To prove the irrationality of $$e = \sum ^\infty _ {n=0} \frac{1}{n!}$$ we can show that $e \lt 3$ by using a suitable geometric series. By Taylor's theorem (applied to $a=0$ and $b=1$) we know that, $$ e-\sum ^n _{k=0} \frac{1}{n!} = R_n = e^\alpha \frac {1}{\bigl(n+1\bigl)!} $$ for some $\alpha$ such that $0 \lt \alpha \lt 1$. Since $e \lt 3$, $R_n \lt \frac{3}{(n+1)!}.$ Now suppose that $e$ is a rational number $\frac{c}d$ where $c$ and $d$ are co-prime. For $n=d$ we see that $d! R_d \in Z$. On the other hand, using the estimate for $R_d$ that we have obtained using Taylor's theorem, $$d! R_d \lt \frac{d! \cdot 3}{(d+1)}! \lt 1$$ if $d \geq 2$
How can I show $\pi$ is irrational using similar ideas?

  • $\begingroup$ There is a method, but kind of complicated, at least more complicated than this, and the method is not using similar idea. You may google it. $\endgroup$ – xbh Aug 12 '18 at 6:04
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    $\begingroup$ You might want to look at Niven's proof that $\pi$ is irrational. $\endgroup$ – Robert Israel Aug 12 '18 at 6:15
  • $\begingroup$ None of these seem to take the approach you want: en.wikipedia.org/wiki/Proof_that_π_is_irrational $\endgroup$ – J.G. Aug 12 '18 at 6:29
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    $\begingroup$ A mild warning: Over 90 per cent of the text in your question is in an image, and hence not searchable. This presses people's buttons here. Do study our guide to new askers. About your question, the Taylor series having values related to $\pi$ often represent inverse trig functions, and don't have those nice factorials in the denominators of the coefficients. Therefore irrationality of $\pi$ needs a different technique, $\endgroup$ – Jyrki Lahtonen Aug 12 '18 at 6:35
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    $\begingroup$ Sorry, I don't visit google drives. Last time I tried I couldn't see anything, because I block cookies from google. I try to keep the information google can collect on my person at a minimum, and (because my cell phone is running on Android) need to be extra careful. $\endgroup$ – Jyrki Lahtonen Aug 12 '18 at 7:48

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