Emu, do you know of any way of proving directly that the digits in the decimal expansion of e are NOT repeating? I'm not saying it can't be done, only that I think it's easier to prove e cannot be written as a fraction.
One standard method is to use the Taylor's series (which may be what emu meant): e= 1+ 1/2 + 1/6+ ...+ 1/n!+ ...
If j is any positve integer then e*j!= integer+ 1/(q+1)+ 1/(q+1)(q+2)+ ... which is not an integer so e cannot be written as a fraction with denominator j for any j.
A theorem I saw years ago was this: If c> 0, and there exist a function f(x), continuous on [0,c], positive on (0,c) and such that f(x) and its iterated anti-derivatives can be taken to be integer valued at both 0 and c, the c is irrational!
Taking f(x)= sin(x) in this theorem shows that pi is irrational.
It can also be used to prove: If c is a positive number other than 1 and ln(c) is rational, then c is irrational.
Since e is a positive number, not equal to 1, and ln(e)= 1 is rational, it follows that e is irrational.
e*j!= integer+ 1/(q+1)+ 1/(q+1)(q+2)+ ... which is not an integer
That's not obvious... I don't see why the infinite sequence there cannot add up to an integral value.
It's quite simple to prove that 1/(n+1)!<e-An<1/(n!*n);
Let's suppose e is rational, so it's equal to p/q, where p and q are integers.
1/(n+1)!<e-(1+1/2!+...+1/n!)<1/(n!*n); | *n!;
But between 1/(n+1) and 1/n is no integer...
n!*p/q must be an integer because for n big enough n! is a multiple of q;
So e is not rational...
that's kind of a neat prove, thanks for posting it. I don't think I would've caught that last part about for n big enough...
That bit isn't obvious either.... but the ordinary taylor remainder formula gives e/(n+1)! for that term which is sufficient for the proof. Don't tell me how to get that end of the inequality, it'd be a good exercise to figure it out myself!