# Thread: N-th Root Irrationality Proof

1. ## 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!

2. 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:
$gcd(r,s)=1 \Rightarrow gcd(r^n,s^n)=1\, for \, all\, n \in \mathbb{Z}$.

Try to prove the above and then:

assume $2^{\frac{1}{n}}=r/s;\, r,s \in \mathbb{Z};\, gcd(r,s)=1$

from the above $2=\frac{r^n}{s^n} \Rightarrow s^n|r^n \Rightarrow \, gcd(r^n,s^n)=s^n=1 \Rightarrow s=1 \Rightarrow 2^{\frac{1}{n}} \in \mathbb{Z}$ which is obviously wrong.

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

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

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

4. The following statement, which is a generalization, is true " $a^{\frac{1}{m}}$ where $a,m\in\mathbb{N}$ is rational if and only if $a^{\frac{1}{m}}$ 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, $2^{\frac{1}{n}}$ is never rational since it's never an integer since $1<2^{\frac{1}{n}}<2$ for $n\geqslant 2$.

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

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

which contradicts Fermat's Last Theorem, and hence the conclusion.
i had never ever thought of it that way!!!

6. 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.