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 $$p^n>n$$ 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:
$$n^{1/n}=\frac{p}{q}\Rightarrow p^{n}=nq^{n}$$
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
$$n|p^{n}\Rightarrow n|p$$
So we can write $$p=\alpha n$$, so...

 

[annoymage]

you know that
$$n|p^{n}\Rightarrow n|p$$

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.