Does all proofs of irrationality are the same way: “Find an integer between $0$ and $1$”?

Question

Excluding square/cubic etc roots of numbers, I've read proofs about the irrationality of $e,\pi,\ln 2,\zeta(2)$ and $\zeta(3)$. In all of them is: assume $x=p/q$, where $x$ is the number trying to proved irrational and $p,q$ integers. Long story short, by assuming $x=p/q$ we found an integer between $0$ and $1$.

Is no other way to prove a number is irrational? Maybe finding a combination of $p,q$ that is not integer. For example, assume $x=p/q$ then $p+q = \text{something not integer}$.

PS. The only proof that I've that does not end this way is the beuker's proof for $\pi$, where he find the estimation $1/p^n<1/n!$ which is not possible thus $\pi$ is irrational.

Answer 1

In general for logarithms there is also another way. Let's prove that given $a,b\in \Bbb{N}$ such that $b$ is not a power of $a$ then $\log_a(b)\notin \Bbb{Q}$.

Premise

First let's prove this lemma:

Given $x,n\in \Bbb{N}$, if $x$ is not a $n$-th power, then $x^{\frac 1n}\notin \Bbb{Q} $

If this wasn't true then:

$$x^{\frac 1n}=\frac ND \ \ \ \ \ \gcd(N,D)=1 \ \ \ D\neq 1$$

This would imply that:

$$x=\frac{N^n}{D^n}$$

Since $x\in \Bbb{N}$, then $D|N$, but this is absurd, because they are coprimes for hyphotesis. So the lemma is true

Proof

Let's suppose that the statemente isn't true, so:

$$\log_a(b)=\frac ND \ \ \ \ \ \gcd(N,D)=1 \ \ \ D\neq 1$$

For the definition of logarithm:

$$a^{\frac ND}=b \Rightarrow a=(b^D)^{\frac 1N} $$

Since $a\in \Bbb{Q}$ , $b^D$ must be a $n$-th power. But this implies that $N|D$ and this is absurd, because they are coprimes for hyphotesis. So the initial statement is true.

Improvement

Now note that given a number $j\in \Bbb{R}/\Bbb{N}$ then:

$\log_a(b)=\frac{\log_j(b)}{\log_j(a)}$

Since $\log_a(b)$ is irrational, at least one among $\log_j(b)$ and $\log_j(a)$ must be irrational. These are some ideas, maybe with some other effort you can prove the more general case(I have proved some other partial results, but I think this is more than sufficient :) )

Answer 2

Every infinite simple [continued fraction] is irrational. This follows from uniqueness of the continued fraction representation plus the fact that the continued fraction of a rational number is finite. I don't see any "integer between $0$ and $1$" there.

Answer 3

Every root of an irreducible polynomial in $\mathbf Q[x]$ with degree greater than 1 is irrational (or just their real roots are if you insist irrational numbers should be real numbers) and irreducibility tests generally do not involve any argument like "there are no integers between 0 and 1": the Eisenstein criterion, reduction mod $p$, and so on.