In my Intro to Proofs course I have an activity (more details below) where the goal is to write a collaborative proof of a statement. What makes it work as a whole class activity is that, to prove the statement, students must first prove 4 independent lemmas. These 4 lemmas can be put together to make the final proof. This means I can initially make 4 groups, each to work on one lemma. Then, after a while, reorganize into groups of 4 where each group member had a different lemma. And these groups are tasked with creating the proof of the statement.

Unfortunately the example below is the only proof I know that has this kind of naturally independent pieces. Most proofs aren't so easily compartmentalized.

So, what are some other statements/proofs that similarly can be naturally broken into separate (but equally difficult) lemmas?

Details of existing activity:

  • Main statement: The constant $e = \sum_{n=0}^\infty \frac{1}{n!}$ is irrational.
  • Lemma 1: For $a,b \in \mathbb{Z}$, $$b!\left(\frac{a}{b} - \sum_{n=0}^b \frac{1}{n!}\right) \in \mathbb{Z}.$$
  • Lemma 2: For integer $b>0$, $$b!\left(e - \sum_{n=0}^b \frac{1}{n!}\right) >0.$$
  • Lemma 3: For integers $0 < b \le n$, $$\frac{b!}{n!} \le \frac{1}{(b+1)^{n-b}}.$$
  • Lemma 4: For $b \ge 1$, $$\sum_{k=b+1}^\infty \frac{1}{(b+1)^{k-b}} = \frac{1}{b}.$$

Note: Please do not post solutions and/or links to solutions for this activity.

  • 1
    $\begingroup$ What topics are you interested in? What tools do your students have available? $\endgroup$ Commented Feb 18, 2022 at 16:55
  • $\begingroup$ I'm going to say "any and all" in the sense that: whatever they don't know but need to know, I will give them and/or if it's too advanced for this class, I can do it in an upper level course. $\endgroup$
    – Aeryk
    Commented Feb 18, 2022 at 18:09
  • 1
    $\begingroup$ Leaving this as a comment since I do not have time to turn it into an answer, but I would imagine that (after defining inversion) you could independently assign many of its properties, and then have the "big theorem" be Steiner's porism. $\endgroup$ Commented Feb 19, 2022 at 12:45

1 Answer 1


The example you have is great! I imagine it would be really hard to actually have all parts with similar difficulty. E.g. in your example, L2 is an immediate observation, while L1 and L3 (particularly the latter) require some manipulations with $n!$. It's still great, but I would expect (I might be wrong, of course) the group solving L2 to finish much faster than other groups.

Here are two examples that I think have a similar spirit, although the division of work between different parts is far from equal.

Irrationality of $\pi$

A proof of $\pi \notin \mathbb{Q}$ found in Bourbaki's book (it might be due to someone else, see one of the comments) is naturally divided into parts, somewhat similarly to your proof of $e \notin \mathbb{Q}$. Take $f_n(x) = \frac{1}{n!} b^n x^n (\pi-x)^n$. Then:

  • L1: $\int_0^\pi f_n(x) \sin(x) dx > 0$
  • L2: $\int_0^\pi f_n(x) \sin(x) dx \to 0$ as $n \to \infty$
  • L3: If $\pi$ is a fraction $a/b$, then derivatives of any order of $f_n$ at $0$ and $\pi$ have integer values. Possible hint: write $b^n x^n (\pi-x)^n$ as $x^n (a-bx)^n$, and use the symmetry between $x=0$ and $x=\pi$.
  • L4: If $f$ is any polynomial of degree $2n$, then $\int_0^\pi f(x) \sin(x) dx$ is a sum of terms $\pm \left[ f^{(k)}(x) \sin x \right]_0^\pi$ and $\pm \left[ f^{(k)}(x) \cos x \right]_0^\pi$ for $k=0,1,\ldots,2n$.

Here L3 and L4 together imply $\int_0^\pi f_n(x) \sin(x) dx \in \mathbb{Z}$, which contradicts L1 and L2.

Liouville's theorem and a construction of a transcendental number

Lioville's theorem states that irrational algebraic numbers don't have efficient rational approximations. More precisely: if $a \in \mathbb{R}$ is an algebraic number of degree $d>1$, then $|a-\frac{p}{q}| \ge \frac{c}{q^d}$. The proof consists of basically three steps, and in the fourth step one can use the theorem to show that a certain number is not algebraic.

  • L1: Assume that $a$ is a root of the polynomial $f(x) = a_0+a_1 x+\ldots+a_d x^d$ with integer coefficients. Assume that $d$ here is the smallest possible. Then $f$ doesn't have rational roots.
  • L2: Assume that $f(x) = a_0+a_1 x+\ldots+a_d x^d$ has integer coefficients and $f(p/q) \neq 0$ (where $p,q \in \mathbb{Z}$). Then $|f(p/q)| \ge q^{-d}$.
  • L3: Let $f$ be any polynomial. Justify that there's a constant $C > 0$ (depending on $a$ and $f$) such that $|f(x)-f(a)| \le C |x-a|$ for $x \in [a-1,a+1]$. Remark: one doesn't even need derivatives to check that.
  • L4: Let $L := \sum_{n=1}^\infty 10^{-n!}$ (Liouville's number). Then for any $d \in \mathbb{N}$ and $c > 0$ there is a rational approximation $p/q$ such that $q^d |a-\frac{p}{q}| < c$.

Here L1, L2, L3 together imply Liouville's theorem mentioned before. If $L$ were algebraic (i.e., a root of a polynomial with integer coefficients), it would contradict this law.

  • 1
    $\begingroup$ I can't tell if this is what the OP wants, but I'm excited about these. I'll be studying this answer myself when I have free time. Maybe this summer. $\endgroup$
    – Sue VanHattum
    Commented Feb 18, 2022 at 22:19
  • 1
    $\begingroup$ Is the thing that you call Bourbaki's proof really Bourbaki's proof, or is it just a proof found in a book of Bourbaki? Mary Cartwright and Ivan Niven both wrote simplifications of Charles Hermite's proof during the 1940s and they look very similar to this one. $\endgroup$ Commented Feb 26, 2022 at 4:35
  • $\begingroup$ Frankly, I don't know (I'm just citing Wikipedia), but it looks like it's the latter. I'll edit the answer. Maybe you could add this nuance to the Wikipedia page? (I see you're already involved there) $\endgroup$ Commented Feb 26, 2022 at 9:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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