The Real And Complex Number Systems

q!
kq1
1
k!
kq1
q!
k!
1
q 1 1
q 1q 2 ...
1
q 1 1
q 1
2
...
1 q
1,
a contradiction. So, we know that e is not a rational number.
n 1
4. Here is an estimate about e k0 k!
, where 0 1. ( In fact, we know
nn!
that e 2. 71828 18284 59045 . . . . )
Proof: Since e k0
1
k!
0 e x n
kn1
1
n 1!
1
n 1!
1
k! ,wherex n
n
k0
1
k!
1 1
n 2 1
n 2n 3 ...
1 1
n 2 1
n 2 2
1
n 1! n n 2
1
1
nn! since n 2 1
n 1 2 n .
So, we finally have
n
e
1
k! , where 0 1.
nn!
k0
Note: We can use the estimate dorectly to show e is an irrational number.
2. For continuous variables, we have the samae result as follows. That is,
x
e.
lim
x
1 1 x
Proof: (1) Since 1 1 n n e as n , we know that for any sequence a n N,
with a n , wehave
lim n
1 a 1 a n
n
e. 5
(2) Given a sequence x n with x n , and define a n x n , then
a n x n a n 1, then we have
a n x
1 1 1 1 n
a n 1
xn
1 a 1 a n1
n
.
Since
a n a
1 1 e and 1 1 n1
a n 1
an e as x by (5)
we know that
...