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\(e \) and \( \pi\)

Theoretically, can \( \pi \) contain the entire sequence of digits of \(e \) ? Also, can \(e \) contain the entire sequence of digits of \( \pi \)?

Note by Hobart Pao
9 months, 2 weeks ago

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This reminds me of Banach–Tarski paradox.

Reshared! Pi Han Goh · 9 months, 2 weeks ago

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Yes! I'll give a constructive argument in a few minutes Agnishom Chattopadhyay · 9 months, 2 weeks ago

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@Agnishom Chattopadhyay ...

  • Digits of pi in e = 2.7182818284590452353602874713526624977572470936999595749669676277240766303535475945713821785251664274274663919320030599218174136

  • Digits of e in pi = 3.1415926535897932384626433832795028841971693993751058209749445923078164062862089986280348253421170679821480865132823066470938446095505822317253594081056239

Agnishom Chattopadhyay · 9 months, 2 weeks ago

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@Agnishom Chattopadhyay I meant seeing 314159265358979323846...somewhere in \(e \) eventually. Without skipping digits. Because the way you did it, of course you could fit all the digits eventually lol Hobart Pao · 9 months, 2 weeks ago

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@Hobart Pao If it did, then the following would hold:

\[ e = x + 10^{-i}\pi \]

where x is a rational number and i is an integer. I do not think this is true. Although, I cannot prove it. Agnishom Chattopadhyay · 9 months, 2 weeks ago

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@Agnishom Chattopadhyay ah! interesting...why does x have to be rational? It could be the sum of two irrationals, right? And since \( \pi \) has infinite digits, will your rational number have decimals? Because you have to then avoid modifying the \( \pi \) portion and maybe have a.0000000000000000000000000000000000000000000000...11111111111111111111111111111111111111111111111111...0000000000000000000000000 just as an example. I could be wrong because I haven't slept much...lol Hobart Pao · 9 months, 2 weeks ago

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@Hobart Pao Let's say that e was 2.72831415...

In this case, x= 2.728 is the rational part Agnishom Chattopadhyay · 9 months, 2 weeks ago

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@Agnishom Chattopadhyay but nonetheless, that is interesting! Would it be okay if \( \pi \) were at the end of \(e \)? Because both are irrational. Would it matter? Hobart Pao · 9 months, 2 weeks ago

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@Agnishom Chattopadhyay so you're talking about putting \( \pi \) at the end of \(e \)? That doesn't make much sense Hobart Pao · 9 months, 2 weeks ago

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@Hobart Pao because it could just be somewhere in the "middle" of \( e\) because \(e \) doesn't end Hobart Pao · 9 months, 2 weeks ago

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@Hobart Pao \(\pi\) is irrational and hence doesn't have a terminating decimal expansion; how can it be in the middle of \(e\)? Ivan Koswara · 9 months, 2 weeks ago

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Not necesary, I think, is like saying that the secuence of digits of \(e\) is a subset of the secuence of digits of \(\pi\) Hjalmar Orellana Soto · 9 months, 2 weeks ago

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@Hjalmar Orellana Soto I mean I've seen these articles where they say "Any/all of your numerical passwords, everyone's birthday, blah blah blah are contained within pi" or something like that. This may be an obvious question, but where's the limit on length of a sequence of digits that can fit into pi? Hobart Pao · 9 months, 2 weeks ago

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A note: Only one of them can be true. I shall prove it, by Contradiction.

Let \(e\) have \(a\) digits and \(\pi\) have \(b\) digits.

Then, \(a \leq b \leq a\).

Since a=a, a=b.

a = b implies that \(\pi = e\).

Contradiction. Aloysius Ng · 9 months, 2 weeks ago

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@Aloysius Ng No, both numbers are irrational and they don't terminate, so you can't say that \(e\) have \(a\) digits and \(\pi\) have \(b\) digits. Pi Han Goh · 9 months, 2 weeks ago

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