# Elementary Number Theory Problems 3.2 Solution (David M. Burton's 7th Edition) - Q4

My Solution for "Establish the following facts: (a) $\sqrt{p}$ is irrational for any prime $p$. (b) If $a$ is a positive integer and $\sqrt[n]{a}$ is rational, then $\sqrt[n]{a}$ must be an integer. (c) For $n \geq 2$, $\sqrt[n]{n}$ is irrational. [Hint: Use the fact that $2^n > n$.]"

This is my solution for Chapter 3.2 Q4 in the book Elementary Number Theory 7th Edition written by David M. Burton.

## Background

All theorems, corollaries, and definitions listed in the book's order:

I will only use theorems or facts that are proved before this chapter. So you will not see that I quote theorems or facts from the later chapters.

## Question

Establish the following facts:
(a) $\sqrt{p}$ is irrational for any prime $p$.
(b) If $a$ is a positive integer and $\sqrt[n]{a}$ is rational, then $\sqrt[n]{a}$ must be an integer.
(c) For $n \geq 2$, $\sqrt[n]{n}$ is irrational.
[Hint: Use the fact that $2^n > n$.]

## Solution

### (a)

We want to prove this by contradiction. Suppose $\sqrt{p}$ is rational. Then $\sqrt{p} = \frac{a}{b}$ for some integers $a$ and $b$, which means:

$$\begin{equation} \begin{split} \sqrt{p} & = \frac{a}{b} \\ p & = \frac{a^2}{b^2} \\ pb^2 & = a^2 \end{split} \nonumber \end{equation}$$

The right-hand side will have an even number of primes because it is square, while the left-hand side will have an odd number of primes because of $p$. This is a contradiction. So, $\sqrt{p}$ must be irrational.

### (b)

Because $\sqrt[n]{a}$ is rational, there exists an integer $m$ and an integer $k$ such that $\sqrt[n]{a} = \frac{m}{k}$ where $gcd(m, k) = 1$ and $k \neq 0$. It means that $a = \frac{m^n}{k^n}$.

From Problems 2.4 Q5(a), we know "If $gcd(a, b) = 1$, $gcd(a^n, b^n) = 1$." Thus $gcd(m, k) = 1$ implies that $gcd(m^n, k ^n) = 1$.

Since $a = \frac{m^n}{k^n}$ is a positive integer and $gcd(m^n, k ^n) = 1$, $k$ must be $1$.

Thererfore, $\sqrt[n]{a} = \frac{m}{k} = m$, which means $\sqrt[n]{a}$ must be an integer.

### (c)

Assume to the contrary, there exists an integer $n \geq 2$ for which $\sqrt[n]{n}$ is rational.

As $n \geq 2$, $\sqrt[n]{n} \gt 0$. From (b), we know $\sqrt[n]{n}$ is an integer.

Since $n \lt 2^{n}$, we have $\sqrt[n]{n} \lt 2$. $\sqrt[n]{n}$ can only be $1$ for positive integers. But it is impossible for $n \geq 2$. This, however, is a contradiction.

Therefore, For $n \geq 2$, $\sqrt[n]{n}$ is irrational.

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