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

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??

• Yes, multiply it by itself seven times. But don't worry about the right hand side. All it matters is that $3-p/q$ is also of the form $m/n$, (rational). Mar 23 '15 at 19:16
• Show first that $\sqrt[7]{2}$ is irrational. A similar proof to the one for the irrationality of $\sqrt{2}$ works.
– mfl
Mar 23 '15 at 19:16
• $3 - \dfrac p q = \dfrac{3q}q - \dfrac p q = \dfrac{3q-p}q = \dfrac m n$, so you have $2^{1/7}=\dfrac m n$. The "3" is just clutter, so get rid of it that way. ${}\qquad{}$ Mar 23 '15 at 19:22
• Nice @MichaelHardy a good explanation... Mar 23 '15 at 19:24
• You just asked a question that hinged on the obvious fact that if $r$ is irrational, then $6-r$ is irrational. Why are you repeating yourself here? Mar 23 '15 at 19:27

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!.

• I really like this answer.
– user207710
Mar 23 '15 at 19:19
• Technically correct, but perhaps a tiny bit of overkill! Mar 23 '15 at 19:25
• Well, Fermat's Last Theorem requires very deep number theory and algebraic geometry. The uniqueness of prime factorization is a much simpler and more direct tool. Mar 23 '15 at 19:33
• Definitely the hard way! Mar 24 '15 at 1:31
• @HussainHalai Fermat's Last Theorem was unsolved for 300 years. When it was finally proven, in 1995, the proof was hundreds of pages long. This is definitely overkill! Apr 13 '15 at 8:42

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.

• You also need to assume $\,\gcd(a,b)=1\,$ to apply the Rational Root Test. Thankfully, the proof is infinitely simpler than that for FLT sledgehammer used elsewhere. Mar 23 '15 at 19:28
• Fixed Now !Thank You @BillDubuque Mar 23 '15 at 19:33

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.