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!

#Algebra #IrrationalNumbers #Exponential #Proof #Numbersystem

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

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.

Michael Mendrin - 6 years, 3 months 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.

Jake Lai - 6 years, 3 months 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.

Michael Mendrin - 6 years, 3 months ago

Absolute ! Millions of salutes to you sir.

Sandeep Bhardwaj - 6 years, 3 months ago

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

Karan Shekhawat - 6 years, 3 months ago

Thanks @Po.

Sanjeet Raria - 6 years, 3 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$