Prove that √3 & ∛3 ∉ ℚ.

Let √3 = a/b where a and b are integers and (a,b) = 1
Then, 3 = a^2 / b^2
a^2 = 3b^2
Therefore a must be a multiple of 3.
Let a = 3c
a^2 = 9c^2 = 3b^2
b^2 = 3c^2
b is also a multiple of 3.
(a,b) > 1, contradictory to the premise that (a,b) = 1
So √3 cannot be written as a fraction of integers and is not a rational number.

Let ∛3 = p/q where p and q are integers and (p,q) = 1
Then, 3 = p^3/q^3
p^3 = 3q^3
It follows that p is a multiple of 3.
Let p = 3r
p^3 = 27r^3 = 3q^3
q^3 = 9r^3
Then q is a multiple of 3.
(p,q) > 1, contradictory to the premise that (a,b)=1
So ∛3 cannot be written as a fraction of integers and is not a rational number.

2012-05-20 22:54:20 補充：
the second last line should be (p,q), not (a,b) =,=