Is the reciprocal of a transcendental number always transcendental?

Question

I know it would always be irrational...

Answer 1

Yes. Note that the set of algebraic numbers forms a field. This means that the set of algebraic numbers is closed under addition, multiplication, taking multiplicative inverses, and taking additive inverses. Thus, if x is transcendental (i.e., not algebraic), then so must its multiplicative inverse. Otherwise, x, being the multiplicative inverse of an algebraic number, would have to be algebraic itself.

Consult any introductory book on Galois theory.

Answer 2

Answer #1 works for me.

OR:

Suppose x is a non-zero number and 1/x is a root of some polynomial p(T) =an*T^n+... +a1*T+a0. Then p(1/x)=0 so an*(1/x)^n+... +a0 =0 then when you multiply by x^n and clean things up you get an+a(n-1)*x+ ...+a1*x^(n-1) +a0*x^n = x^n*0 = 0. So x is a solution to a polynomial.

Therefore, if 1/x is algebraic then x is algebraic. The contrapositive of this is: if x is not algebraic then 1/x is not algebraic. Substitute the definition: if x is tanscendental then 1/x is transcendental.

The 1/x business is used for continued fractions if you end up studying them.