Is there a real number $r$ such that all $r^n$ are irrational for all integers $n\ge1$ but it is not transcendental?
Asked
Active
Viewed 77 times
0
-
3Sure, $\sqrt 2 +1$ for example. (proof by induction) – lulu Nov 12 '16 at 00:34
-
You can share it as an aswer if you want. – student forever Nov 12 '16 at 00:39
-
1Most algebraic numbers have this property. The only exceptions are of the form $\sqrt[n]{q}$ (for $q$ rational) and these are relatively uncommon amongst all algebraic numbers. – Nov 12 '16 at 00:40
2 Answers
0
Yes: $1+\sqrt 2$, for instance. $$(1+\sqrt2)^n=1+2\binom n2+2^2\binom n 4+\dots+\biggl(\binom n1+2\binom n3+2^2\binom n5+\dotsm\biggr)\sqrt2.$$
Bernard
- 175,478
0
A number is transcendental if it is not root of any polynomial having coefficients in $\mathbb Q$. A number such that $r^n$ is irrational $\forall n$ is only not a root of $x^n-q$, $\forall n \forall q\in \mathbb Q$, but it could be not transcendental anyway, by being root of a polynomial coprime with the last one.
Bargabbiati
- 2,271