Dedicated to my best friend Sanjeet Raria :

Prove that \(e\) is irrational number.

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## Comments

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TopNewestWe start with the series expansion of \(e\)

\(e=1+\dfrac { 1 }{ 2! } +\dfrac { 1 }{ 3! } +...\)

which, for any \(n\), we can restate as

\(e=\left( 1+\dfrac { 1 }{ 2! } +\dfrac { 1 }{ 3! } +...+\dfrac { 1 }{ n! } \right) +\left( \dfrac { 1 }{ (n+1)! } +\dfrac { 1 }{ (n+2)! } +... \right) \)

\(e=\left( 1+\dfrac { 1 }{ 2! } +\dfrac { 1 }{ 3! } +...+\dfrac { 1 }{ n! } \right) +\dfrac { 1 }{ n! } \left( \dfrac { 1 }{ (n+1) } +\dfrac { 1 }{ (n+1)(n+2) } +... \right) \)

But, because we know that

\(\dfrac { 1 }{ (n+1) } +\dfrac { 1 }{ (n+1)(n+1) } +\dfrac { 1 }{ (n+1)(n+1)(n+1) } +...=\dfrac { 1 }{ n } \)

we can write

\(0<e-\left( 1+\dfrac { 1 }{ 2! } +\dfrac { 1 }{ 3! } +...+\dfrac { 1 }{ n! } \right) \le \dfrac { 1 }{ n! } \dfrac { 1 }{ n } \)

or, combining the fractions and multiplying all sides by \(n!\)

\(0<en!-N\le \dfrac { 1 }{ n } \)

where N is an integer. If \(e\) can be expressed as a fraction \(a/b\) where \(a\) and \(b\) are integers, then we can choose \(n\) large enough so that \(en!\) is an integer, which means that \(en!-N\) is an integer. But it’s impossible for there to be an integer between \(0\) and \( \dfrac { 1 }{ n }. \) Hence, \(e\) must be irrational.

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When I read the last line of the proof, I was a bit puzzled as to why it worked. Sure you followed these steps through, but shouldn't work for just about anything?

Then I realised that \(e \equiv \sum n!^{-1}\)! As Hardy once said, "A chess player may offer the sacrifice of a pawn or even a piece, but a mathematician offers the game." Great example.

Like Sandeep Bhardwaj said, a million salutes.

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That was a really astute point you brought up, as to "why should this work for this particular instance?" Thanks, it's worth deeper analysis. So maybe it wasn't just hand-waving, after all.

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Absolute ! Millions of salutes to you sir.

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I have no idea to prove it . But Just For a Joke : Type 'e' in any Advanced calculator .

Hence Proved . Lol ! :)

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Thanks @Po.

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