This is going to be hard to make precise, because it depends on details of the proof theory, but here I'm considering a foundation of ZFC just after constructing the real numbers as a complete ordered field, and the rational numbers as a subset of the reals. From such a basis, what is the shortest proof of that there exists an irrational number?

Here are some other statements from which the result follows easily:

  • The rational numbers are not complete.
  • The rational numbers are countable and the real numbers are not.
  • There is a real number $x$ such that $x^2=2$, and no rational number can have this property.

If there are methods other than the countable/uncountable proof or $\sqrt 2\not\in\Bbb Q$ that you think can be done with a shorter proof from the axioms, I would also be interested to hear.

Remember not to assume too much here. All of the proofs mentioned here are hiding some significant complexity: proving that the real numbers are uncountable requires constructing a cantor-set-like family of infinite series; proving that $\sqrt 2$ exists requires the construction of the square root function via the babylonian method or similar. I'm curious if there is some clever manipulation of the complete ordered field axioms that directly leads to the existence of an irrational number without the detours.

  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – Xander Henderson
    Aug 28, 2021 at 23:05
  • 2
    $\begingroup$ I'm concerned about an inherent "unanswerability" of this question. How is the shortest proof to be determined? $\endgroup$
    – Lee Mosher
    Aug 29, 2021 at 2:17
  • $\begingroup$ @LeeMosher While a formal definition can be given, I'm willing to settle for a big-list style collection of candidate proofs and a popularity contest between them to judge which seems shortest. $\endgroup$ Aug 29, 2021 at 10:04
  • $\begingroup$ Does this qualify? en.wikipedia.org/wiki/Proof_that_e_is_irrational $\endgroup$
    – lhf
    Aug 30, 2021 at 20:01

4 Answers 4


I'm not sure if this is the shortest approach, but proving that $\sqrt 2$ is irrational doesn't require too much effort. First, prove that no rational squares to $2$. Then, define the set $$ E=\{x\in\Bbb Q:x<0\text{ or }x^2<2\} \, . $$ (Those familiar with the construction of the reals will recognise $E$ as the Dedekind cut of $\sqrt 2$.)

This set does not have a supremum in the rationals: if $r\in\Bbb Q$ is an upper bound of $E$, then $s=(2r+2)/(r+2)$ is a smaller upper bound of $E$. Proof that $s$ is an upper bound of $E$: $$ \left(\frac{2r+2}{r+2}\right)^2-2=\frac{2(r^2-2)}{(r+2)^2}>0 \, . $$ Proof that $s<r$: $$ \frac{2r+2}{r+2}=r-\frac{r^2-2}{r+2}<r \, . $$ Since the reals have the least upper bound property, and $E$ is obviously non-empty and bounded above, $E$ must have a supremum in $\Bbb R$. And since $E$ does not have a supremum in $\Bbb Q$, this supremum must be irrational.

Remark: this approach doesn't actually prove that $(\sup E)^2=2$, but since we are given the ordered field axioms for free, it is not too much work to prove this. In the comments, user21820's outlines one possible approach. Below, I offer another way of showing that there is a real number which squares to $2$.

Using the least upper bound property of $\mathbb R$, we can prove that if a sequence which is increasing and bounded above, then its supremum is its limit. Now consider the sequence $(a_n)_{n\in\mathbb N}$ given by $a_0=1$, and $$ a_{n+1}=\frac{2a_n+2}{a_n+2} \, . $$ It can easily be proven by induction that this sequence is increasing and bounded above. Let $l=\lim_{n\to\infty}a_n$. We have $$ l=\lim_{n\to\infty}a_{n+1}=\lim_{n\to\infty}\frac{2a_n+2}{a_n+2}=\frac{\lim_{n\to\infty}2a_n+2}{\lim_{n\to\infty}a_n+2}=\frac{2l+2}{l+2} \, , $$ and so $l^2+2l=2l+2$, meaning that $l^2=2$. Since $(a_n)_{n\in\mathbb N}$ is a positive sequence of terms, we see that $l$ equals the positive square root of $2$.

  • $\begingroup$ You have the ordered field axioms, so you can use that to prove that the square of the supremum is two. $\endgroup$
    – user21820
    Aug 28, 2021 at 9:21
  • $\begingroup$ Hi @user21820. How would you use the ordered field axioms to prove that $(\sup E)^2=2$? It's not immediately clear to me. $\endgroup$
    – Joe
    Aug 28, 2021 at 9:28
  • 2
    $\begingroup$ Let that sup be $t$. If $t^2>2$ then find a better upper bound on $E$. If $t^2 < 2$ then find suitable $k,m∈ℕ$ such that $(t+1/m)^2 < 2$ and $k ≤ t·m < k+1$, so $(k+1)/m > t$ but $((k+1)/m)^2 ≤ (t+1/m)^2 < 2$. This does use the archimedean property (which can be proven using completeness again), but now I'm wondering whether there is a single-supremum proof... $\endgroup$
    – user21820
    Aug 28, 2021 at 9:37
  • 1
    $\begingroup$ The same could said for $\sqrt{p}$ for $p$ prime. $\endgroup$ Aug 29, 2021 at 0:07
  • $\begingroup$ Incidentally, you don't need the $x < 0$ disjunct in $E$. $\endgroup$ Aug 29, 2021 at 10:11

Suppose $\log_2 3 = \dfrac m n$ for some $m,n\in\{1,2,3,\ldots\}.$

Then $2^m = 3^n,$ so an even number equals an odd number.

Now consider the set of all positive fractions $m/n$ for which $2^m<3^n$ and the complementary set for which $2^m>3^n.$

  • 3
    $\begingroup$ We still need to argue that $\log_23\in\mathbb{R}$, so this isn't quite as short as it looks. $\endgroup$ Aug 29, 2021 at 1:51
  • $\begingroup$ @NoahSchweber : True, but I'm not entirely sure of the entire intent of the question. $\endgroup$ Aug 29, 2021 at 1:52
  • $\begingroup$ @NoahSchweber : ok, I've added a third paragraph. $\endgroup$ Aug 29, 2021 at 1:55
  • $\begingroup$ Clearly the best proof. I do not think that the proof that this number is real is necessary. $\endgroup$
    – Peter
    Mar 25 at 14:55

I realized that in your case there is a better approach than Joe's answer. Since you have the full completeness axiom, you should not apply it to a set of rationals. Rather, you should apply it to $E = \{ x : x∈ℝ ∧ x^2 < 2 \}$. Then you can much more easily prove that its supremum $t$ satisfies $t^2 = 2$, without any further use of the completeness axiom in any form (including the archimedean property). $ \def\lfrac#1#2{{\large\frac{#1}{#2}}} $

Sketch: Clearly $1 ∈ E$ and $E ≤ 2$, so let $t = \sup(E)$. If $t^2 > 2$, then $\big(t-\lfrac{t^2-2}{2t}\big)^2$ $=2+\big(\lfrac{t^2-2}{2t}\big)^2 > 2$, so $t-\lfrac{t^2-2}{2t}$ is a smaller upper bound for $E$ than $t$. $t^2 < 2$, then $\big(t+\lfrac{2-t^2}{3t}\big)^2$ $= t^2+\lfrac{2-t^2}{3t}·\big(2t+\lfrac{2-t^2}{3t}\big)$ $< t^2+\lfrac{2-t^2}{3t}·3t = 2$ because $\lfrac{2-t^2}{3t}$ $< \lfrac{4t^2-t^2}{3t}$, so $t+\lfrac{2-t^2}{3t} ∈ E$ contradicting definition of $t$.

  • $\begingroup$ I take it you still need the "prove no rational squares to 2" part of Joe's proof? Since that doesn't seem to fall out of this part of the proof. Besides that, I don't see how this proof differs from Joe's; the algebra looks about the same and everything in sight is rational functions so it doesn't matter if $E$ is a set of reals or rationals. $\endgroup$ Aug 31, 2021 at 1:07
  • $\begingroup$ @MarioCarneiro: Yes to your first question, but no to your second. It is more difficult to prove that the supremum $t$ of the set satisfies $t^2 = 2$ when it is a set of rationals, because to get contradiction from $t^2 < 2$ you need to find a member of the set greater than $t$, and as already shown in my answer it is easier to do so when the set is not restricted to rationals. $\endgroup$
    – user21820
    Sep 1, 2021 at 14:01
  • $\begingroup$ @MarioCarneiro: You missed the fact that $t$ is not rational so all the algebraic expressions here are not rational. The omitted details of Joe's proof requires an extra use of the completeness axiom as I stated in a comment there. $\endgroup$
    – user21820
    Sep 1, 2021 at 14:03
  • $\begingroup$ We can assume without loss of generality that $t$ is rational, because the whole point is to prove there is an irrational number so if $t$ isn't rational then we're done. (Obviously this assumption is false, but we don't know it is until we reach the contradiction at the end of the proof.) $\endgroup$ Sep 1, 2021 at 16:58
  • $\begingroup$ @MarioCarneiro: Ah ok you're working under that overarching assumption. Then you're right that you can use my algebraic argument for Joe's set as well. I had been assuming that you actually wanted to construct the square-root of two, and for that specific purpose my approach is shorter. $\endgroup$
    – user21820
    Sep 1, 2021 at 17:26

I really like this short elementary proof that $e$ is irrational given by A. R. G. MacDivitt and Yukio Yanagisawa (The Mathematical Gazette , Volume 71 , Issue 457 , October 1987 , pp. 217, DOI: https://doi.org/10.2307/3616765)

Step 1: Suppose $e$ is rational; $e$ being defined by the infinite sum, $\sum_{k=0}^\infty \frac{1}{k!}$; that is:

$$\sum_{k=0}^\infty \frac{1}{k!}=\frac{m}{n}\tag{1},$$

where $m$ and $n$ are positive integers.

Step 2: Multiply both sides of (1) by $n!$, thus

$$n!\left(1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+...+\frac{1}{n!} \right)+\frac{1}{(n+1)}+\frac{1}{(n+1)(n+2)}+...=n!\,\frac{m}{n}\tag{2}$$

so that $S=\frac{1}{(n+1)}+\frac{1}{(n+1)(n+2)}+...$ is a positive integer, with $nS\ge1$.

Step 3: Construct contradiction.

$nS\ge1$ can be written $(n+1)S-1\ge S$, implying that

$$\frac{1}{(n+2)}+\frac{1}{(n+2)(n+3)}+...\ge \frac{1}{(n+1)}+\frac{1}{(n+1)(n+2)}+...;$$

which after comparing both sides of the inequality term by term can immediately seen to be a contradiction.


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.