Question

HELP !!

any simple proof for proving "e" is irrational ?

Answer 1

lets use the definition of e ?

Answer 2

okay

Answer 3

\(\Huge e=\sum_{n=0}^{\infty }\frac {1}{n!}\)

Answer 4

can we use this instead ?

\(\large e= \lim \limits_{n\to \infty} \left(1+\frac{1}{n}\right)^n\)

Answer 5

because I did not work the the earlier series definition yet

Answer 6

mmm i'm attended to expanding e and then assume its rational to reach a contradiction


for the one u mintion i never proved e by limit but if u have an idea i would like to see

Answer 7

okay lets use the series definition then :)

Answer 8

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

Answer 9

lets assume e is rational
\( \Huge e=\frac {a}{b}\) s.t both a,b bositive integers and a<b

Answer 10

im wondering what Co is ryping

Answer 11

\[\large \dfrac{a}{b} = 1+\frac {1}{1!}+\frac {1}{2!}+\frac {1}{3!}+...

\]

Answer 12

hence
be=a
so be is integer

Answer 13

. Observe that we can write any real number r as
\[a + \frac{ b _{2} }{ 2 } + \frac{ b _{3} }{ 3! } + \frac{ b _{4} }{ 4! } + \frac{ b _{5} }{ 5! } + ....\]
where a∈Z and each bn∈{0,1,…,n−1}. This is the expansion of r in the factorial number system, where bn is the n'th "digit". In particular,
a=⌊r⌋ and bn=⌊n!(r−sn−1)⌋
for each n, where sn denotes the n'th partial sum of the above series.

It is easy to see that r is rational if and only if this expansion terminates. Then e must be irrational, since its expansion does not terminate:
\[e = 2 + \frac{ 1 }{ 2 }+ \frac{ 1 }{ 3! }+ \frac{ 1 }{ 4! }+ \frac{ 1 }{ 5! } + ...\]

Answer 14

also b!e is integer as well ok ?

Answer 15

stop wondering now :P @ikram002p

Answer 16

lol @CO_oLBoY first time to know that i can write any real number r as that xD

Answer 17

whoa ! interesting, how this factorial number system works - is it like positional number system ?

Answer 18

we can represent 1.237 as 1 + a/2! + b/3! + c/4! is it

Answer 19

its like using
some F(x) tylor expantion

Answer 20

bn=⌊n!(r−sn−1)⌋
this looks bit hard to interpret to me >.<

Answer 21

since it an infinite series of fractions there must exist a fraction in the series that has a lower decimal value that a few previous fractions

Answer 22

Therefore it must be irrational done

Answer 23

:D

Answer 24

what about other infinite series of fractions that converge to integers ?

Answer 25

Oh you're claiming the entire p-series family converges to irrational numbers ?

Answer 26

mmm im using NT to prove
idk how to start to prove with limits

Answer 27

okay true xD ill try again when something else hits me

Answer 28

looks cool_boy's method is bit advanced for my level,

can we finish this method :
\(\large \dfrac{a}{b} = 1+\frac {1}{1!}+\frac {1}{2!}+\frac {1}{3!}+... \)
next what

Answer 29

okay wait no!! its not posible for it to converge

Answer 30

because there constantly exists a new prime number multiplication in the denominator

Answer 31

so that previous statement is valid coupled with this

Answer 32

\(\large e = 1+\frac {1}{1!}+\frac {1}{2!}+\frac {1}{3!}+... \)
as i said before b!e is integer , so lets multibly by b!

Answer 33

\(\large b!e = b!+\frac {b!}{1!}+\frac {b!}{2!}+\frac {b!}{3!}+... \)

Answer 34

also if we continue expantion it would be like this
\(\large b!e = b!+\frac {b!}{1!}+\frac {b!}{2!}+\frac {b!}{3!}+...+\frac {b!}{b!}+\frac {b!}{(b+1)!}+... \)

Answer 35

b!e is integer so the expantion is integer too
also this sumation
\( \large S = b!+\frac {b!}{1!}+\frac {b!}{2!}+\frac {b!}{3!}+...+\frac {b!}{b!}.\)

Answer 36

wid you so far :)

Answer 37

\(\large b!e = S +\frac {b!}{(b+1)!}+\frac {b!}{(b+2)!}+\frac {b!}{(b+3)!}...\)

Answer 38

let
\(\large M=\frac {b!}{(b+1)!}+\frac {b!}{(b+2)!}+\frac {b!}{(b+3)!}...\)
so
b!e=S+M

Answer 39

\[\large b!e = S +\frac {1}{b+1} + \frac{1}{(b+1)(b+1)} + ...
\]

Answer 40

mmm

Answer 41

u mean b+2 in next term right :D

Answer 42

if yes then we are approching

Answer 43

yes typo

Answer 44

cool :D

Answer 45

so
n+1 <n+2
1/n+1>1/n+2
\(\large M=\frac {1}{b+1} + \frac{1}{(b+1)(b+2)} + \frac{1}{(b+1)(b+2)(b+3)}... <\frac {1}{b+1} + \frac{1}{(b+1)^2} + \frac{1}{(b+1)^3}... \)

Answer 46

yes \(\large \lt \dfrac{1/(b+1)}{1 - 1/(b+1)}\)

Answer 47

yeah :D
so
M<1/b+1 (b+1 / b )
ok ?
so
M<1/b+1

Answer 48

which is a contradiction since
b!e=S+M
b!e integer and so S

Answer 49

oh so that's a fraction

Answer 50

M<1/b+1
since b positive integer
so M cant be integer

Answer 51

sry typo its
M<1/b

Answer 52

how did they get that upper estimate

Answer 53

look at the steps dan :O

Answer 54

b!e - S < 1/(b+1)

so this contradiction ends the proof, really cool xD

Answer 55

ikr :D

Answer 56

dan the Qn u asked for will have the same steps as this one

Answer 57

oh ofcourse silly question xD

Answer 58

1/n+1 >1/n+2
1/(n+1)^2>(n+1) (n+2)
1/(n+1)^3>(n+1) (n+2)(n+3)
1/(n+1)^4>(n+1) (n+2)(n+3)(n+4) ...

Answer 59

\(\large \dfrac{1}{3} \lt \dfrac{1}{2}\)

Answer 60

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

Answer 61

ya iwas reading that its nice :)

Answer 62

cool ideas :)

Answer 63

lolz
same proof :P

Answer 64

yeah

Answer 65

lol @ikram002p but now u do :P