# We know e (exponential) is a irrational number

• newton1
In summary, we know that e (exponential) is an irrational number and we can prove it through various methods. One way is by calculating e using an infinite series of non-repeating rational numbers, which results in an irrational sum. Another way is by proving that e cannot be written as a fraction through the use of Taylor's series. It can also be shown using a theorem that if there exists a function that is continuous and positive on a certain interval and its iterated anti-derivatives can be taken to be integer valued at the endpoints, then the number is irrational. Finally, we can also use a proof similar to the one used for proving pi is irrational to show that e is irrational. In conclusion, e cannot be written

#### newton1

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

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The way we calculate e can be similar 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.

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.

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

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...:)

e-An<1/(n!*n)

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!

Hurkyl