數學知識交流--無理數(2)

Let ³√3=a/b,
which a and b is natural number,
and there is no common factor between a and b ---[#]

³√3=a/b
3=a³/b³
a³=3b³

Since 3 is a prime number, it has no factor.
So, besides 3, no integer's cube will equal to 3, or 3's multiples.
So, a has a factor:3.

Let a=3c, which c is a natural number.
a³=3b³
(3c)³=3b³
27c³=3b³
9c³=b³

From the above,
So, besides 3, no integer's cube will equal to 3, or 3's multiples.
So, b has a factor:3.

So, a and b has a common factor:3. ---[@]

Since [#] and [@] can't come out at the same time,
³√3≠a/b, ³√3 is a irrational number.