Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

How to prove that the number $e=2.718281...$ is a transcendental number? The truth is I have no idea how to do it.

If I can recommend a book or reference on this topic thank you.

There are many tests on the transcendence of $ e $?

I'd read several shows on the transcendence of $ e $

share|improve this question
cs.toronto.edu/~yuvalf/… –  kush Nov 9 '12 at 17:25

3 Answers 3

up vote 6 down vote accepted

Try Michael Spivak's Calculus. I find it amusing that he would prove the transcendence of $ e $ in a calculus textbook.

share|improve this answer
I thought the same exact thing! And he also devotes an entire chapter to the mere transcendence of $\pi$. –  Taylor Martin Nov 9 '12 at 18:41
Oh... I didn't realize that he also proved the transcendence of $ \pi $ in the same book. It is a beautiful book. –  Haskell Curry Nov 9 '12 at 19:33

Your might be interested in the Lindemann-Weierstrass-theorem, which is useful for proving the transcendence of numbers, e.g., $\pi$ and $e$. If you read further, you'll see that the transcendence of both $\pi$ and $e$ are direct "corollaries" of the Lindemann-Weierstrass theorem.

Indeed, $e^x$ is transcendent if $x$ is algebraic and $x \neq 0\,$ (by the Lindemann–Weierstrass theorem).

A sketch of a (much) more elementary proof is given here.

share|improve this answer

I realize that amWhy already gave one answer: the Lindemann-Weierstrass theorem. I dont know how to prove this... to be frank it is over my head. But in general it states that in the equation: $$e^a = b$$ If $a\ne 0$ is algebraic then $b$ is transcendental. The converse is also true: if $b$ is algebraic then $a$ must be transcendental. Obviously $a=0,b=1$ has to be an exception.

Take that on faith for a moment. Like I said, I cannot prove the L-W theorem but I welcome you to research it.

If you ask whether or not $\ln(2)$ is algebraic or transcendental, take note that $e^{\ln(2)}=2$ is algebraic, and you can conclude that $\ln(2)$ is transcendental.

Similarly, you asked if $e$ itself was algebraic or transcendental. That is easy. $e^1 = e$ by definition. The power is algebraic therefore $e$ is transcendental. This is not circular reasoning. It rests squarely on the truth of the Lindemann-Weierstrass theorem: for any non-zero algebraic power, $e^a$, whatever that may be, is transcendental.

If you ever see a proof for the transcendence of $e$ more complicated than the one I just gave, I assure you it is just a proof for a special case of the L-W theorem. The general case proof would suit you better.

Likewise, take Euler's Identity: $e^{i\pi} = -1$. The result is algebraic, therefore the power $i\pi$ is transcendental. But we already know $i$ to be algebraic, so it is the $\pi$ that is transcendental. This is how you prove the transcendence of $\pi$.

I would also like to mention a theorem that hasnt been mentioned yet: the Gelfond-Schneider theorem. It states that: $$a^b$$ is always transcendental if:

  • $a$ is algebraic and $a\ne 0,1$, and
  • $b$ is algebraic and irrational

Numbers such as $2^{\sqrt{2}}$ are proven transcendental in this way.

We can also prove that $\sqrt{2}^{\sqrt{2}}$ is transcendental as a consequence of the G-S theorem, because it is simply the square root of $2^{\sqrt{2}}$. The square root of a transcendental number is still transcendental.

I, personally, do not know of any other helpful theorems. But then I am new to transcendental number theory. Good luck.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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