# N-th Root Irrationality Proof

• May 5th 2011, 09:37 AM
jstarks44444
N-th Root Irrationality Proof
Hey all, any help with the following proof would be appreciated:

The real number n-th root of 2 is irrational.

A similar proof, which says that the square root of 2 is irrational, is proved in the following way by contradiction:

Assume sqroot(2) = m/n for some m,n in Z (integers)
Since it is rational, you can assume m and n have no common factors.
2 = m^2 / n^2 implies m / n = 2n / m
This means n divides m, but that means sqroot(2) is an integer, which is a contradiction.

At our disposal, we have the fact that:
-The real numbers sqroot(2) is irrational
-If r in the Naturals is not a perfect square, then sqroot(r) is irrational
-Let m and n be nonzero integers. Then (m/n)*sqroot(2) is irrational

Thanks for the help!
• May 5th 2011, 10:11 AM
abhishekkgp
Quote:

Originally Posted by jstarks44444
Hey all, any help with the following proof would be appreciated:

The real number n-th root of 2 is irrational.

A similar proof, which says that the square root of 2 is irrational, is proved in the following way by contradiction:

Assume sqroot(2) = m/n for some m,n in Z (integers)
Since it is rational, you can assume m and n have no common factors.
2 = m^2 / n^2 implies m / n = 2n / m
This means n divides m, but that means sqroot(2) is an integer, which is a contradiction.

At our disposal, we have the fact that:
-The real numbers sqroot(2) is irrational
-If r in the Naturals is not a perfect square, then sqroot(r) is irrational
-Let m and n be nonzero integers. Then (m/n)*sqroot(2) is irrational

Thanks for the help!

The following result is very useful:
http://latex.codecogs.com/png.latex?...\in \mathbb{Z}.

Try to prove the above and then:

assume http://latex.codecogs.com/png.latex?...;\, gcd(r,s)=1

from the above http://latex.codecogs.com/png.latex?...\in \mathbb{Z} which is obviously wrong.
• May 5th 2011, 12:28 PM
TheCoffeeMachine
Here is an overkill! You have already proven that √2 is irrational.
Now suppose there exist $\displaystyle p,q$ such that $\displaystyle 2^{\frac{1}{m}} = \frac{p}{q}$, for $\displaystyle m \ge 3$ then:

$\displaystyle 2^{\frac{1}{m}} = \frac{p}{q} \Rightarrow 2 = \frac{p^m}{q^m} \Rightarrow 2q^m = p^m \Rightarrow q^m+q^m = p^m, ~~ m \ge 3$

which contradicts Fermat's Last Theorem, and hence the conclusion.
• May 5th 2011, 01:00 PM
Drexel28
The following statement, which is a generalization, is true "$\displaystyle a^{\frac{1}{m}}$ where $\displaystyle a,m\in\mathbb{N}$ is rational if and only if $\displaystyle a^{\frac{1}{m}}$ is an integer[/tex]" this can be seen as a corollary of the fact that the only rational algebraic integers are integers themselves. The proof is simple and basically follows from the rational root theorem. So, $\displaystyle 2^{\frac{1}{n}}$ is never rational since it's never an integer since $\displaystyle 1<2^{\frac{1}{n}}<2$ for $\displaystyle n\geqslant 2$.
• May 5th 2011, 08:41 PM
abhishekkgp
Quote:

Originally Posted by TheCoffeeMachine
Here is an overkill! You have already proven that √2 is irrational.
Now suppose there exist $\displaystyle p,q$ such that $\displaystyle 2^{\frac{1}{m}} = \frac{p}{q}$, for $\displaystyle m \ge 3$ then:

$\displaystyle 2^{\frac{1}{m}} = \frac{p}{q} \Rightarrow 2 = \frac{p^m}{q^m} \Rightarrow 2q^m = p^m \Rightarrow q^m+q^m = p^m, ~~ m \ge 3$

which contradicts Fermat's Last Theorem, and hence the conclusion.

i had never ever thought of it that way!!!
• May 5th 2011, 09:25 PM
Drexel28
Quote:

Originally Posted by abhishekkgp
i had never ever thought of it that way!!!

Yes, because that was the first corollary of FLT that I thought of too.