Proof : e is irrational
FOREWORD
It is well known that :
$e =sum _(n=0)^(oo) 1/(n!) $ and also
$e = lim_(n rarr oo) (1+1/n)^n $.
The partial sum , $ s_n $ of the series above is defined as $s_n = sum_(k=0)^n 1/(k!) = 1+1+1/(2!) +1/(3!)+…. +1/(n!) $
Some calculations bring to the inequality :
$ 0 < e-s_n < 1/(n!n) $ (1)
which is useful to evaluate the error in computing $e $ if we consider $s_n $ as approximate value of $ e.$
For instance $s_(10) $ approximates $e $ with an error less than $10^(-7) $ .
E IS IRRATIONAL
The inequality (1) permits to prove the irrationality of $ e $ .
Let’s assume for absurd that $e $ is rational . Then $ e = p/q $ , with $p,q $ positive integers.
From (1 ) we get
$ 0 < e-s_q < 1/(q! q) $ (2) [ we choose for $n $ the integer $q$].
Consequently :
$0 < q! ( e-s_q) < 1/q $
Our initial assumption leads to :
$ q!e $ is an integer ( $q!p/q $is an integer indeed )
$q! s_q =q!(1+1+1/2!+…1/q! ) $ is also an integer ,
then $ q! (e-s_q ) $ is an integer too.
Since $q >=1 $ , inequality (2) implies the existence of an integer between $0$ and $1 $ .
Therefore our initial assumption : $e $ is rational , leads to a contradiction and it is so proved that $e $ is an irrational number.
It is well known that :
$e =sum _(n=0)^(oo) 1/(n!) $ and also
$e = lim_(n rarr oo) (1+1/n)^n $.
The partial sum , $ s_n $ of the series above is defined as $s_n = sum_(k=0)^n 1/(k!) = 1+1+1/(2!) +1/(3!)+…. +1/(n!) $
Some calculations bring to the inequality :
$ 0 < e-s_n < 1/(n!n) $ (1)
which is useful to evaluate the error in computing $e $ if we consider $s_n $ as approximate value of $ e.$
For instance $s_(10) $ approximates $e $ with an error less than $10^(-7) $ .
E IS IRRATIONAL
The inequality (1) permits to prove the irrationality of $ e $ .
Let’s assume for absurd that $e $ is rational . Then $ e = p/q $ , with $p,q $ positive integers.
From (1 ) we get
$ 0 < e-s_q < 1/(q! q) $ (2) [ we choose for $n $ the integer $q$].
Consequently :
$0 < q! ( e-s_q) < 1/q $
Our initial assumption leads to :
$ q!e $ is an integer ( $q!p/q $is an integer indeed )
$q! s_q =q!(1+1+1/2!+…1/q! ) $ is also an integer ,
then $ q! (e-s_q ) $ is an integer too.
Since $q >=1 $ , inequality (2) implies the existence of an integer between $0$ and $1 $ .
Therefore our initial assumption : $e $ is rational , leads to a contradiction and it is so proved that $e $ is an irrational number.
Risposte
In my book of Analysis 1, written by my professor Marco Degiovanni,
And so Degiovanni has been your professor. He was Assistant Professor in the Analisi II course for physicists (prof. A. Marino), which I attended at the University of Pisa, a long time ago. I remember him as a brilliant teacher.
It's enough to have good students, like Milnor. They would solve nice exercises:
http://www.math.utah.edu/vigre/annual-r ... 01-02.html
http://www.ams.org/notices/199510/tucker.pdf
http://www.math.utah.edu/vigre/annual-r ... 01-02.html
http://www.ams.org/notices/199510/tucker.pdf
"Luca.Lussardi":
In my book of Analysis 1, written by my professor Marco Degiovanni, the irrationality of $e$ is given as an exercise for the students of the first year...... no words....
You are joking, aren't you?
In my book of Analysis 1, written by my professor Marco Degiovanni, the irrationality of $e$ is given as an exercise for the students of the first year...... no words....
"giuseppe87x":
I know this proof, it is in my book; Camillo where did you find it?
It is an interesting proof because it shows that
1) $e-sum_(n=0)^q1/(n!)<=1/(q!q) $ and
2) $e-(1+1/n)^n>=1/(2n)$
which means that the sequence ${(1+1/n)^n}$ converges to number $e$ slower than $sum_(n=0)^(infty)1/(n!)$.
Much slower indeed

I found the proof in W.Rudin Principles of Mathematical Analysis.
I know this proof, it is in my book; Camillo where did you find it?
It is an interesting proof because it shows that
1) $e-sum_(n=0)^q1/(n!)<=1/(q!q) $ and
2) $e-(1+1/n)^n>=1/(2n)$
which means that the sequence ${(1+1/n)^n}$ converges to number $e$ slower than $sum_(n=0)^(infty)1/(n!)$.
It is an interesting proof because it shows that
1) $e-sum_(n=0)^q1/(n!)<=1/(q!q) $ and
2) $e-(1+1/n)^n>=1/(2n)$
which means that the sequence ${(1+1/n)^n}$ converges to number $e$ slower than $sum_(n=0)^(infty)1/(n!)$.
I don't know who first had this idea , I rearranged a little bit the proof from the text .......
Let see who knows where I took it from .
Let see who knows where I took it from .

of course!
Otherwise, shame on many generations of mathematicians (and friends)!
But is is nice anyway. Nice to have posted it, also.
And thanks for having "forced" me to read it.
And, for transitivity, thanks also ti Luca.Lussardi who "obliged" you to do that.
And to Tipper who started the topic.
And so on...
Otherwise, shame on many generations of mathematicians (and friends)!
But is is nice anyway. Nice to have posted it, also.
And thanks for having "forced" me to read it.
And, for transitivity, thanks also ti Luca.Lussardi who "obliged" you to do that.
And to Tipper who started the topic.
And so on...
It's not mine

nice proof