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Let $\mu$ be the irrationality measure defined as such: The irrationality measure of a real number $\alpha$ is the supremum of real numbers $\mu$ such that the inequality

$$\left| \alpha-\frac{p}q \right| < \frac{1}{q^\mu}$$

has infinitely many solutions in integers $(p, q)$ with $q>0$.

For example:

$$\mu(\sqrt2)=2,\quad\mu(e)=2,\quad\mu(\pi) \leq 7.103,\quad\mu(\zeta(3)) \leq 5.513.$$

Note also that for rational numbers $\mu(\alpha)=1$, that for some numbers called Liouville numbers $\mu(\alpha)=\infty$, and that most irrational numbers have an irrationality measure of $2$ (Roth’s theorem).

Then, here are some conjectures:

Conjecture 1. Let $\alpha$ be an irrational number (which isn’t a Liouville number and which isn’t a solution to $\mu(\alpha)=\alpha$), then $\mu(\mu(\dots(\mu(\alpha))\dots))$ always falls to $1$ after a finite number of iterations.

Conjecture 2. Let $\alpha>2$ be an irrational number, then there exist infinitely many solutions to the equation $\mu(x)=\alpha$.

Also, depending on 2:

Conjecture 2.1. There exist no finite interval contained in $]2, +\infty[$ such that there exist infinitely many solutions to the equation in this interval.

Then

Conjecture 3. Let $\alpha$ be an irrational number (which isn’t a Liouville number) such that $\mu(\alpha) \neq 2$, then $\mu(\alpha)$ is irrational.

Your comments are welcome!

Let $\mu$ be the irrationality measure defined as such: The irrationality measure of a real number $\alpha$ is the supremum of real numbers $\mu$ such that the inequality

$$\left| \alpha-\frac{p}q \right| < \frac{1}{q^\mu}$$

has infinitely many solutions in integers $(p, q)$ with $q>0$.

For example:

$$\mu(\sqrt2)=2,\quad\mu(e)=2,\quad\mu(\pi) \leq 7.103,\quad\mu(\zeta(3)) \leq 5.513.$$

Note also that for rational numbers $\mu(\alpha)=1$, that for some numbers called Liouville numbers $\mu(\alpha)=\infty$, and that most irrational numbers have an irrationality measure of $2$ (Roth’s theorem).

Then, here are some conjectures:

Conjecture 1. Let $\alpha$ be an irrational number (which isn’t a Liouville number and which isn’t a solution to $\mu(\alpha)=\alpha$), then $\mu(\mu(\dots(\mu(\alpha))\dots))$ always falls to $1$ after a finite number of iterations.

Conjecture 2. There exist infinitely many solutions to the equation $\mu(\alpha)=\alpha$ with $\alpha>2$.

Also, depending on 2:

Conjecture 2.1. There exist no finite interval contained in $]2, +\infty[$ such that there exist infinitely many solutions to the equation in this interval.

Then

Conjecture 3. Let $\alpha$ be an irrational number (which isn’t a Liouville number) such that $\mu(\alpha) \neq 2$, then $\mu(\alpha)$ is irrational.

Your comments are welcome!

Let $\mu$ be the irrationality measure defined as such: The irrationality measure of a real number $\alpha$ is the supremum of real numbers $\mu$ such that the inequality

$$\left| \alpha-\frac{p}q \right| < \frac{1}{q^\mu}$$

has infinitely many solutions in integers $(p, q)$ with $q>0$.

For example:

$$\mu(\sqrt2)=2,\quad\mu(e)=2,\quad\mu(\pi) \leq 7.103,\quad\mu(\zeta(3)) \leq 5.513.$$

Note also that for rational numbers $\mu(\alpha)=1$, that for some numbers called Liouville numbers $\mu(\alpha)=\infty$, and that most irrational numbers have an irrationality measure of $2$ (Roth’s theorem).

Then, here are some conjectures:

Conjecture 1. Let $\alpha$ be an irrational number (which isn’t a Liouville number and which isn’t a solution to $\mu(\alpha)=\alpha$), then $\mu(\mu(\dots(\mu(\alpha))\dots))$ always falls to $1$ after a finite number of iterations.

Conjecture 2. Let $\alpha$ be an irrational number, then there exist infinitely many solutions to the equation $\mu(\alpha)=\alpha$ with $\alpha \in \mathopen]2, +\infty\mathclose[$.

