingroup$ @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. $\endgroup$
    – Asaf Karagila
    Commented Jul 4, 2011 at 5:20

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 6-year-old. I explain it at a level that is hopefully comprehensible to his 38-year-old 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 well-defined, viz. since polynomials have only a finite number of roots, they eventually have constant sign, so if $\rm\,f\ne g$ then eventually $\rm\,f-g\,$ is $>0\,$ or $<0,\,$ thus eventually $\rm\,f>g\,$ or $\rm\,g>f$. In fact it is easy to see that this is equivalent to defining the sign of a polynomial to be the sign of its leading coefficient (the leading term eventually dominates lower-degree 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). [Note: more generally see function germs at $\infty$, e.g. Hardy fields].

This ordered 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 $\rm\,f(x) = f(1) + (x-1)\ g(x)\,$ for some $\rm\,g(x)\in \mathbb Z[x]\,.\,$ Since $\rm\,x-1\,$ 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, rational numbers (fractions) writable 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 complex integers $\rm\ m + n\ {\it i}\:\!,\ m,n\in \mathbb Z\,,\ {\it 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.

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" (since they are not rings, e.g. addition and multiplication may not be commutative). Based on a few decades teaching closely related concepts, I suspect that you'll have much more luck teaching novices 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).

An accessible 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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    $\begingroup$ The concept of an "educated layperson" seems to me contradictory in nature. $\endgroup$
    – dberm22
    Commented Jun 30, 2015 at 19:02
  • $\begingroup$ @dberm22 If you search on this and closely related phrases you will find that they are commonly used, e.g. this Google search. $\endgroup$ Commented Jun 30, 2015 at 19:10
  • $\begingroup$ You are also welcome to comment to my answer to this question that I posted in December. $\endgroup$
    – Anixx
    Commented Apr 28, 2020 at 6:30

JH Conway's Surreal Numbers have a well-defined 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 6-year-old, 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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    $\begingroup$ +1, esp. for the remark that it is not recommend for 6-year-olds. I find it quite bizarre that anyone thinks that a 6-year-old has any hope of comprehending ordinals (let alone surreals). This would be a challenge even for most prodigies at age 6. $\endgroup$ Commented Jul 3, 2011 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.


Look at this my answer. In that approach the "numerocity" ("refined cardinality") of natural numbers is denoted $\omega_-$ and the numerocity of non-negative 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 value-like measure of $\omega_+$ is $e^{-\gamma}$. So, there is a reason why we better consider $\omega_+e^\gamma$ as an odd number...

  • $\begingroup$ @JDH your comment? $\endgroup$
    – Anixx
    Commented Dec 22, 2019 at 13:20

This is intended mostly for the "38 year old father". The most straightforward way of formalizing the even/odd distinction for infinite numbers is in nonstandard analysis, which works with a number system that naturally includes infinite (more precisely, unlimited) numbers, which have the same properties as the ordinary (standard) numbers, such as the following properties: even $\pm1=$ odd, odd $\pm1=$ even. Cantorian infinities fail at least some of those properties. Meanwhile, in nonstandard analysis they are satisfied by the transfer principle.


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