Let us define the function $f(m,n,a)=\pi \cdot a^{\frac {m}{n}}$ where $a,m,n \in \mathbb N$.
Is $f(m,n,a)$ always irrational for every triple $(m,n,a) \in \mathbb N^3$?
Is $\pi$ multiplied by roots of naturals always irrational?
1
$\begingroup$
real-analysis
irrational-numbers
pi
2 Answers
4
If $f(m,n,a)$ is a rational number then $f(m,n,a)^{n}/a^m$ would be rational. But this latter number is a power of $pi$. This will contradict that $\pi$ is a transcendental number.
Leave you to consider edge cases where some parameters are zero or 1.
1
If you allow $a=0$, then the answer is no.
If you disallow $a=0$, then the answer is yes, since $\pi$ is transcendental: $a^{m\over n}$ is always algebraic for $a, m, n\in\mathbb{N}$, and the quotient of two (nonzero) algebraic numbers is algebraic; so if $f(a, m, n)$ were rational (remember that rational numbers are algebraic), we'd have $\pi={f(a, m, n)\over a^{m\over n}}$ is the quotient of two nonzero (since $a\not=0$ algebraic numbers, and hence $\pi$ would be algebraic.