I tried googling for simple proofs that some number is transcendental, sadly I couldn't find any I could understand.

Do any of you guys know a simple transcendentality (if that's a word) proof?

E: What I meant is that I wanted a rather simple proof that some particular number is transcendental ($e$ or $\pi$ would work), not a method to prove that any number is transcendental, sorry for the confusion.

Or even a proof about transcendental numbers being 'as common' as algebraic numbers?

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    $\begingroup$ There are only countably many algebraic numbers... $\endgroup$ May 6, 2015 at 22:54
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    $\begingroup$ Well, given an algebraic, non-zero number $a,$ we have that $a\pi$ is transcendental because $\pi$ is. Hence, there are at least as many transcendentals as there are nonzero algebraics. $\endgroup$ May 6, 2015 at 22:56
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    $\begingroup$ Do you mean a proof that some specific number is transcendental or that there is at least one transcendental number? The counting argument of celtschk shows there are uncountably many but does not identify a single one. $\endgroup$ May 6, 2015 at 22:58
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    $\begingroup$ The Gelfond-Schneider and Lindemann-Weierstrass Theorems are extremely helpful in establishing trascendentality (assuming the numbers you're considering are of the appropriate forms) $\endgroup$
    – Hayden
    May 6, 2015 at 23:00
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    $\begingroup$ @RossMillikan : Can you really prove uncountability of $\mathbb R$ without constructing, for a given sequence, a member of $\mathbb R$ that is not in it? How? Cantor's arguments do give specific numbers not in the arbitraty sequence he starts with. ${}\qquad{}$ $\endgroup$ May 6, 2015 at 23:14

4 Answers 4


The easiest proof is via Liouville's criterion.

Lemma. Suppose $\alpha$ is an irrational algebraic number. There exist integers $C,n$ such that for all integers $p/q$, $$ \left| \alpha - \frac{p}{q} \right| \geq \frac{C}{q^n}. $$

You can find a proof of the lemma on Wikipedia, and in many other sources. Now consider the number $$ \alpha = \sum_{m=0}^\infty \frac{1}{10^{m!}}, $$ sometimes known as Liouville's number. Since the decimal expansion of $\alpha$ is aperiodic, it is irrational. On the other hand, for all $r$ we have $$ \left| \alpha - \frac{\sum_{m=0}^r 10^{r!-m!}}{10^{r!}} \right| \leq \sum_{m=r+1}^\infty \frac{1}{10^{m!}} \leq \sum_{m=r+1}^\infty \frac{1}{10^{(r+1)!+m-(r+1)}} \leq \frac{2}{10^{(r+1)!}}. $$ Since $2/10^{(r+1)!}$ is so much smaller than $10^{r!}$ for large $r$, it is not hard to check that $\alpha$ doesn't satisfy the conclusion of the lemma. Indeed, suppose that it did, for some $C,n$. Then for all $r$ we must have $$ \frac{2}{10^{(r+1)!}} \geq \frac{C}{10^{r!n}} \Longleftrightarrow 10^{r!n} \geq (C/2) 10^{(r+1)!} \Longleftrightarrow 1 \geq (C/2) 10^{r! (r+1-n)}, $$ which fails for large enough $r$.

Since the conclusion of the lemma doesn't hold, one of its premises must be false. Since $\alpha$ is definitely irrational, it must be not algebraic. In other words, $\alpha$ is transcendental.

  • $\begingroup$ @NealYoung Thanks for noticing! $\endgroup$ Feb 5 at 11:45

It's possible to argue fairly elementarily that Liouville's number $$ L = \sum_{k=1}^\infty 10^{-k!} $$ is transcendental, by showing directly that for every integer polynomial $$p(x) = a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0$$ there is a decimal position in $p(L)$ that must be nonzero.

The powers of $L$ have the form $$ L^k = c_1 10^{-b_1} + c_2 10^{-b_2} + c_3 10^{-b_3} + \cdots $$ where each of the (all different) $b_i$ is a number that can be written as the sum of $k$ (not necessarily different) factorials, and $c_i$ is some integer $\le k!$ that depends on whether some of the factorials are different.

