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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??

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  • $\begingroup$ 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). $\endgroup$
    – Nathanson
    Mar 23 '15 at 19:16
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    $\begingroup$ Show first that $\sqrt[7]{2}$ is irrational. A similar proof to the one for the irrationality of $\sqrt{2}$ works. $\endgroup$
    – mfl
    Mar 23 '15 at 19:16
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    $\begingroup$ $3 - \dfrac p q = \dfrac{3q}q - \dfrac p q = \dfrac{3q-p}q = \dfrac m n$, so you have $2^{1/7}=\dfrac m n$. The "3" is just clutter, so get rid of it that way. ${}\qquad{}$ $\endgroup$ Mar 23 '15 at 19:22
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    $\begingroup$ Nice @MichaelHardy a good explanation... $\endgroup$ Mar 23 '15 at 19:24
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    $\begingroup$ 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? $\endgroup$
    – TonyK
    Mar 23 '15 at 19:27
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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!.

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  • $\begingroup$ I really like this answer. $\endgroup$
    – user207710
    Mar 23 '15 at 19:19
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    $\begingroup$ Technically correct, but perhaps a tiny bit of overkill! $\endgroup$ Mar 23 '15 at 19:25
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    $\begingroup$ 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. $\endgroup$ Mar 23 '15 at 19:33
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    $\begingroup$ Definitely the hard way! $\endgroup$ Mar 24 '15 at 1:31
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    $\begingroup$ @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! $\endgroup$ Apr 13 '15 at 8:42
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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.

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  • $\begingroup$ 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. $\endgroup$ Mar 23 '15 at 19:28
  • $\begingroup$ Fixed Now !Thank You @BillDubuque $\endgroup$ Mar 23 '15 at 19:33
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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.

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