Philosophically why should proving that $\gamma$ is irrational (let alone transcendental) be so much harder than proving $\pi$ or $e$ are irrational?

45$\begingroup$ Because philosophers are not likely to provide a key idea? $\endgroup$ – KConrad May 2 '13 at 2:26

16$\begingroup$ As funny as it may be to pretend something else was meant, asking what the philosophical justification is does not ask why philosophers aren't proving theorems for us. $\endgroup$ – Douglas Zare May 2 '13 at 2:39

49$\begingroup$ I feel like this question points in the wrong direction. As far as I understand (I'm not an expert), there's nothing special about $\gamma$ that makes it particularly hard. Instead, just about all irrationality questions are hard by default, and it's $e$ and $\pi$ that are special in being unusually tractable. $\endgroup$ – Henry Cohn May 2 '13 at 3:31

$\begingroup$ I can, at the least, say that the irrationality of $e$ is not so tough to prove given its Maclaurin Expansion. See, e.g., mathoverflow.net/questions/103129/… $\endgroup$ – Benjamin Dickman May 4 '13 at 20:21

$\begingroup$ +1 because the question produced a great answer. I think it might have been a perfectly good question if it asked what is the philosophy behind proving transcendence, why have we been able to prove what we proved, and why the limits we have exist, etc. $\endgroup$ – David Corwin May 5 '13 at 5:13
There are number theorists who understand this subject much better than I do. However, I feel obliged to post an incomplete answer quickly before people have a chance to close this question.
There are a lot more connections known between $\pi$ and $e$ and other numbers than between $\gamma$ and other numbers. We can get proofs of their irrationality by using some of these connections, such as continued fraction expansions for both.
$\gamma$ may be thought of as a renormalized version of $\zeta(1)$, where $\zeta$ is the Riemann zeta function $\zeta(s) = \sum_{n=1}^\infty n^{s}$.
$$\gamma = \lim_{s\to 1} \bigg(\zeta(s)  \frac{1}{s1} \bigg)$$
At even integers, $\zeta(s)$ may be rewritten as a sum over nonzero integers, not just the positive integers. That's one explanation for why it is easier to get a handle on $\zeta(s)$ at even values (where it is a rational times $\pi^s$) than at positive odd integer values. See the answers to "Establishing zeta(3) as a definite integral and its computation."
There is some hope. ApĂ©ry proved that $\zeta(3)$ is irrational, and this can be related to proofs that other well known numbers are irrational. There are expressions for $\pi$, $\log 2$, $\zeta(3)$ as periods, definite integrals of algebraic functions on $[0,1]$. These can be used in a unified way to prove all of these are irrational (although it's still tricky for $\zeta(3)$), and there are conjectures about the possible rational or algebraic relations between periods. However, so far, $\gamma$ isn't known to be a period although it is an exponential period (as is $e$). No other values of $\zeta$ at positive odd integers are individually known to be irrational.

3$\begingroup$ Note, though, that the fact that $\gamma$ is not known to be a "period" does not exclude an irrationality proof from some other direction; the irrationality of numbers such as $\log_2 3$ is even easier to prove than the irrationality of $\pi$, and $\log_2(3)$ is not expected to be a period (though it's the ratio of the periods $\log 3$ and $\log 2$). $\endgroup$ – Noam D. Elkies May 2 '13 at 13:46

4$\begingroup$ For a general number field $K$, Ihara introduced in 2006 an EulerKronecker constant $\gamma_K$, which is best defined from the Laurent expansion at $s=1$ not of $\zeta_K(s)$, but rather of $\zeta_K'(s)/\zeta_K(s)$: $\zeta_K'(s)/\zeta_K(s) = 1/(s1) + \gamma_K + O(s1)$. In terms of the Laurent expansion $\zeta_K(s) = R/(s1) + c + O(s1)$, we have $\gamma_K = c/R$. For $K = {\mathbf Q}$, $R = 1$ and therefore $c = \gamma_{\mathbf Q}$. But for general $K$ the number $R$ is not 1 and the constant term of $\zeta_K(s)$ at $s = 1$ is the "wrong" object of interest. $\endgroup$ – KConrad May 2 '13 at 22:32

2$\begingroup$ "There are a lot more connections known between Ď€ and e and other numbers than between Îł and other numbers." This may not entirely true. The numbers $e^{\gamma}$ pops up every now and then in the theory of primes. For example Merten's Theorems, CramerGranville's conjecture etc to name a few. But I do agree that the connection between $\gamma$ and other numbers that has nothing to do with primes directly or indirectly is far less common. $\endgroup$ – Nilotpal Kanti Sinha May 3 '13 at 5:25

