Question

show that \(e\) is irrational

Answer 1

http://en.wikipedia.org/wiki/Proof_that_e_is_irrational

Answer 2

need help understanding the Fourier's proof

Answer 3

I got the bounded part : 2 < e < 3

Answer 4

i feel lost from the step of letting e = a/b

Answer 5

dont listein to her wikipedia is not a good site to use most of it stuff is wrong

Answer 6

the proof is correct...what exactly are you confused on?

Answer 7

\[e = b! \left(e - S_b\right)\]

i have managed to digest this far,
after this, why are they assuming e = a/b instead of some other different fraction like p/q ?

Answer 8

\[e = b! \left(\color{red}{\frac{a}{b}} - S_b\right)\]

we cannot conclude the proof if we assume e = p/q right ?

Answer 9

because we wont know if `q | b!` or not

Answer 10

@Zarkon

Answer 11

** tail of series \[x = b! \left(e - S_b\right)\]

Answer 12

this is a proof by contradiction...we assume e=a/b and reach a contradiction. Therefore e cant be written as a fraction. therefore it is irrational

Answer 13

a/b or p/q...what does it matter what we call the fraction?

Answer 14

yeah it doesn't matter but why are they using same variable for both n'th partial sum and e ?

Answer 15

the denominator of `e` could be something different from the index in `b`'th partial sum right ?

Answer 16

they can define x anyway they want. with this version of x we get our contradiction

Answer 17

Oh they are comparing partial sum based on the denominator of e is it ?

Answer 18

assuming e=a/b they are constructing a number x that is both an integer and not an integer at the same time. this is a contadiction

Answer 19

Wow! I see.. got it! xD thanks @Zarkon

Answer 20

\[ 0 \lt b!(e-S_b) \lt \dfrac{1}{b} \]

Answer 21

\[ 0 \lt b!(\frac{a}{b}-S_b) \lt \dfrac{1}{b} \]

\[ 0 \lt \text{integer}\lt \dfrac{1}{b} \]

contradiction

Answer 22

and \[\frac{1}{b}<1\]

Answer 23

Indeed, b/c the index b is >=1

Answer 24

\[0 \lt \text{integer}\lt \dfrac{1}{b}<1\]

so \[0 \lt \text{integer}\lt 1\]

Answer 25

i think we can let b > 1 in e = a/b since e was "proven" to be between 2 and 3 and so is not an integer

Answer 26

that is in the proof...
"Note that b couldn't be equal to one as e is not an integer"

Answer 27

yes! I see showing 2 < e < 3 is also the key step in the proof

Answer 28

i like this proof xD

Answer 29

do u still trying to understand ?

Answer 30

understood it in bits and pieces.. need to go over it one more time

Answer 31

e = a/b

the difference between b'th partial sum and e when multiplied by b! gives an integer eventhough it is bounded between 0 and 1/b. really a nice argument :)

Answer 32

yeah it is ^_^

Answer 33

**
e = a/b

the difference between b'th partial sum and e when multiplied by b! gives an integer eventhough it is bounded between 0 and 1/b. which is a contradiction because there are no integers between 0 and 1/b