Because a and p are in the field 𝐹( − 1) , 3 is in the field 𝐹( − 1) . This contradicts with what we supposed at the beginning in which 𝐹( ) is the lowest field which has a constructible root. So if a cubic has constructible roots, at least one root is rational; if a cubic has no rational roots, it has no constructible roots.
How can we use this theorem?
To prove that we cannot trisect angle with a straight-edge and a compass, we need to show that there exists an angle that cannot be trisected using a straightedge and a compass. For example, a 120˚ angle cannot be trisected, because we can prove that (40) is not constructible. To do this we need to put (40) into a cubic function.
Using the trigonometric identity:
(3 ) = 4 3 ( ) − 3( )
in which =40˚:
(120) = 0.5
0.5 = 4 3 (40) − 3(40)
If we let (40) = then the cubic will become:
8 3 − 6 − 1 = 0
Which we can prove has no rational roots by typing it into a calculator. So we can say that it has no constructible roots. Because (40) is a root of this function, (40) is not constructible, therefore an angle of 120˚ which we can construct cannot be trisected.
In conclusion, we cannot trisect an angle in a straightedge-compass construction.
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