Algebraic and analytic irrationality proofs


At the behest of mathematicians wiser than myself, I have been thinking about number theory generally and irrationality proofs in particular. I think that there are two flavors of irrationality proofs available to us: algebraic and analytic. Here I will sketch three irrationality proofs which naturally present clear examples of these two flavors.

Roughly speaking, analytic proofs of irrationality construct infinite approximations of a number which are “too good,” while algebraic proofs use only number-theoretic tools, such as primality. This is not such a clear definition. We shall prove the following claim in three ways to demonstrate this:

Proposition. $\sqrt{a^2 + 4}$ is irrational for all nonzero integers $a$.

Algebraic Proof 1. Suppose that $a^2 + 4 = b^2$ for some integer $b$. Without loss of generality suppose that $a$ and $b$ are nonnegative. Since $4 = (b - a)(b + a)$, we have only three possible cases:

  1. $b - a = 4$ and $b + a = 1$. This implies $2b = 5$, which is impossible.

  2. $b - a = 1$ and $b + a = 4$. This also implies $2b = 5$, which is still impossible.

  3. $b - a = b + a = 2$. Then $b = 2$ and $a = 0$.

Therefore $(a, b) = (0, 2)$ is the only integer solution. $\blacksquare$

Algebraic Proof 2. Without loss of generality, suppose that $a$ and $b$ are nonnegative. If $a = 0$ then $b = 2$, so suppose that $a \geq 1$ (and $b \geq 1$ follows). If $a = 1$ then $b^2 = 5$, which is impossible. Thus suppose $a \geq 2$. We clearly have $b > a$, so $b^2 \geq (a + 1)^2 > a^2 + 4$. This shows that $(a, b) = (0, 2)$ is the only integer solution. $\blacksquare$

The analytic proof requires a bit of work.

Lemma. For every nonzero integer $a$, the polynomial $1 - az - z^2$ has distinct nonzero roots, one inside the unit circle, and one outside.

Proof. The discriminant is $\sqrt{a^2 + 4} \neq 0$, so there are two roots. The minus root which comes from the quadratic formula lies inside the unit circle, and the fact that the roots multiply to $1$ in absolute value shows that the other is outside the unit circle. $\blacksquare$

Lemma. Let $p_n$ and $q_n$ be the numerator and denominator, respectively, of the $n$th convergent to the infinite continued fraction $[a, a, \dots]$. Then

\[\frac{p_n}{q_n} = (a + r) (1 + O(\epsilon)^n),\]

where $r$ is the root inside the unit circle of $1 - az - z^2$ and $0 < \epsilon < 1$.

I won’t prove this lemma, but here is how a proof would go: Derive generating functions for $p_n$ and $q_n$ from their well-known recurrences. These are rational and give us a closed form. You can factor out a $1 / r$ times some constants from both closed forms, and dividing them gives the asymptotic expansion above with $\epsilon < 1$ since $r$ is the smaller root.

(I am too lazy to type this proof, but it really is a routine computation.)

Analytic conclusion. By the above lemmas, the continued fraction $[a, a, \dots]$ equals $a + r$ when $a \neq 0$, so this number is irrational since it has an infinite continued fraction. By the quadratic formula $a + r$ is a rational linear combination of $1$ and $\sqrt{a^2 + 4}$, so it is irrational iff $\sqrt{a^2 + 4}$ is irrational.

Is either method better? I’m not sure. The algebraic proofs are shorter here, but the analytic proof is pretty interesting. I really like generating functions, so I’m biased. It seems good to know both.