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I read on MathOverflow the following:

"The number of proofs that we have of showing some numbers are irrational are very limited. We either show a number $α$ is irrational because it is algebraic of degree greater than one (by exhibiting an irreducible polynomial $f$ of degree greater than one $f(α)=0$)."

An algebraic number is a number that is a root for a polynomial with integers coefficients. What is an algebraic number of degree greater than one?

How do we show an "irreducible polynomial $f$ of degree greater than one $f(α)=0$" (a non-constant polynomial such that $\alpha$ is a root for it?)

Is there some literature that show that some number is irrational with this approach? Maybe a link on the internet or a book that show exactly this?

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    $\begingroup$ For example, let $f(x)=x^2-2$ then $f$ is an irreducible polynomial of degree greater than one and $f(\pm\sqrt{2})=0$ hence $\pm\sqrt{2}$ are irrational numbers. $\endgroup$ Oct 17, 2019 at 21:02
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    $\begingroup$ I thought that the definition of an algebraic number was any number that is a root of a polynomial with integer coefficients. Why did you write "can't be?" $\endgroup$ Oct 17, 2019 at 21:33
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    $\begingroup$ @MichaelTuchman I've fixed. Thanks! $\endgroup$
    – Pinteco
    Oct 17, 2019 at 21:48
  • $\begingroup$ Normally, I'm not such a nitpicker, but in this case the entire meaning was changed to its opposite. $\endgroup$ Oct 18, 2019 at 1:18

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An algebraic number is a number that is a root of a polynomial with integer coefficients (and not all coefficients $0$).

If it is a root of such a polynomial that has degree $d$ and is irreducible over the rationals, then it can't be the root of any polynomial with integer coefficients (not all $0$) and degree less than $d$. This is because if $z$ is a root of two polynomials $f$ and $g$, then it is a root of $\gcd(f,g)$, which is a polynomial that divides both $f$ and $g$. In that case we say the number is algebraic of degree $d$.

In particular, $z$ is rational if and only if it is the root of a polynomial of degree $1$ with integer coefficients (namely $z = a/b$ is a root of $b z - a$). So if it is algebraic of degree $>1$, it can't be rational.

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When you say a number $z$ is algebraic, all you know is that there is some polynomial $P$ with rational coefficients such that $P(z)=0$. The notion of "degree" tells you more: An algebraic number of degree $d$ is an algebraic number such that there is some polynomial of degree $d$ such that $P(z)=0$ and no polynomial of lesser degree having $z$ as a root. This polynomial $P$ is called the minimal polynomial and has the property that if $Q(z)=0$ then $P$ divides $Q$ evenly.

"Algebraic of degree one" therefore means that $z$ satisfies $az+b=0$ for some rational $a,b$ or, otherwise put, $z=-b/a$. This is synonymous with the statement that $z$ is rational.

"Algebraic of degree two" turns out to mean - if you examine the quadratic formula - that the number is of the form $a\pm \sqrt{b}$ where $a,b$ are rational and $b$ is not a perfect square. If you want to prove, for instance, that $\sqrt{2}$ is an algebraic of degree two, you first note that it is a root of $x^2-2$ and then, by whatever machinery you have, show that $x^2-2$ is not divisible by any polynomials of lesser degree - so must be the minimal polynomial of $\sqrt{2}$. While you could probably show that $x^2-2$ is irreducible without any special machinery, in general, it can be fairly difficult to decide if a polynomial is irreducible, though algorithms to do that do exist. Note that since all the rationals have degree one, this implies that $\sqrt{2}$ is irrational.

I don't think there's many references to this method of proof in particular, because it's usually taken as a corollary of the theory of field extensions - the comment you refer to is pointing out that algebraic numbers are quite well understood and once we can write a polynomial that a number satisfies, we can situate it inside of a well-developed theory to immediately see that it is irrational.

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