# We know e (exponential) is a irrational number

## Main Question or Discussion Point

we know e (exponential) is a irrational number....
how can we prove it??

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emu
The way we calculate e can be similiar to Pi.
One way yields an infinite series of non-repeating rational numbers. The sum is therefore irrational.

Try proving the sqrt(5) is irrational.

HallsofIvy
Homework Helper
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.

While e cannot be written as a fraction, e to its first 2 million decimal places can. I'm just not going to.

Hurkyl
Staff Emeritus
Gold Member
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.

Hurkyl

Let An=1+1/2!+...+1/n!;
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!;
1/(n+1)<n!*p/q-n!*(1+1/2!+...+1/n!)<1/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...
QED

climbhi
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...

thank you....:)

Hurkyl
Staff Emeritus