1$\begingroup$ You claim e known to be a period; AFAIK, but I might be wrong, this is neither known (nor much conjectured). Could you please provide a reference for this claim. Thanks in advance. $\endgroup$ – user9072 May 4 '13 at 18:52

11$\begingroup$ Does this mean the OiledMacaroni constant is a renormalized feta function? Mmm... Greek food... $\endgroup$ – Brian Rushton May 4 '13 at 20:33
Philosophically, there is essentially only one way to prove that a number is irrational/transcendental, which is to use the fact that there is no integer between 0 and 1. That is, one assumes that the number in question is rational/algebraic, and constructs some quantity that can be shown to be bounded away from 0, less than 1, and also an integer. To get these estimates, one typically needs some rapidly converging series expansion that is closely related the number of interest. For example the reason that Fourier's proof that $e$ is irrational is so simple is that we have a readymade rapidly converging series $\sum 1/n!$ for $e$ that allows us to construct an integer between 0 and 1 from the assumption that $e$ is rational. As for numbers like $\pi$ or $e^\pi$, they basically piggyback on the rapidly converging series $\exp z = \sum z^n/n!$ because $\exp(i\pi) = 1$.
In general, the less obvious it is how to relate your number to a suitable rapidly converging series, the harder it will be to prove irrationality/transcendence. ApĂ©ry's dramatic success with $\zeta(3)$ was based on the highly nonobvious rapidly converging series representation $$ \zeta(3) = \frac{5}{2} \sum_{n=1}^\infty \frac{(1)^{n1}}{n^3\binom{2n}{n}}. $$ But $\zeta(5)$ doesn't yield to exactly the same method because there is ample numerical evidence that the obvious analogous series that you would conjecture for it after seeing the above identity does not exist. Later progress on $\zeta(2n+1)$, particularly by Rivoal and Zudilin, relies on taking various subtle linear combinations of them in order to eventually deduce the nonexistent integer between 0 and 1, and so are only able to prove that at least one of a set of zeta values is irrational.
So could there be another proof that Euler missed out there, based on an elementary but nonobvious identity for $\gamma$ that has the right properties for the usual machinery to grind through? Maybe, but currently there is no method in sight for relating $\gamma$ in the right way to a suitable rapidly converging series. If you like, that is the "philosophical" reason why we're stuck.
By the way, I'd highly recommend Making Transcendence Transparent by Burger and Tubbs if you want an accessible treatment of transcendental number theory. They do an excellent job of showing how the same basic ideas underlie all the results in the area, while introducing the technical complications one at a time in digestible chunks.

8$\begingroup$ As an afterthought, I figure I should mention how the irrationality of $\sqrt 2$ can be reconciled with my first sentence above. Given a positive integer $q_i$ with the property that $q_i\sqrt 2$ is also an integer, we can find a smaller positive integer $q_{i+1}:=q_i(\sqrt 21)$ with the same property. Iterating, we must eventually get some positive integer $q_n<1$. Of course this example is simple enough that we don't need a rapidly converging series expansion. $\endgroup$ – Timothy Chow May 6 '13 at 14:24

5$\begingroup$ Using the fact that $~\displaystyle\sum_{n=1}^\infty\frac{(2x)^{2n}}{n^2\displaystyle{2n\choose n}}=2\arcsin^2x$, the series in question can be rewritten as $$\zeta(3)=10\int_0^\tfrac12\frac{\text{arcsinh}^2x}xdx.$$ $\endgroup$ – Lucian Sep 16 '14 at 6:56

3$\begingroup$ Philosophically, there is essentially only one way to prove that a number is irrational/transcendental, which is to use the fact that there is no integer between 0 and 1 In a sense, yes, but then what (nonexistence) statement about integers does not fall into that category? I mean, what you say is true, but how far is it from a tautology? $\endgroup$ – fedja Dec 28 '14 at 17:29

2$\begingroup$ @DouglasZare: It's true that diagonalization is another technique, just as diagonalization can be used to construct Boolean functions of high circuit complexity or statements unprovable from a given set of axioms. Implicitly, I was talking about the limited number of methods of proving irrationality/transcendence of "natural" constants, analogously to the limited number of ways of proving high circuit complexity of "natural" (or "explicit") Boolean functions or coming up with "natural" unprovable statements. $\endgroup$ – Timothy Chow Dec 17 '15 at 21:46

1$\begingroup$ Diagonalization can be used to prove things about Chaitin's $\Omega$, but I wouldn't say it is part of the definition. I accept that it might not be viewed as natural, although there are variations that are more geometric, such as the probability that you can tile the plane with some randomly chosen tiles. I had hoped that you would turn the proof into something about an integer between $0$ and $1$, but it might not look like an important part of the proof. $\endgroup$ – Douglas Zare Dec 21 '15 at 0:27