Proof That Root of Primes is Irrational

Let $p$ be a prime number
Prove: $p^{1/n}\ne \frac{a}{b}\forall a\in \mathbb{Z},b\in \mathbb{Z}-\left\{0,1\right\},n\in 2,3,4\dots$
Assume for contradiction that $p^{1/n}=\frac{a}{b}$ and is reduced

$$\begin{array}{rl}{p}^{1/n}& =\frac{a}{b}\\ p& =\frac{a^n}{b^n}\end{array}$$

• Then $a^n=p{b}^n$, and so $a^n$ is divisible by a prime
• But $a$ must also be divisible by a prime because otherwise, squaring $a$ wouldn’t get the factor of $p$ ($p$ is only divisible by itself and $1$)
• We write $a=kp,k\in \mathbb{Z}$
• Then $p=\frac{k^n{p}^n}{b^n}$
• Then $b^n=p^{n-1}{k}^n$
• So then $b^n$ is divisible by $p$, and so must $b$ be
• But this contradicts our assumption that $\frac{a}{b}$ was reduced
  $\therefore$ by Proof by Contradiction, the root of a prime number must be irrational
  $Q.E.D.$