1
$\begingroup$

I should mention that I still haven't taken Calculus or even Pre-Calculus, which is why I want to ask this. I've seen proofs $e$ is irrational, but not this one. Is this correct, and if it isn't, why not?

Prove $e$ is irrational:

One definition for $e$ is $$ e = \lim\limits_{n\to \infty} (1+\frac{1}{n})^n $$

Let's assume $e$ is rational. Because of that, we can set: $$ (1+\frac{1}{n})^n = \frac{a_n}{b_n}$$ $n$-root both sides: $$ 1+\frac{1}{n} = \sqrt[n]{\frac{a_n}{b_n}} $$ Distribute the $n$-root: $$ 1+\frac{1}{n} = \frac{\sqrt[n]{a_n}}{\sqrt[n]{b_n}} $$ Multiply by $\sqrt[n]{b_n}$ on both sides: $$ \sqrt[n]{b_n}(1+\frac{1}{n}) = \sqrt[n]{a_n} $$ Distribute the $\sqrt[n]{b_n}$: $$ \sqrt[n]{b_n}+\frac{\sqrt[n]{b_n}}{n} = \sqrt[n]{a_n} $$ Add the fractions: $$ \frac{n\sqrt[n]{b_n}}{n}+\frac{\sqrt[n]{b_n}}{n} = \sqrt[n]{a_n} $$ $$ \frac{n\sqrt[n]{b_n}+\sqrt[n]{b_n}}{n} = \sqrt[n]{a_n} $$ Factor out $\sqrt[n]{b_n}$: $$ \frac{\sqrt[n]{b_n}(n+1)}{n} = \sqrt[n]{a_n} $$ Divide by $\sqrt[n]{b_n}$ on both sides: $$ \frac{(n+1)}{n} = \frac{\sqrt[n]{a_n}}{\sqrt[n]{b_n}} $$ Factor the $n$-root: $$ \frac{(n+1)}{n} = \sqrt[n]{\frac{a_n}{b_n}} $$ Raise both sides to the power $n$: $$ (\frac{(n+1)}{n})^n = (\sqrt[n]{\frac{a_n}{b_n}})^n $$ Distribute the power $n$: $$ \frac{(n+1)^n}{n^n} = \frac{a_n}{b_n} $$

So, by the transitive property of equality, we have:$$ \frac{a_n}{b_n} = (1+\frac{1}{n})^n = \frac{(n+1)^n}{n^n} $$

Substituting that in, we have: $$ e = \lim\limits_{n\to \infty} \frac{(n+1)^n}{n^n}$$ We can just put infinity into the equation: $$ e = \frac{(\infty + 1)^\infty}{\infty^\infty} $$ $$ e = \frac{\infty^\infty}{\infty^\infty} $$ $$ e = \frac{\infty}{\infty} $$ Which is indeterminate. Because of this, out original assumption that $e$ is rational was wrong, therefore, $e$ is irrational.

$\endgroup$
  • 1
    $\begingroup$ There's no need for all that juggling, $1+1/n=\frac {n+1}n$... This proves nothing. I recommend you learn a bit more about limits. $\endgroup$ – YoTengoUnLCD Jun 8 '16 at 17:13
  • $\begingroup$ You could write $(1+\frac{1}{n})^n = \frac{(n+1)^n}{n^n}$ straight from the begining $\endgroup$ – Jennifer Jun 8 '16 at 17:13
  • $\begingroup$ (1) Even if $e$ were rational, it could still be a limit of irrational numbers. (2) The work to go from $(1+1/n)^n$ to $(n+1)^n/n^n$ is not necessary. (3) "Just put infinity into the equation" isn't meaningful. $\endgroup$ – mjqxxxx Jun 8 '16 at 17:14
  • 2
    $\begingroup$ You have not studied calculus. So you do not know that "put $\infty$ in the equation" is not a legitimate step. But keep studying! $\endgroup$ – GEdgar Jun 8 '16 at 17:18
  • 1
    $\begingroup$ That being said, I admire the spirit. I just think you need to learn more about limits and real numbers to make more genuine progress. Using that definition of $e$ to proceed toward a proof of its irrationality is going to be pretty difficult, otherwise (and needlessly so, honestly). $\endgroup$ – Brian Tung Jun 8 '16 at 17:21
7
$\begingroup$

Your proof is not correct.

First note that (toward the end) $$ \lim_{n\to \infty} \frac{(n+1)^n}{n^n} = \lim_{n\to \infty} \left(1 + \frac{1}{n}\right)^n $$ So you haven't really done anything.

--

Also, you say that since $e$ is rational $$ \left(1 + \frac{1}{n}\right)^n $$ is rational ($n\in \mathbb{Z}\setminus \{0\}$). But that is true even is $e$ is not rational.

--

Just cause $\lim_{x\to \infty} f(x) = \infty$ and $\lim_{x\to \infty} g(x) = \infty$, you can't say anything about $\lim_{x\to \infty} \frac{f(x)}{g(x)}$.

For example:

$$ \lim_{n\to \infty} \frac{n}{n} = 1 $$ and $$ \lim_{n\to \infty}\frac{n^2}{n} = \infty $$

$\endgroup$
5
$\begingroup$

No, this is not correct at all. In fact, you never use the assumption that $e$ is rational, since $1+\frac{1}{n} = \frac{n+1}{n}$ is perfectly true.

When you have an indeterminate form, it does not mean that someting is wrong, but simply that you must make further calculations to conclude.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy