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As $\phi$ is a real solution of the equation $x^2-x-1=0$ and it is irrational, I derived the following generalization:

Proposition: Let it be some polynomial of the form $x^n-x^{n-1}-...-x-1=0$ such that $n>1$. Then, every real solution of the polynomial is an algebraic irrational number.

The restriction $n>1$ is needed; $x-1=0$ has a real solution ($1$) which is not an irrational number.

Proof

Any polynomial of the form described can be rewritten as $$x^n=x^{n-1}+x^{n-2}+...+x+1$$

As the RHS is a geometric progression, we can transform the above expression to get $$x^n=\frac{x^n-1}{x-1}$$

Operating, we get that $$x^{n+1}=2x^n-1$$

Thus, $x<2$. Now, suppose $x=\frac{p}{q}$, where $p$ and $q$ are positive integers. Substituting, we have that $$\frac{p^{n+1}}{q^{n+1}}=2\frac{p^{n}}{q^{n}}-1$$

Multiplying both sides by $q^n$, we get that $$\frac{p^{n+1}}{q}=2{p^{n}}-q^n$$

As the RHS is an integer, then we have that the LHS is an integer; and that is only possible if $q\mid{p}$,where $\mid$ means "divides". However, as $x<2$, then $\frac{p}{q}<2$, and thus it is impossible that $q\mid{p}$ unless $p=q$, and thus $x=1$.

As every rational solution to the polynomial of the form $x^n-x^{n-1}-...-x-1=0$ such that $n>1$ is greater than $1$, we can conclude that every real solution of the polynomial is an algebraic irrational number. Subsequently, we can define an infinite set of algebraic irrational numbers containing the real solutions of the polynomials of the form $x^n-x^{n-1}-...-x-1=0$ (with the exception mentioned).

I want to know if the proposition and proof exposed are correct. Any comment/correction would be welcomed!

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    $\begingroup$ You could use the "rational root theorem". $\endgroup$ Oct 6, 2020 at 17:42

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Yes, this is fine and it's a nice proof. If you keep following this line of reasoning you'll arrive at a more general result: if $p(x) = x^n + a_{n-1} x^{n-1} + \dots + a_0$ is a monic polynomial with integer coefficients, then a root of $p$ is rational iff it's an integer, and in fact it must be an integer dividing $a_0$. This is a (useful and important) special case of the rational root theorem, and it lets you write down lots and lots of irrational algebraic numbers, namely the algebraic integers.

In this case the rational root theorem tells us that the only possible rational roots are $\pm 1$ and it's easy to rule out both of these. But this very last conclusion

Subsequently, we can define an infinite set of algebraic irrational numbers containing the real solutions of the polynomials of the form $x^n-x^{n-1}-...-x-1=0$

needs a little more work. You need to rule out the possibility that this infinite sequence of polynomials has only finitely many distinct roots. (And in any case if this is all you want to prove then you can take the numbers $\sqrt[n]{2}, n \ge 2$.)

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  • $\begingroup$ Thanks @QiaochuYuan! Really useful answer! $\endgroup$ Oct 6, 2020 at 18:21

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