How can i prove that $\sqrt[12]{2}$ is irrational number?
I'm trying:
$$\sqrt[12]{2} = \frac{p}{q}$$ where $p$, $q$ are integers
it follows that :
$$p^{12} = 2q^{12} $$
What is argument of irrationality in this case? From what we know that the right-hand side has an even number of 2s and the right hand side has an odd number of 2s?
I will be grateful for your help Best regards
On
If $\sqrt[12]{2} = \frac{p}{q}$, then $\sqrt[]{2}=(\sqrt[12]{2})^6=\frac{p^6}{q^6}$.
Your turn !
On
You did everything fine!
When you get to $p^{12} = 2q^{12}$ you use the argument you mentioned. The LHS has an even number of 2s in its factorization whilst the RHS has an odd number of 2s, thus the two sides are not equal.
On
A (late) proof not using unique factorization theorem, but only Euclid's lemma.
Suppose that $p,q$ are coprime integers. As $p^{12} = 2q^{12}$, you get that $2 \ | \ p^{12}$. As $2$ is a prime number $2 \ | \ p$. Therefore $p = 2 p^\prime$ and $$2^{12} \left(p^\prime \right)^{12} = 2q^{12} \Leftrightarrow 2^{11}\left(p^\prime \right)^{12} = q^{12}$$
From this you get that $2 \ | \ q^{12}$ and therefore $2 \ | \ q$ (again as $q$ is a prime). You finally get the contradiction that $p,q$ are not coprime integers. Therefore the initial assumption that $\sqrt[12]{2}$ is a rational number is wrong.
The basic argument that $\sqrt 2$ is irrational is generalized using the fundamental theorem of arithmetic.
Supose $(\frac p q)^{12} = 2$. Then $p^{12} = 2 q^{12}$. The prime factorization on the left hand side contains a number of $2$-s which is a multiple of $12$, but the prime factorization on the right hand side contains a number of $2$-s which is one more than a multiple of $12$.