# Proof that $\sqrt[3]{17}$ is irrational [duplicate]

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.

• 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. – Arthur Sep 28 '18 at 10:38
• This made me chuckle. – Wesley Strik Sep 28 '18 at 10:59
• @Arthur: that argument is circular (subtly), which has been discussed on MathOverflow. – JavaMan Sep 28 '18 at 11:30
• @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? – Arthur Sep 28 '18 at 11:40
• @Arthur: Top answer from this thread: mathoverflow.net/questions/42512/… – JavaMan Sep 30 '18 at 3:24

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?

• $17|p$ means there must exist some number, say $l\in \mathbb{Z}$ such that $p=17l$. – Wesley Strik Sep 28 '18 at 10:48
• @WesleyGroupshaveFeelingsToo Indeed. – José Carlos Santos Sep 28 '18 at 10:49
• 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. – Wesley Strik Sep 28 '18 at 10:53
• @WesleyGroupshaveFeelingsToo No. It's just fine. – José Carlos Santos Sep 28 '18 at 10:56
• muito obrigado. – Wesley Strik Sep 28 '18 at 10:58

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.

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.

$$17|p^3\iff 17|p$$ 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 $$17\nmid p^3.$$

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.

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.

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 $$\sqrt 2$$.