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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.

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    $\begingroup$ From a purely first order logic perspective, all proofs by contradiction (say, in number theory) are equivalent to finding an integer between $0$ and $1$. For, if $C$ is the contradiction you want, then $C\iff 0=1$. $\endgroup$ – Jason DeVito Apr 10 at 17:27
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    $\begingroup$ You might be interested in transcendental numbers. Lindemann gave a way to prove a number is transcendental thereby proving it is irrational. Admittedly, I have not read Lindemann’s proof; it could be that he reduces to case of an integer belonging to $(0,1)$, though as Jason accurately points out, any contradiction is equivalent to any other false statement. $\endgroup$ – Clayton Apr 10 at 17:36
  • $\begingroup$ Presumably your exclusion of roots also excludes radicals structured from them such as $1+\sqrt{2}$. $\endgroup$ – J.G. Apr 10 at 17:44
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    $\begingroup$ I'm not very interested in roots of numbers. @Clayton " any contradiction is equivalent to any other false statement". But why most of these contradictions are integers in $(0,1)$?. Beuker's proved that $\pi$ is irrational by showing that the exponential grows faster than the factorial, which is a contradiction. But this is the only exception I found. $\endgroup$ – Pinteco Apr 10 at 18:00
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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 :) )

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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.

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  • $\begingroup$ I think I messed up on the question. I've should asked why almost all proof by $contradiction$ ends in a integer in $(0,1)$. $\endgroup$ – Pinteco Apr 15 at 16:53
  • $\begingroup$ Suppose (enter your favourite infinite continued fraction) is rational. Then its continued fraction is finite. But this continued fraction is infinite. Contradiction. $\endgroup$ – Robert Israel Apr 15 at 20:55
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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.

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