Re: Irrational numbers questions

On Wed, 03 Jan 2007 12:57:34 +0100, Han de Bruijn
<Han.deBruijn@xxxxxxxxxxxxxx> wrote:

The following is from an old poster at 'Irrational Numbers Proof':

http://www.newton.dep.anl.gov/askasci/math99/math99119.htm

Quote:

We write ln(N) = a, which means what power "a" must I raise the number
"e" in order to obtain the number "N". The answer is always irrational
for integers "N" except for N=1, because ln(1) = 0. The function ln(N)
may even be transcendental.

May be true. But I wonder how to prove this.

Let's start somewhere. Let it be known that e is irrational. A rather
concise proof is found in:

http://en.wikipedia.org/wiki/Proof_that_e_is_irrational

Can it be proven that, with e irrational, also e^N (with N natural) is
irrational?

It's true that e^N is irrational for natural numbers N. That
doesn't follow from the fact that e is irrational (for example
if x = sqrt(2) then x is irrational although x^2 is rational),
but it's a fact, a special case of the fact that e is transcendental.

Guess so, but why .. And is the sum of an irrational and a
rational number also irrational? Guess so, but why ..

The fact that the sum of an irrational and a rational must
be irrational is on the other hand completely trivial.

Assume now that ln(N) is rational = m/n . Then e^m = N^n where the left
hand side is irrational and the right hand side is rational. Giving a
contradiction. Thus ln(N) is irrational. Right?

Yes.

Does it follow now that sum(k=1..N)(1/k) - ln(N) is irrational?

Yes, because a rational plus an irrational is irratinoal.

Note: this is an approximation of the Euler-Mascheroni constant.

Han de Bruijn


************************

David C. Ullrich
.

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