Proving irrationality of $\sqrt[3]{3}+\sqrt[3]{9}$

Question

I need to prove $$\sqrt[3]{3}+\sqrt[3]{9}$$ is irrational, I assumed $$\sqrt[3]{3}+\sqrt[3]{9} = \frac{m}{n}$$ I cubed both sides and got $$\sqrt[3]{3}+\sqrt[3]{9} = \frac{m^3-12n^2}{9n^3}$$

I tried setting $$\frac{m^3-12n^2}{9n^3} = \frac{m}{n}$$ but that led me nowhere. so what can I do?

Please use descriptive titles. You have 150 characters to use, and you're more than welcomed to use $\rm\LaTeX$ as well. "Proving the irrationality of a number" is a meaningless title. Same goes to your other post from earlier. — Asaf Karagila, Jan 28 '19 at 18:43
Closely related: https://math.stackexchange.com/questions/1542708 — Watson, Jan 29 '19 at 09:06
It may be that this exercise was intended to reinforce specific material covered in your textbook or other course materials, but your Readers will not be aware of that context unless you share it. It is certainly a problem that is amenable to various attacks, so knowing what you are studying will help Readers respond in a way you find useful. — hardmath, Jan 29 '19 at 17:18

Michael Rozenberg · Accepted Answer · 2019-01-28T09:07:14.333

16

Let $\sqrt[3]3+\sqrt[3]9=r$.

Thus, since for all reals $a$, $b$ and $c$ we have: $$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-ac-bc),$$ we obtain: $$3+9-r^3+9r=0.$$

Now, let $r=\frac{m}{n},$ where $m$ and $n$ are naturals with $gcd=1.$

Thus, $$m^3-9n^2m-12n^3=0,$$ which says that $m$ is divisible by $3$.

Let $m=3m'$, where $m'$ is a natural number.

Thus, $$9m'^3-9n^2m'=4n^3,$$ which says that $n$ is divisible by $3$, which is a contradiction.

edited Jan 28 '19 at 09:07

answered Jan 28 '19 at 09:01

Michael Rozenberg

194,933

3

What did you set as a, b, and c? – Guysudai1 Jan 28 '19 at 09:07
1

@Guysudai1 They are any real numbers. – Michael Rozenberg Jan 28 '19 at 09:07
4

@Guysudai1 Michael seems to have chosen $a=\sqrt[3]{3}$, $b=\sqrt[3]{9}$, and $c=-r$. Then $a+b+c=0$ and one can study the left-hand side. – Joonas Ilmavirta Jan 28 '19 at 13:43

lhf · Answer 2 · 2019-01-28T11:13:46.377

8

Here are two other takes.

Take 1

Let $\alpha = \sqrt[3]{3}+\sqrt[3]{9}$. Then $\alpha^3 = 9 \alpha + 12$.

By the rational root theorem, if $\alpha$ is rational, then $\alpha$ is an integer.

Now $1 < \sqrt[3]{3} < 2 $ and $2 < \sqrt[3]{9} < 3 $, and so $3 < \alpha < 5$.

Since $x=4$ is not a root of $x^3 = 9 x + 12$, $\alpha$ is not an integer and so $\alpha$ is irrational.

Take 2

We have $\alpha = \beta+\beta^2$, where $\beta=\sqrt[3]{3}$.

If $\alpha$ were rational, then $\beta$ would be a root of a quadratic polynomial with rational coefficients. However, the polynomial with rational coefficients with least degree having $\beta$ as a root is $x^3-3$.

edited Jan 28 '19 at 11:13

answered Jan 28 '19 at 11:07

lhf

216,483

It is fine what $\beta$ is a root of $x^3 - 3$ but how do you know that it is of least degree polynomial with rational coefficients. Are you using some standard result? – prashant sharma Feb 04 '19 at 07:26
@prashantsharma, $x^3 - 3$ is a cubic with no rational roots and so is irreducible. – lhf Feb 04 '19 at 07:58
@Ihf Means are you suggesting that any equation with no rational roots is irreducible? – prashant sharma Feb 05 '19 at 09:24
@prashantsharma, not any equation, but cubics, yes. – lhf Feb 05 '19 at 10:12
@Ihf Can you please supply a proof? – prashant sharma Feb 05 '19 at 10:34
@prashantsharma, $f=gh$. What are the possible degrees of $g$ and $h$? – lhf Feb 05 '19 at 10:48
@Ihf The possible degrees of g,h are 1, 2. Since all the coefficients of f are rational so are of g, h otherwise we get a contradiction. Is this your argument? – prashant sharma Feb 11 '19 at 07:04

score 5 · Answer 3 · answered Jan 28 '19 at 09:11

5

Let $a=\sqrt[3]{3}$ and $b=\sqrt[3]{9}$, and suppose that $a+b=r$ is rational. Then $$r^3 =(a+b)^3 = a^3+3ab(a+b)+b^3 = 12+9(a+b)=12+9r$$

So $r$ is a solution of equation $x^3-9x-12=0$ which has no rational root.

answered Jan 28 '19 at 09:11

nonuser

90,026

Proving irrationality of $\sqrt[3]{3}+\sqrt[3]{9}$

3 Answers3

Take 1

Take 2