The Real And Complex Number Systems

So,
and
[
∑k−1
[k!x] = k!
j=1
∑k−1
= k!
j=1
a j
j! + a k + k!
a j
j! + a k since k!
[
∑k−1
k [(k − 1)!x] = k (k − 1)!
j=1
∑k−1
= k (k − 1)!
∑k−1
= k!
j=1
a j
j!
j=1
n∑
j=k+1
n∑
a j
j!
j=k+1
]
a j
j! < 1
a j
n∑
j! + (k − 1)!
j=k
]
a j
j! .
a j
n∑
j! since (k − 1)!
j=k
a j
j! < 1
which implies that
a k = [k!x] − k [(k − 1)!x] for k = 2, ..., n.
Last, in order to show that n is the smallest integer such that n!x is an
integer. It is clear that
n∑ a k
n!x = n!
k! ∈ Z.
In addition,
So, we have proved it.
k=1
(n − 1)!x = (n − 1)!
n∑
k=1
n−1
∑
= (n − 1)!
k=1
/∈ Z since a n
n
12
a k
k!
a k
k! + a n
n
/∈ Z.