- #1
Math100
- 770
- 219
- Homework Statement
- Establish the following fact:
For ## n\geq 2 ##, ## \sqrt[n]{n} ## is irrational.
[Hint: Use the fact that ## 2^{n}>n ##.]
- Relevant Equations
- None.
Proof:
Suppose for the sake of contradiction that ## \sqrt[n]{n} ## is rational for ## n\geq 2 ##.
Then we have ## \sqrt[n]{n}=\frac{a}{b} ## for some ## a,b\in\mathbb{Z} ## such that
## gcd(a,b)=1 ## where ## b\neq 0 ##.
Thus ## (\sqrt[n]{n})^{n}=(\frac{a}{b})^{n} ##
## n=\frac{a^{n}}{b^{n}} ##
## b^{n} n=a^{n} ##.
Note that ## gcd(a,b)=1 ## implies ## gcd(a^{n}, b^{n})=1 ##.
This means ## b^{n}=1 ## because ## n\geq 2 ## is an integer.
Now we have ## a^{n}=n ##.
Since ## n\geq 2 ##, it follows that ## a\neq 1 ##.
Thus ## a^{n}\geq 2^{n} ## for ## a\geq 2 ##.
But the fact that ## 2^{n}>n ## contradicts our assumption of ## a^{n}\geq 2^{n} ## for
## a\geq 2 ##.
Therefore, for ## n\geq 2 ##, ## \sqrt[n]{n} ## is irrational.
Suppose for the sake of contradiction that ## \sqrt[n]{n} ## is rational for ## n\geq 2 ##.
Then we have ## \sqrt[n]{n}=\frac{a}{b} ## for some ## a,b\in\mathbb{Z} ## such that
## gcd(a,b)=1 ## where ## b\neq 0 ##.
Thus ## (\sqrt[n]{n})^{n}=(\frac{a}{b})^{n} ##
## n=\frac{a^{n}}{b^{n}} ##
## b^{n} n=a^{n} ##.
Note that ## gcd(a,b)=1 ## implies ## gcd(a^{n}, b^{n})=1 ##.
This means ## b^{n}=1 ## because ## n\geq 2 ## is an integer.
Now we have ## a^{n}=n ##.
Since ## n\geq 2 ##, it follows that ## a\neq 1 ##.
Thus ## a^{n}\geq 2^{n} ## for ## a\geq 2 ##.
But the fact that ## 2^{n}>n ## contradicts our assumption of ## a^{n}\geq 2^{n} ## for
## a\geq 2 ##.
Therefore, for ## n\geq 2 ##, ## \sqrt[n]{n} ## is irrational.