Now let $n$ be the degree of $p$. Among the $b_i$s we find numbers of the form $B_{h,n}=(h+1)!+(h+2)!+\cdots (h+n)!$, and by choosing $h$ large enough, these numbers can be arbitrarily far from anything that can be written as a sum of fewer than $n$ factorials (which are the ones the appear in the expansion of lower powers of $L$).

So if we choose $h$ large enough, we find a $B_{h,n}$ such that the only terms in $p(L)$ that contribute to the digits around decimal position $B_{h,n}$ is the product of $a_n$ with $C_{h,n}$, which is nonzero.

(Everything to the right of this is the sum of products of some $10^{-b_j}$ with a factor that is at most $n!a_n+(n-1)!a_{n-1}+\cdots+a_1$. The bound of the factor depends only on $p$, so if only we choose $h$ such that that the first $b_j$ after $B_{h,n}$ is separated by at least the length of this bound their contribution cannot reach position $B_{h,n}$.

(Similarly, we can arrange for there to be enough space to the left of $B_{h,n}$ to make room for all of $a_nC_{h,n}$ before the previous $b_j$).

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    $\begingroup$ That's not a "simple" proof as currently written. It might be if you explained the pieces as you went along. But some parts are just plain confusing: "Now let $n$ be the degree of $p$." The way you defined $p$, specifies that it is a polynomial of degree $n$. $\endgroup$
    – Wildcard
    Dec 9, 2016 at 6:30

There is a BIG difference between showing that a particular number that naturally occurs (like $e$ or $\pi$) is transcendental and showing that some number is.

The existence of transcendental numbers was first shown in 1844 by Liouville. In 1851 he proved that $\sum_{k=1}^{\infty} \frac1{10^{k!}} $ is transcendental.

In 1873, Hermite proved that $e$ is transcendental. The next year, Cantor showed that the algebraic numbers were countable, so that almost all numbers are transcendental. In 1882, Lindemann showed that $\pi$ is transcendental.

Look up more if you want.

  • $\begingroup$ Yes, that's what I meant, a simple proof that a particular number, such as $\pi$ or $e$ is transcendental. $\endgroup$ May 6, 2015 at 23:34
  • $\begingroup$ Do a search for "prove e is transcendental". You get hits like this: sixthform.info/maths/files/pitrans.pdf $\endgroup$ May 6, 2015 at 23:39

Cantor showed a simple way to list the real algebraic numbers in a sequence $a_1,a_2,a_3,\ldots$. Every term has only finitely many terms before it, and every algebraic number will be reached after only finitely many steps. Now let us seek a transcendental number in the interval $(0,1)$. Find the first number in the above sequence that is between $0$ and $1$; call it $b_1$, and narrow down the interval from $(0,1)$ to $(b_1,1)$. Find the first term in the sequence that comes after $b_1$ and falls in our now narrower interval; callit $c_1$, and narrow down the interval from $(b_1,1)$ to $(b_1,c_1)$. Then find the first number in the sequence that comes after $c_1$ and falls in our now yet narrower interval; call it $b_2$, and narrow down the interval to $(b_2,c_1)$. And so on: the next narrowing will be to $(b_2,c_2)$ then $(b_3,c_2)$, then $(b_3,c_3)$. We have $0<b_1<b_2<b_3<\cdots < c_3<c_2<c_1<1$. Some number must be between the $b$s and the $c$, and it can't be algebraic because our process has eliminated all of those.

This argument involving narrowing was in Cantor's very first paper on uncountability, published in 1874. His "diagonal argument" came later.


  • $\begingroup$ "that comes after $b_1$", "that comes after $c_1$": these conditions are unnecessary, aren't they? $\endgroup$
    – TonyK
    May 6, 2015 at 23:39
  • $\begingroup$ @TonyK : I'm not sure${}\,\ldots\qquad{}$ $\endgroup$ May 7, 2015 at 3:56
  • $\begingroup$ It's unfortunate that the question was edited to outlaw this answer soon after the answer was posted. $\endgroup$ May 7, 2015 at 7:45
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    $\begingroup$ I don't see that it necessarily "outlaws" it. I do give a particular transcendental number between $0$ and $1$ (once you have the enumeration, which I was not explicit about). ${}\qquad{}$ $\endgroup$ May 7, 2015 at 15:01

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