Prove that $3 - 2 ^ {1/7}$ is Irrational

Question

How to prove that $3 - 2 ^ {1/7}$ is irrational?

If I do

$$\frac p q = 3 - 2 ^ {1/7}$$

$$2 ^ {1/7} = 3 - \frac p q $$

Hint needed

Should I multiply by $7$ times??

Answer 1

Suppose $3-2^{1/7}=\frac{m}{n},$ for some $m,n\in\Bbb Z,$ $n\neq0.$ Then $(3n-m)^7=n^7+n^7,$ which contradicts Fermat's Last Theorem!.

Answer 2

Suppose $3-2^{1/7}$ is rational,Hence $2^{1/7}=3-(3-2^{1/7})$ is rational.we will prove that $2^{1/7}$ is irrational.

Direct Proof:Just use the rational root test on the polynomial equation $x^7-2=0$ (note that $\sqrt[7]{2}$ is a solution to this equation). If this equation were to have a rational root $\frac{a}{b}$ with gcd$(a,b)=1$ (with $a,b\in \mathbb{Z}$ and $b\not=0$), then $b\vert 1$ and $a\vert 2$. Thus, $\frac{a}{b}\in\{\pm 1,\pm 2\}$. However, none of $\pm 1,\pm 2$ are solutions of $x^7-2=0$. Therefore the equation $x^7-2=0$ has no rational solutions and $\sqrt[7]{2}$ is irrational.Hence We are done!

Alternatively, suppose we have $\sqrt[7]{2}=\frac{a}{b}$ for some $a,b\in \mathbb{Z}$, $b\not=0$, and $\gcd(a,b)=1$. Then, rearranging and cubing, we have $2b^7=a^7$. Therefore $a^7$ is even....what does that say about $a$? What, in turn, does that say about $b$? It's really not that different from the classic proof that $\sqrt{2}$ is irrational.

Answer 3

You can safely ignore the term $3$.

Let $p^7=2q^7$, with $p$ and $q$ relative primes. Then $p^7$ is even, so that $p$ is even, so that $q^7$ is a multiple of $2^6$ so that $q$ is even, a contradiction.