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Tangential fact (har har): The Taylor series for e^x, combined with the uniqueness of representing a real number in base-factorial, immediately shows that e is irrational.

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jeremyscanvic 4 days ago | parent [-]

I know that the irrationality of e can be proven using the theory of continued fractions or by showing that finite Taylor sums for e give out rational approximations that are too good for e to be rational. What you're saying does not ring a bell though. Can you elaborate on that? This sounds really interesting!

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ogogmad 3 days ago | parent [-]

Base-factorial is a representation for real numbers between 0 and 1. The representation has the form A/2! + B/3! + C/4! +..., where A is in {0,1}, B is in {0,1,2}, C is in {0,1,2,3}, etc. Every irrational number in [0,1) has only 1 such representation, while every rational number in [0,1) has exactly 2 such representations (where one of those is finite). Notice that e-2 is in [0,1), and the Taylor series of e^x produces a representation of e-2 in base-factorial which corresponds to neither possible forms that a rational number can take. It follows that e is irrational.

Some details here: https://en.wikipedia.org/wiki/Factorial_number_system#Fracti...

Ultimately, the above argument resembles Fourier's proof that e is irrational. But I find it more intuitive. A lot of other special constants can be proved irrational using similar techniques.

	▲	jeremyscanvic 2 days ago \| parent [-]
		What a wild number system! That's very cool. Thanks!