Proving that n-th root of n is irrational

[annoymage]

Homework Statement

if n>1, prove that \sqrt[n]{n} irrational

Homework Equations

n/a

The Attempt at a Solution

so suppose it is rational, so i know \sqrt[n]{n} must be integer, say p

then n=p^n and how do i prove this is not true??

i can prove if p>1 then p^n&gt;n for all n>1 by induction, and when p=1 its clearly contradiction, so i don't know the case when p<1

or rather someone have other easy way to prove the question? clue please T_T

 

[LCKurtz]

Homework Statement

if n>1, prove that \sqrt[n]{n} irrational

Homework Equations

n/a

The Attempt at a Solution

so suppose it is rational, so i know \sqrt[n]{n} must be integer, say p

I assume by your notation that n is a positive integer.

What you know is that \sqrt[n]{n} can be written as the quotient of two integers p and q having no common factors.

 

[annoymage]

What you know is that \sqrt[n]{n} can be written as the quotient of two integers p and q having no common factors.

i know, but i also know that if \sqrt[n]{n}=\frac{p}{q}\ ,\ gcd(p,q)=1 then q=1

but heyyyyyy, i realized something

I assume by your notation that n is a positive integer.

when i assume \sqrt[n]{n}=p i know p must be positive

so i already prove it right? since n=p^n is a contradiction because p^n is always greater than n if p>1, and if p=1 then n \neq 1^n is it right?

 

[hunt_mat]

What he's saying that, suppose that n^{1/n} is rational, then there is a p and a q (with no common factors) such that:
<br /> n^{1/n}=\frac{p}{q}\Rightarrow p^{n}=nq^{n}<br />
Now you look for how p^{n} is made up, what factors does it have?

 

[annoymage]

ahhhhh i see so i have to find common factor other than 1,

n|p^n , etc etc... , then n|q^n, then since gcd(a,b)=1, then gcd(a^n,b^n)=1 so n|1 so n=1, still can't find common factor T_T more clue

 

[hunt_mat]

you know that
<br /> n|p^{n}\Rightarrow n|p<br />
So we can write p=\alpha n, so...

 

[annoymage]

you know that
<br /> n|p^{n}\Rightarrow n|p<br />

wait is that true? i got counter example n=4, p=2
right?

 

[hunt_mat]

Then there is a factor of n which is also factor of p, and factors of p are want you want. You're correct of course, well done for noticing, If n is prime, does my solution work? If n isn't prime then it can be broken down into prime factors.

 

[annoymage]

ahh i see i see

if n is prime then n is the new common factor

If not, n has a prime factor, say d, d l n, then bla bla bla d is the new common factor right? thankssssssssss

 

[hunt_mat]

You can get a great deal of the information of how to prove should go by examining the proof the square root of 2 is irrational.