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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 $

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3  
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.

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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.

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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.

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