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# Dedicated to my best friend.

Dedicated to my best friend Sanjeet Raria :

Prove that $$e$$ is irrational number.

Go through more proofs via Proofs - Rigorous Mathematics and enhance your mathematical growth!

Note by Sandeep Bhardwaj
3 years ago

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We 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.

- 3 years ago

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.

- 3 years ago

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.

- 3 years ago

Absolute ! Millions of salutes to you sir.

- 3 years ago

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

- 3 years ago

Thanks @Po.

- 3 years ago