# Someone help me understand how e is irrational proof

1. Nov 25, 2007

### rock.freak667

http://web01.shu.edu/projects/reals/infinity/irrat_nm.html

Well there is the proof i am reading and trying to understand...

can someone tell me how they knew that $0<R_n<\frac{3}{(n+1)!}$

2. Nov 25, 2007

### morphism

Taylor's theorem.

3. Nov 25, 2007

### rock.freak667

Well I didn't really learn Taylor's theorem in much depth..so I don't really understand how they got the remainder to be less than or equal to that

4. Nov 26, 2007

### Gib Z

Taylor's theorem states that any function that is n-times differentiable by a Taylor series, plus a remainder term, which with some functions goes to zero as the number of terms of the series increase. We can find a taylor series around a certain value of x, but 0 is the easiest and what is required here.

So, basically the theorem states an n-times differentiable function can be expressed as such:

$$f(x) = f(0) + f'(0)x + f''(0) \frac{x^2}{2} + f'''(0) \frac{x^3}{6} ..... + R_n = \sum_{k=0}^n f^k (0) \frac{x^k}{k!} + R_n$$

There are several, and actually equivalent ways of expressing the remainder term (for those with watchful eyes, will regard it as a form of the mean value theorem) but for this one, it looks like you'll need the Cauchy form, search "Cauchy Remainder term" in google.

Basically that form states that $$R_n = \frac{(x - t)^n}{n!} x f^{n+1} (t)$$ where t is some number in the closed interval [0, x]. Note this is for around the point zero again.

Hopefully you can construct an inequality with the Remainder term?

EDIT: O I do forget to state, in this notation f^k denotes the k-th derivative of f, not an exponent. f^0 should be interpreted as f.

5. Nov 26, 2007

### uart

Hi rockfreak. Remember that in this problem all we're trying to do is to show that b*R_n is between zero and one (for a given positive integer b and for suitably large n) and therefore that b*R_n cannot be an integer. The exact bound we find is not all that important so long as we can make b*R_n less than one. What I'm saying is that the bound doesn't necessarily have to be lowest one we can find and it doesn't even have to be a good approximation to the actual series remainder, it just has to be larger than the series remainder while still being able to make b*R_n less than one.

Sometimes just making a simple comparison with the series under question and another known series is a very easy way to get a bound. In this case R_n is :

$$R_n = 1/(n+1)! + 1/(n+2)! + 1/(n+3)! + ...$$

$$R_n = \frac{1}{(n+1)!}\, \{\, 1 + \frac{1}{(n+2)} + \frac{1}{(n+2)(n+3)} + \frac{1}{(n+2)(n+3)(n+4)} + \, \ldots \}$$

$$R_n < \frac{1}{(n+1)!} \, \{\, 1 + 1/2 + 1/4 +1/8 + \, \ldots\}$$

Therefore,

$$R_n < \frac{2}{(n+1)!}$$

Can you see how I put put the expression in curley brackets {} in comparision with a simple geometric series to find an appropriate bound.

Last edited: Nov 26, 2007