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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?

Digits of pi in e = 2.7182818284590452353602874713526624977572470936999595749669676277240766303535475945713821785251664274274663919320030599218174136

Digits of e in pi =
3.1415926535897932384626433832795028841971693993751058209749445923078164062862089986280348253421170679821480865132823066470938446095505822317253594081056239

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

@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

@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?

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## Comments

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TopNewestThis reminds me of Banach–Tarski paradox.

Reshared!

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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 $\pi$

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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?

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Yes! I'll give a constructive argument in a few minutes

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...

Digits of pi in e = 2.7182818284590452

353602874713526624977572470936999595749669676277240766303535475945713821785251664274274663919320030599218174136Digits of e in pi = 3.14159

26535897932384626433832795028841971693993751058209749445923078164062862089986280348253421170679821480865132823066470938446095505822317253594081056239Log in to reply

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

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

where x is a rational number and i is an integer. I do not think this is true. Although, I cannot prove it.

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$\pi$ 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

ah! interesting...why does x have to be rational? It could be the sum of two irrationals, right? And sinceLog in to reply

In this case, x= 2.728 is the rational part

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$\pi$ at the end of $e$? That doesn't make much sense

so you're talking about puttingLog in to reply

$e$ because $e$ doesn't end

because it could just be somewhere in the "middle" ofLog in to reply

$\pi$ is irrational and hence doesn't have a terminating decimal expansion; how can it be in the middle of $e$?

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$\pi$ were at the end of $e$? Because both are irrational. Would it matter?

but nonetheless, that is interesting! Would it be okay ifLog in to reply

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.

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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.

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