Proof by contradiction $\left(P^B\right)^{\frac 1A}$

Question

Let $P=\text{a prime number}$

$A=\text{an even positive integer}$

$B=\text{an odd positive integer}$

Prove by contradiction that $\left(P^B\right)^{\frac 1A}$ is irrational

Answer 1

Let $\,q^a = p^b\,$ for $\,q\in\Bbb C,\ p\,$ prime, $\ 0 < a,b\in \Bbb Z.\ $ If $\,\color{#c00}{q\in\Bbb Q}\,$ then $\,q\in\Bbb Z\,$ by the Rational Root Test, so comparing unique prime factorizations $\Rightarrow q=\pm p^n,\ a\mid b.\,$ But in OP: $ $ even $\,a\nmid b\,$ odd, so $\,\color{#c00}{q\not\in \Bbb Q}$

Answer 2

Let $\left(P^B\right)^{\frac 1A} = \frac xy$ where $x$ and $y$ are co-prime integers.

So, $$P^B = \frac{x^A}{y^A}$$

Taking the positive square root of both sides we have

$$P^\frac B2 = \frac{x^\frac A2}{y^\frac A2}$$

Let $B = 2n+1$ and $A = 2m$, where $m$ and $n$ are whole numbers. Note that the LHS is $P^n\sqrt P$. Since the root of a prime is irrational , the LHS is irrational. However, the RHS is merely $\left(\frac xy\right)^m$ which is a rational raised to an integer, which is rational. Hence, this is a contradiction, and $\left(P^B\right)^{\frac 1A}$ is irrational.

Answer 3

Start with the same definitions as Certainly not a dog.

Let $\left(P^B\right)^{\frac 1A} = \frac xy$ where $x$ and $y$ are co-prime integers. Let $B = 2n+1$ and $A = 2m$, where $m$ and $n$ are positive integers.

Then $y^{2m}P^{2n+1}=x^{2m}$

By FTA, $y=\prod q_i;\ x=\prod r_j$ where $q,r$ are primes, and since $x,y$ are co-prime, $q_i\ne r_j$

So $P^{2n+1}\prod q_i^{2m}=\prod r_j^{2m}$

Now we consider whether any of $(q_i,r_j)=P$. Plainly, if any of $q_i=P$, then since $q_i\ne r_j$, $r_j\ne P$. This is impossible, as the LHS would have prime factors of $P$ and the RHS would have none. In fact, the RHS must have some factors of $P$, i.e. $r_j=P$ for one or more $j$. Let us say that in $k$ instances, $r_j=P$. Then the RHS has a factor of $P^{2km}$

Dividing through, the LHS will have a factor of $P^{2n+1-2km}=P^{2(n-km)+1}$ and the RHS will have no factors of $P$, or effectively a factor of $P^0$. Since $2(n-km)+1$ cannot equal $0$, we arrive at a contradiction. The assumption that we can find suitable $x,y$ is false, and the original expression $\left(P^B\right)^{\frac 1A}$ must be irrational.