# How to show a number is irrational with this approach?

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?

• 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. Oct 17, 2019 at 21:02
• 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?" Oct 17, 2019 at 21:33
• @MichaelTuchman I've fixed. Thanks! Oct 17, 2019 at 21:48
• Normally, I'm not such a nitpicker, but in this case the entire meaning was changed to its opposite. Oct 18, 2019 at 1:18

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.

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.