# Approximation of $e$ by a rational number

I am having a problem with the final question of this exercise.

Show that $e$ is irrational (I did that). Then find the first $5$ digits in a decimal expansion of $e$ ($2.71828$).

Can you approximate $e$ by a rational number with error $< 10^{-1000}$ ?

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Yes, you can. Do you need to give such a number? Do you know about continued fractions? –  Gregor Bruns Nov 14 '12 at 12:10
I don't think I need to find the actual number. But I think I have to just prove its existence. I never actually manipulated with continued fractions –  user43418 Nov 14 '12 at 12:12
This question is very unclear. It could mean one of three things: 1: Can e be approximated by such a rational? Answer: Yes, obviously -- every real number can. 2: Do you know how to approximate it? Answer: Yes, in principle. 3: Find such a rational. Answer: Give me a few moments... –  TonyK Nov 14 '12 at 14:31
I think that a more interesting question is whether there is a good short rational approximation to e, analogous to 355/113 for $\pi$. Approximating e to absurd accuracy is just donkeywork for a computer. What puzzles me here is: if you are a good enough mathematician to prove that e is irrational, why would you ask such a naive question? –  John Bentin Nov 14 '12 at 14:45

In the standard proof that $e$ is irrational, one first proves that $$0 < e -s_n < \frac1{n!n} \qquad\mbox{where}\qquad s_n = \sum_{k=0}^n \frac1{k!}$$ So you only need to find $n$ such that $\frac1{n!n}< 10^{-1000}$ or $n!>10^{1000}$. You can use Stirling's approximation for that I guess. Wolfram Alpha says $n=450$ suffices.

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I would think that stating that such an n exists would probably suffice for this question. One could show the Taylor expansion for e has an Nth term such that all n > N terms will be smaller than 10^{-1000}. –  user12345121212 Nov 14 '12 at 12:16

I don't think we have to really know anything about $e$ to say that we can approximate it with a rational number with an error less then $10^{-1000}$.

Say $e=a_0.a_1 a_2 \ldots$

There is clearly a rational number $q=a_0.a_1 \ldots a_{1001}$ and $|q-e|\leq10^{-1000}$ (note that the difference is bounded by a geometric sum)

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Note that the question asks for an error less than $10^{-1000}$. –  robjohn Nov 15 '12 at 14:36
@robjohn - thanks for pointing that out, I edited accordingly. –  Belgi Nov 16 '12 at 23:25

If 2.71828 are the first few digits of $e$ the we have $$2.71828 < e < 2.71829$$ Put $q = 271828/100000$ we deduce $$0 < e-q < 0.00001 = 10^{-5}.$$

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What does this answer ? –  Belgi Nov 14 '12 at 12:51
@Belgi, think a bit. it shows how to solve the question asked –  sperners lemma Nov 14 '12 at 15:51
@spernerslemma: does it give any more of a hint than do either of the previous answers? –  robjohn Nov 14 '12 at 16:42
@Belgi: extrapolate by changing 5 to 1000, but don't write $e$ to 1000 places :-) –  robjohn Nov 15 '12 at 14:33
@Belgi: the question is "Can you approximate $e$..." not "Approximate $e$..." No answer actually computes a rational approximation of $e$ to an error of less than $10^{-1000}$. –  robjohn Nov 17 '12 at 5:11