Please prove that e is irrational.?

Question

Good luck.

Answer 1

Perhaps this is more intuitive than deductive... but if you look at the Taylor expansion, centered at 0, of 'e' it seems obvious
     ∞
e = ∑ [ 1 / n! ]
    n=0

I, for one, see the continuous summation of fractions who do not share a common denominator, and grows at more than a geometric rate.

I dont see this expansion ever being simplified into anything other than a letter denoting a constant.

Answer 2

Suppose that e is rational. Then e=p/q, with p and q positive integers. Then q!*e is an integer. But:

q!*e = [k=0, ∞]∑q!/k! = [k=0, q]∑q!/k! + [k=q+1, ∞]∑q!/k!

Now, since for k≤q, k! | q!, it follows that each term in [k=0, q]∑q!/k! is an integer, so [k=0, q]∑q!/k! is itself an integer, thus we have that:

q!*e - [k=0, q]∑q!/k! = [k=q+1, ∞]∑q!/k!

is an integer. However, we know that:

[k=q+1, ∞]∑q!/k!
= [k=q+1, ∞]∑q!/[n=1, k]∏n
= [k=q+1, ∞]∑q!/([n=1, q]∏n * [n=q+1, k]∏n)
= [k=q+1, ∞]∑q!/(q! * [n=q+1, k]∏n)
= [k=q+1, ∞]∑1/[n=q+1, k]∏n
≤ [k=q+1, ∞]∑1/[n=q+1, k]∏(q+1)
= [k=q+1, ∞]∑1/(q+1)^(k-q)
= (1/(q+1))/(1 - 1/(q+1)) (by the geometric series formula)
= 1/(q+1 - 1)
= 1/q

Now, since e is clearly not an integer (manual computation is sufficient to establish this), q≠1, so 1/q < 1. Thus we have that [k=q+1, ∞]∑q!/k! ≤ 1/q < 1. Also, since every term in [k=q+1, ∞]∑q!/k! is positive, so is [k=q+1, ∞]∑q!/k!, thus we have 0 < [k=q+1, ∞]∑q!/k!. But [k=q+1, ∞]∑q!/k! is supposed to be an integer, and there are no integers between 0 and 1 -- a contradiction! Therefore, e must in fact be irrational. Q.E.D.

Answer 3

We know that IF e is rational, we can write,

e = a/b = 1+1+1/2!+1/3!+1/4!+........+1/b!+1/(b+1)!+...

where a and b are integers b>1. Multiply by b! to get,

b! (1+1+1/2!+1/3!+...+1/b!) +(1/b+1/(b(b+1))+1/(b(b+1)(b+2))+.... = (a/b)(b!)

The left term is an integer, the next term is less than 1.

Their sum cannot possibly be an integer , (a/b)(b!).

Therefore the assumption is abandoned and e is irrational.

Answer 4

Using continuous fractions it is possible to show that:
e = [2,1,2,1,1,4,1,1,6,1,1,8,1,1,10,1,1,12,...]
And as this never terminates then e is not rational.

The notation above means:
e = 2 + 1/( 1 + 1/( 1 + 1/( 2 + 1/ (1 + 1/( 1 + 1/( 4 + ...

Also e can be written as:
e = 1 + 1/2! + 1/3! + 1/4! + 1/5! + 1/6! + ...
And similarly as this never ends then e is not rational.

If you ask to prove it is transcendental I will hit you with a large trout.

Answer 5

Wikipedia has a nice proof.

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