Show that \sqrt[3]{\pi} is transcendental

Jakayla Benton 2022-04-24 Answered

Show that π3 is transcendental

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Answers (1)

ritmesysv
Answered 2022-04-25 Author has 12 answers

Step 1
Suppose a=πn is the root of a polynomial P(x) with rational coefficients of positive degree.
Writing P(a)=0 and using that an=π, powers ak with k>n can be replaced with ak=πkn·πnkmodn
It follows that there exist rational polynomials in π aj,j=0,1,,n1 such that:
1) an-1πnn-1+an-2πnn-2++a1πn+a0=0
Multiply the above by πn successively n1 times:
2) an-2πnn-1+an-3πnn-2++a0πn+an-1π=0
an-3πnn-1+an-2πnn-2++an-1ππn+an-2π=0
a0πnn-1+an-1ππnn-2++a2ππn+a1π=0
Considering the n equations (1) and (2) as a linear system in 1, πn, πn2, ,πnn1 it is a homogeneous system with a non-trivial solution, so its determinant must be zero. But all coefficients are rational polynomials in π which implies π is algebraic. The contradiction means that the original assumption cannot hold true, so n{π} is transcendental.
The same line of proof works for the nth root of any transcendental number, not just π.
(The above is essentially proving that P(πn)=0R(π)=0 where R is the resultant R(z)=res(P(x),,xnz,,x), only without assuming prior knowledge of resultants or other higher algebra results.)

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