- #1

Math100

- 771

- 219

- Homework Statement
- If ## a ## is a positive integer and ## \sqrt[n]{a} ## is rational, then ## \sqrt[n]{a} ## must be an integer.

- Relevant Equations
- None.

Proof:

Suppose ## a ## is a positive integer and ## \sqrt[n]{a} ## is rational.

Then we have ## \sqrt[n]{a}=\frac{b}{c} ## for some ## b,c\in\mathbb{Z} ##

such that ## gcd(b, c)=1 ## where ## c\neq 0 ##.

Thus ## \sqrt[n]{a}=\frac{b}{c} ##

## (\sqrt[n]{a})^{n}=(\frac{b}{c})^n ##

## a=\frac{b^n}{c^n} ##,

which implies that ## ac^n=b^n ##.

This means ## c^{n}\mid b^{n} ##.

Since ## gcd(b, c)=1 ##, it follows that ## c=1 ##.

Thus ## \sqrt[n]{a}=\frac{b}{c} ##

## =\frac{b}{1} ##

## =b ##,

which implies that ## b ## is an integer.

Therefore, if ## a ## is a positive integer and ## \sqrt[n]{a} ## is rational,

then ## \sqrt[n]{a} ## must be an integer.

Suppose ## a ## is a positive integer and ## \sqrt[n]{a} ## is rational.

Then we have ## \sqrt[n]{a}=\frac{b}{c} ## for some ## b,c\in\mathbb{Z} ##

such that ## gcd(b, c)=1 ## where ## c\neq 0 ##.

Thus ## \sqrt[n]{a}=\frac{b}{c} ##

## (\sqrt[n]{a})^{n}=(\frac{b}{c})^n ##

## a=\frac{b^n}{c^n} ##,

which implies that ## ac^n=b^n ##.

This means ## c^{n}\mid b^{n} ##.

Since ## gcd(b, c)=1 ##, it follows that ## c=1 ##.

Thus ## \sqrt[n]{a}=\frac{b}{c} ##

## =\frac{b}{1} ##

## =b ##,

which implies that ## b ## is an integer.

Therefore, if ## a ## is a positive integer and ## \sqrt[n]{a} ## is rational,

then ## \sqrt[n]{a} ## must be an integer.