showing that the Euler's number is irrational Our teacher wants us to do the following:
Suppose that e is rational i. e $e=\frac{a}{b}$ where $a,b\in\mathbb{N}$. Choose $n\in\mathbb{N}$ such that $n>b$ and $n>3$. Use the following inequality
$0<e-(1+\frac{1}{1!}+\frac{1}{2!}+....+\frac{1}{n!})<\frac{3}{(n+1)!}$ and replace e with $\frac{a}{b}$. Then multiply both sides of the inequality with $n!$ to see that the inequality leads to existence of an integer N that satisfies $0<N<\frac{3}{4}$ which leads to a contradiction.  
I have tried to to it but I am not getting there...
I choose $n=4$ and $b=3$ and 
$0<\frac{a}{b}-1-1-\frac{1}{2}-\frac{1}{6}-\frac{1}{24}<\frac{3}{5!}$
$0<8a-\frac{17}{24}<\frac{3}{5}$
$0<8a<\frac{157}{120}$
$0<a<\frac{157}{960}$.....
How should I do it???
 A: You’re not supposed to pick a specific value of $n$, like $4$; you’re supposed to look at arbitrary integers $n>\max\{b,3\}$. Suppose that $n>b$ and $n>3$. Multiplying the given inequality by $n!$, you get
$$0<en!-n!\left(1+\frac1{1!}+\frac1{2!}+\ldots+\frac1{n!}\right)<\frac{3n!}{(n+1)!}\;,$$
and if you now replace $e$ in this inequality by $\frac{a}b$, you get
$$0<\frac{an!}b-n!\left(1+\frac1{1!}+\frac1{2!}+\ldots+\frac1{n!}\right)<\frac{3n!}{(n+1)!}\;.$$
Now multiply out and do some simplifying:
$$0<\frac{an!}b-\left(n!+\frac{n!}{1!}+\frac{n!}{2!}+\ldots+\frac{n!}{n!}\right)<\frac3{n+1}\;.$$
Use the fact that $n>b$ to explain why $\frac{an!}b$ is an integer. All of the fractions $\frac{n!}{k!}$ with $1\le k\le n$ are integers (why?), so the whole middle expression is an integer lying strictly between $0$ and $\frac3{n+1}$. But $n>3$, so this is impossible; why?
A: By assumption, $n\ge b$. Therefore, it is possibly to bring the fraction $\frac ab$ to the denominator $n!$ as $n!$ is a multiple of $b$ (which is one of the factors in $1\cdot2\cdots n$). Likewise, all the other fractions can be braought to denominator $n!$ and hence the sum of all these fractions can be written with denominator $n!$. After multiplication with $n!$, only an integer - the numerator - remains.
Specifically, if you multiply
$$ 0<\frac a3-1-1-\frac12-\frac16-\frac1{24}<\frac3{120}$$
with $4!=24$, you should get
$$0<8a-24-24-12-4-1<\frac35. $$
Whatever $a$ is, it is an integer, hence so is the expression $8a-24-24-12-4-1$. But there is no integer strictly between $0$ and $\frac 35$.
Of cours, it is not suffucuent to consider this for just a specifically chosen $b=3$ (and $n=4$) because we don't know what the correct denomianator in a fraction for $e$ might be. It is quite clear that $b=3$ is a bad choice already because $2.\bar 6<2.7128\ldots<3$ rules that out. The argument above works for any integer $b$.
A: 
"how can we conclude that an!/b is an integer whenever n>b? Could you elaborate a bit more?" 

If n>b then b is in n x (n-1) x (n-2) x ... x (n-n+1)
therefore an!/b = a(n x (n-1) x ... x (n-b+1)) which is an integer.
Sorry about the formating, new to the forum.
