1. Introduction
The continued fraction expansion of
can be calculated as follow(cf. [1] -[3] ).

Because

hence

When one calculates the continued fraction expansion of
it is important to determine the integer part of
For square root, its continued fraction expansion can be obtained easily because it is circled while there is no obvious method to do so for cube root. In this note, we will determine the integer part of cube root and its combination. So we can achieve the continued fraction expansion of cube root according to the integer part of the cube root.
2. Main Results
Let N be a positive integer and not a cube. Denote
as its cube root. For
set

Then these numbers are satisfied the identity:
(1)
We achieve two interesting properties on
and
.
Theorem 1. If
, then
. Therefore the number
has the same sign as M.
We consider two cases respectively.
1) If
.
a) If
. Substituting
into the expression
, then we have
![]()
since every term in the last expression is nonnegative.
b) If
. We have also the similar expression as case a):
![]()
The above inequality holds because ![]()
2) If ![]()
a) If
Then we have
![]()
We also have
![]()
![]()
Hence by the identity (1)
![]()
b) If
Then
Because of
therefore
![]()
So we show that
for both cases. According to identity (1), both
and M have the same sign.
Remark 1. The result is very amazing. Because the quotient ring
is a filed, where
is the maximal ideal generated by the irreducible polynomial
. For any
there exists
such that
![]()
But it is surprising that the number
defined in Theorem 1 is always positive.
Theorem 2. If
and
then
![]()
That is to say,
is the biggest integer less than or equal to ![]()
According to Theorem 1,
![]()
Hence
![]()
So
![]()
The proof is completed.
Remark 2. Applying the Theorem 2, we can design an algorithm to calculate the continued fraction expansion of the cube root
.
Acknowledgements
The authors wish to thank Prof. Xiangqin Meng for her some helpful advices.