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Consider $\sqrt[3]{17}$. Like the famous proof that $\sqrt2$ is irrational, I also wish to prove that this number is irrational. Suppose it is rational, then we can write:

$$ 17 = \frac{p^3}{q^3}.$$ and then $$ 17q^3 = p^3$$

With the proof of $\sqrt2$ we used the fact that we got an even number at this step in the proof and that $p$ and $q$ were in lowest terms. However, 17 is a prime number, somehow we could use this fact and the fact that every number has a unique prime factorisation to arrive at a contradiction, but I don't quite see it yet.

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    $\begingroup$ If it had been $16$ instead of $17$, you could've written $16p^3 = (2p)^3 + (2p)^3 = q^3$, and Fermat's last theorem would tell you that it is impossible. Unfortunately, FLT isn't strong enough to show it for $17$, as $17$ is not the sum of two cubes. $\endgroup$ – Arthur Sep 28 '18 at 10:38
  • $\begingroup$ This made me chuckle. $\endgroup$ – Wesley Strik Sep 28 '18 at 10:59
  • $\begingroup$ @Arthur: that argument is circular (subtly), which has been discussed on MathOverflow. $\endgroup$ – JavaMan Sep 28 '18 at 11:30
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    $\begingroup$ @JavaMan I agree with you in the case of infinite descent proofs specifically for the exponent $3$, as (at least the one on Wikipedia explicitly excludes that case). But for the full FLT, by way of the modularity conjecture? I don't know enough about it to say. Do you have a link? $\endgroup$ – Arthur Sep 28 '18 at 11:40
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    $\begingroup$ @Arthur: Top answer from this thread: mathoverflow.net/questions/42512/… $\endgroup$ – JavaMan Sep 30 '18 at 3:24
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The argument that works with $2$ also works with $17$. Since $17q^3=p^3$, $17\mid p^3$ and therefore $17\mid p$. Can you take it from here?

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  • $\begingroup$ $17|p$ means there must exist some number, say $l\in \mathbb{Z}$ such that $p=17l$. $\endgroup$ – Wesley Strik Sep 28 '18 at 10:48
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    $\begingroup$ @WesleyGroupshaveFeelingsToo Indeed. $\endgroup$ – José Carlos Santos Sep 28 '18 at 10:49
  • $\begingroup$ Then $17q^3=17^3 l^3 \implies q^3 =17^2 l^3$, so $17^2 |q^3$ but then also $17|q^3$ which means we can write $q=17k$. If we had assumed from the start that the the fraction $\frac{p}{q}$ was in lowest terms, then we can now write $\frac{p}{q}=\frac{17l}{17k}=\frac{l}{k}$, we arrive at a contradiction. $\endgroup$ – Wesley Strik Sep 28 '18 at 10:53
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    $\begingroup$ @WesleyGroupshaveFeelingsToo No. It's just fine. $\endgroup$ – José Carlos Santos Sep 28 '18 at 10:56
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    $\begingroup$ muito obrigado. $\endgroup$ – Wesley Strik Sep 28 '18 at 10:58
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If you are allowed to use the uniqueness of prime factorisations then an equivalent argument is as follows:

In the prime factorisation of any cube, the exponent of each prime must be a multiple of $3$ i.e. it is $0, 3, 6, 12$ etc. In particular, the exponents of $17$ in the prime factorisations of $p^3$ and $q^3$ must each be a multiple of $3$.

But since $17q^3 = p^3$, the exponents of $17$ in the prime factorisations of $p^3$ and $q^3$ must differ by $1$ (otherwise $p^3$ would have two different prime factorisations). Two multiples of $3$ cannot differ by $1$, so $p$ and $q$ such that $17q^3 = p^3$ do not exist.

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Using same idea have used here. It's clear that, $17^{1/3}$ is root of the monic polynomial $x^3-17=0 $. Now, if $17^{1/3}$ is an rational algebraic number, it need to be an integer. But, $2^3=8<17<3^3=27$; so, $2<\sqrt[3]{17}<3$. Hence, it is a irrational number.

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$$17|p^3\iff 17|p$$ just like in the proof for $\sqrt2$ and the rest follows.


Proof of $\implies$:

If $$17\nmid p$$ then $$p\bmod17=r,1\le r\le16.$$

Then by the binomial expansion

$$p^3\equiv r^3\mod17$$ and $$17\nmid p^3.$$

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Assume $17^{\frac{1}{3}}$ is rational.

Then there exists integers $c$ and $d$ such that $$\frac{c}{d}=17^{\frac{1}{3}}$$

Now, by Zorn's lemma there exists a greatest common divisor of $c, d$ say $k$. Then $ak=c$ and $bk=d$. Then by definition of the rationals \begin{align} \frac{c}{d} &= \frac{ak}{bk} \\ &=\frac{a}{b} \\ &= 17^{\frac{1}{3}} \end{align} Then $17$ divides $a^3$. If we assume $17$ divides $a$ then we can use the unique factorisation theorem to conclude that $a=a_{1}^{m_{1}}a_{2}^{m_{2}} \cdots$ to which $a^3=a_{1}^{3m_{1}}a_{2}^{3m_{2}} \cdots$ which is a contradiction since it obviously has no factors which are $17$ since $17$ is prime. Thus $17 |a$.

Since $17|a \,\exists\, h \,\text{s.t}\, a=17h$ hence $17^2 h^3=b^3$ and by similar logic to above, $17^2 |b$. Since this is true we can use the prime factorisation theorem to conclude that $17|b$. Hence $a$ and $b$ have a common factor. This is a contracdiction.

Hence $17^{\frac{1}{3}}$ is irrational.

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Assume that $\sqrt[3]{17}$ is rational. Since it is not a integer, you can find $a,b$ co-prime integers so that $$\sqrt[3]{17}=\frac ab\to 17=\frac{a^3}{b^3}$$

But we assumed $a,b$ to be co-prime, so $\frac{a^3}{b^3}$ isn't a integer, a contradiction.

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HINT

We have that

$$17q^3 = p^3 \implies q^3 = 17^2p_1^3$$

then we can conclude by contradiction as for the proof for the irrationality of $\sqrt 2$.

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