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Using wolframalpha.com we get that the real root of this polynomial is $-0.68233$

The only way that I have found how to prove it is using the Rational Root theorem.

Using that theorem the possible rationals roots are 1 and -1, but none of them work.

Then the real root has to be irrational.

My questions are:

  1. Is there other way to prove it without using the Rational Root Theorem?

  2. Is my proof written well?

Beginner
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  • As a general rule, we like questions here to be self-contained, because they are supposed to be a record for later people who ask similar questions. It's quite possible the URL you've linked to won't exist in one or five or ten years. – Thomas Andrews Jan 01 '15 at 05:36
  • When you say "Is my proof written well," I don't see any proof of yours, only Dr. Rob's proof on the answer forum. As for whether you can copy from that site, I have no idea, but that doesn't resolve the problem of posting links here. – Thomas Andrews Jan 01 '15 at 05:55
  • you have the main idea right. i don't think you should worry too much about writing formal proofs. – abel Jan 02 '15 at 04:50
  • @BillDubuque I added the tag elementary number theory, as you requested. – Beginner Mar 08 '19 at 04:20

3 Answers3

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Here is an easy proof. Suppose it has a rational root say $a/b$ and $\gcd (a,b)=1$. Then putting $x=a/b$, we obtain $$a^3+ab^2+b^3=0.$$ Note that $|a|>1$ or $|b|>1$, since $\pm 1$ are not roots. Let $p$ be a prime such that $p\mid a$ (assume $|a|>1)$. Then $p\mid a^3$ and $p\mid ab^2$ and hence $p\mid a^3+ab^2$. Now, from above equation we have $b^3=-(a^3+ab^2)$. Therefore $p\mid b^3,$ i.e. $p\mid b$. That is $p\mid \gcd (a,b)$, which is a contradiction.

pritam
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  • @Beginner: Have you noted that this proof necessarily is the Rational Root Theorem? –  Jan 01 '15 at 05:31
  • @Beginner: Note that $p\mid a^3$ and $p\mid ab^2$. Then $p\mid (a^3+ab^2)=-b^3$ and hence $p\mid b$. – pritam Jan 01 '15 at 05:37
  • @Beginner: I have added the explanation in the answer. – pritam Jan 01 '15 at 06:07
  • @Beginner: I think it depends on the reviewer/examiner of your proof. If the reviewer is willing to accept the 'Rational Root Theorem' as a standard result (i.e. allow you to use the Theorem) then your proof is fine. Otherwise, you have to provide a proof (not just a link) of the Theorem. – pritam Jan 01 '15 at 06:43
  • That's not really a proof without RRT since you have simply the proof of RRT in this special case. – Bill Dubuque Mar 08 '19 at 02:22
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If it were rational it would be $m/n$, where $m$ and $n$ are integers. Then we would have $$ \left(\frac m n \right)^3 + \frac m n + 1 =0. $$ From that we get $$ m^3 + mn^2 + n^3 = 0. $$ If the fraction $m/n$ is in lowest terms, then $m$ and $n$ have no prime factors in common. But $$ n^3 = -m(m^2+n^2) = (-m\cdot\text{something}) $$ so $n^3$ is divisible by the prime factors of $m$. That can happen only if $m$ has no prime factors, i.e. $m=\pm1$. In the same way, we have $$ m^3 = - n^2(m+n) = (-n^2\cdot\text{something}), $$ and so $n=\pm1$. But $1$ and $-1$ are not roots of the equation.

Michael Hardy
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First Question

For another approach without using the Rational Root Theorem is to solve the cubic.

The roots of a cubic equation of the form $x^3+px+q=0$ are given by,

$\sqrt[3]{-\dfrac{q}{2}+\sqrt{\dfrac{q^2}{4}+\dfrac{p^3}{27}}}+\sqrt[3]{-\dfrac{q}{2}-\sqrt{\dfrac{q^2}{4}+\dfrac{p^3}{27}}}\tag{1}$$ \omega\left(\sqrt[3]{-\dfrac{q}{2}+\sqrt{\dfrac{q^2}{4}+\dfrac{p^3}{27}}}\right)+\omega^2\left(\sqrt[3]{-\dfrac{q}{2}-\sqrt{\dfrac{q^2}{4}+\dfrac{p^3}{27}}}\right)\tag{2}$$\omega^2\left(\sqrt[3]{-\dfrac{q}{2}+\sqrt{\dfrac{q^2}{4}+\dfrac{p^3}{27}}}\right)+\omega\left(\sqrt[3]{-\dfrac{q}{2}-\sqrt{\dfrac{q^2}{4}+\dfrac{p^3}{27}}}\right)\tag{3}$

What's left is solving the cubic $x^3+x+1=0$ using the above formulas and finding which among them is real and whether it is rational or irrational. I leave that to you.

Second Question

What you have described as your proof is really not a proof. It can be said to be a sketch of the proof. For example, you said that,

...Using that theorem the possible rationals roots are $1$ and $-1$, but none of them work.

It isn't illogical to ask,

How exactly does the theorem lets you conclude that the only possible rational roots are $1$ and $-1$?

To answer this you have to prove your conclusion. To be precise, what your argument lacks is elaboration. Otherwise, your argument is fine.

In short, when you are writing proofs, imagine that you are writing it for someone who wants to have a clear picture of your argument from your proof. Your job will be to make it as clear and as rigorous as possible.

For example, you may have written the proof as follows,

A real root always exists since the degree of the equation is odd.

The only way that I have found how to prove it is using the Rational Root theorem. Which tells us that...(insert a brief statement of the Rational Root Theorem here).

Using this theorem the possible rationals roots are $1$ and $-1$, but none of them work. Since $f(x)=x^3+x+1\implies f(1)=3\ne 0$ and $f(-1)=1\ne 0$.

Hence the real root has to be irrational.

Community
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