Jonathan Sondow elegantly proves the irrationality of e in his aptly titled A Geometric Proof that e Is Irrational and a New Measure of Its Irrationality (The American Mathematical Monthly, Vol. 113, No. 7 (Aug. - Sep., 2006), pp. 637, http://www.jstor.org/stable/27642006).

In his argument, he constructs a sequence of nested intervals $I_n$ for every $n \geq 1$, each of the form $[k/n!, (k+1)/n!]$, such that $\bigcap I_n = \{e\}$, with $e$ lying strictly between the endpoints of each $I_n.$ From this, we conclude that $e$ cannot be written as a fraction with denominator $n!$ for any $n \geq 1.$

Fact: Every rational number $p/q$ can be written as a fraction with a factorial in its denominator: $p/q = p(q-1)!/q!$.

Thus, we conclude that $e$ is irrational.

The reason this proof technique works so well with $e$ is, of course, related to the Maclaurin series for the exponential function, $e^x.$

That any rational number can be written in lowest terms is employed in other irrationality proofs (e.g., the classic proof for that of $\sqrt{2}$) but I had not seen the above fact drawn upon before reading this particular paper.

My question is: are there other examples of real numbers (which are not related to $e$ in some trivial way) whose irrationality can be proved using the Fact above?

  • $\begingroup$ Trivia: the Fact is related to an old sequence due to Lucas and Kempner: oeis.org/A002034 $\endgroup$ – Charles Jul 26 '12 at 6:38

The same proof technique, for modified versions of the Fact, proves that some values of some hypergeometric functions are irrational. For example, the Bessel functions of the first kind have the following power series:

$$J_n(x) = \sum_{i=0}^\infty \frac{(-1)^i}{i! (i+n)!} \bigg(\frac x 2\bigg)^{2i+n} $$

For any choice of integers $m\ne 0$ and $n$, every rational number can be written as a quotient of integers so that the denominator is of the form $i!(i+n)!m^{2i+n}$. Since $J_n(2/m)$ can't, it is irrational.

| cite | improve this answer | |

For any interval $I_n$ of the form you define, there are $n+1$ intervals that could be $I_{n+1}$ nested in it. Choose one of those intervals, then repeat the process. As long as you choose an interval other than the first infinitely many times and an interval other than the last infinitely many times, you get an irrational number by this proof.

Or in other words, we have:

$$\sum_{m=n+1}^\infty \frac{m-1}{m!} = \sum_{m=n+1}^\infty \left(\frac{1}{(m-1)!} - \frac{1}{m!}\right) = \frac{1}{n!}$$

so in user22202's answer, for any sequence of integers $a_m$ such that $0 \leq a_m \leq m-1$, and $a_m>0$ infinitely often and $a_m < m-1$ infinitely often, we have

$$ \sum_{m=0}^\infty \frac{a_m} {m!}$$

is irrational.

| cite | improve this answer | |
  • 1
    $\begingroup$ It is worth noticing that the analogous conclusion holds even with changing sign coefficients: if $a_m\in\mathbb{Z}$ and $|a_m|\le m-1$, the sum of the corresponding series is a rational number if and only if $a_m$ is either eventually equal to $0$, or eventually equal to $m-1$, or eventually equal to $-(m-1)$. $\endgroup$ – Pietro Majer Jan 7 '16 at 9:37

Nice proof! Although now I wonder just how different it really is from this one. In Proofs from the Book that technique is to show that $e^2$ and with some adjustments, $e^4$ are irrational.

One can certainly take other sums where the proof you mention applies, although one has to be more careful than I initially thought:

Flawed conjecture (see the comments for a counterexample) You could take any convergent infinite sum $\sum \frac{n_i}{d_i}$ where $\gcd(n_i,d_i)=1$ , and for every integer $q \gt 1$ there is a $j$ so that $q$ divides $d_i$ for all $i \gt j$.

Strengthen the conditions to

  • $d_1 | d_2 | \cdots$
  • for every integer $q \gt 1$ there is an $i$ so that $q$ divides $d_i$
  • $0 \lt |\frac{n_i}{d_i}| \lt \frac{1}{d_{i-1}}$

and the proof does seem to go through. A question is which such sums give interesting real numbers.

$\cos(1)=1-\frac{1}{2}+\frac{1}{24}-\frac{1}{720}\cdots$ works, although $\cos(1)=\frac{e^i+e^{-i}}{2}$ so it depends what counts as a trivial relation. Also $\cos(\frac{1}{k}).$

| cite | improve this answer | |
  • 2
    $\begingroup$ It doesn't work that generally. Consider $\sum_{j=1}^\infty \dfrac{n_j}{j!}$ where $n_j$ is defined recursively as the greatest integer $x$ with $\gcd(x,j!) = 1$ and $\dfrac{x}{j!} < 1 - \sum_{i=1}^{j-1} \dfrac{n_i}{i!}$. The sum of this series is a rational number, namely $1$. $\endgroup$ – Robert Israel Jul 26 '12 at 1:59
  • $\begingroup$ (those sums should have started at $2$, not $1$) $\endgroup$ – Robert Israel Jul 26 '12 at 2:33
  • $\begingroup$ Good point, I'll revise it. $\endgroup$ – Aaron Meyerowitz Jul 26 '12 at 3:35

If one writes the real number $x$ as

$$ x=\sum_{m=0}^{\infty}\frac{a_m}{m!} $$

where $a_m\in \mathbb{Z}$, so that for each $n\in \mathbb{Z}$, $n>0$, one can write

$$ x=\frac{c_n}{n!} + \sum_{m=n+1}^{\infty}\frac{a_m}{m!} $$

for some $c_n\in \mathbb{Z}$, perhaps one can try to control the $a_m$ so that

$$ \sum_{m=n+1}^{\infty}\frac{a_m}{m!} < \frac{1}{n!} $$

for all $n$?

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.