Also, depending on 2:

Conjecture 2.1. There exist no finite interval contained in $]2, +\infty[$ such that there exist infinitely many solutions to the equation in this interval.

Then

Conjecture 3. Let $\alpha$ be an irrational number (which isn’t a Liouville number) such that $\mu(\alpha) \neq 2$, then $\mu(\alpha)$ is irrational.

Your comments are welcome!

Let $\mu$ be the irrationality measure defined as such: The irrationality measure of a real number $\alpha$ is the supremum of real numbers $\mu$ such that the inequality

$$\left| \alpha-\frac{p}q \right| < \frac{1}{q^\mu}$$

has infinitely many solutions in integers $(p, q)$ with $q>0$.

For example:

$$\mu(\sqrt2)=2,\quad\mu(e)=2,\quad\mu(\pi) \leq 7.103,\quad\mu(\zeta(3)) \leq 5.513.$$

Note also that for rational numbers $\mu(\alpha)=1$, that for some numbers called Liouville numbers $\mu(\alpha)=\infty$, and that most irrational numbers have an irrationality measure of $2$ (Roth’s theorem).

Then, here are some conjectures:

Conjecture 1. Let $\alpha$ be an irrational number (which isn’t a Liouville number and which isn’t a solution to $\mu(\alpha)=\alpha$), then $\mu(\mu(\dots(\mu(\alpha))\dots))$ always falls to $1$ after a finite number of iterations.

Conjecture 2. Let $\alpha>2$ be an irrational number, then there exist infinitely many solutions to the equation $\mu(x)=\alpha$.

Also, depending on 2:

Conjecture 2.1. There exist no finite interval contained in $]2, +\infty[$ such that there exist infinitely many solutions to the equation in this interval.

Then

Conjecture 3. Let $\alpha$ be an irrational number (which isn’t a Liouville number) such that $\mu(\alpha) \neq 2$, then $\mu(\alpha)$ is irrational.

Your comments are welcome!

Let $\mu$ be the irrationality measure defined as such  : "The irrationality measure of a real number $\alpha$ is the supremum of real numbers $\mu$ such that the inequality

$$\left| \alpha-\frac{p}q \right| < \frac{1}{q^\mu}$$

has infinitely many solutions in integers $(p, q)$ with $q>0$.

For example  :

$\mu(\sqrt2)=2$, $\mu(e)=2$, $\mu(\pi) \leq 7.103$, $\mu(\zeta(3)) \leq 5.513$

Note also that for rational numbers $\mu(\alpha)=1$, that for some numbers called Liouville numbers $\mu(\alpha)=\infty$ and that most irrational numbers have an irrationality measure of 2 (Roth's theorem).

Then, here are some conjectures

Conjecture 1. Let $\alpha$ be an irrational number (which isn't a Liouville number and which isn't a solution to $\mu(\alpha)=\alpha$), then $\mu(\mu(...(\mu(\alpha))))...$ for a finite number of iteration always falls to 1.

Conjecture 2. Let $\alpha$ be an irrational number, then, there exist infinitely many solutions to the equation $\mu(\alpha)=\alpha$ with $\alpha \in ]2, +\infty[$

Also, depending on 2  :

Conjecture 2.1. There exist no finite interval contained in $]2, +\infty[$ such that there exist infinitely many solutions to the equation in this interval.

Then

Conjecture 3. Let $\alpha$ be an irrational number (which isn't a Liouville number) such that $\mu(\alpha) \neq 2$, then, $\mu(\alpha)$is irrational.

Your comments are welcome  !

Let $\mu$ be the irrationality measure defined as such: The irrationality measure of a real number $\alpha$ is the supremum of real numbers $\mu$ such that the inequality

$$\left| \alpha-\frac{p}q \right| < \frac{1}{q^\mu}$$

has infinitely many solutions in integers $(p, q)$ with $q>0$.

For example:

$$\mu(\sqrt2)=2,\quad\mu(e)=2,\quad\mu(\pi) \leq 7.103,\quad\mu(\zeta(3)) \leq 5.513.$$

Note also that for rational numbers $\mu(\alpha)=1$, that for some numbers called Liouville numbers $\mu(\alpha)=\infty$, and that most irrational numbers have an irrationality measure of $2$ (Roth’s theorem).

Then, here are some conjectures:

Conjecture 1. Let $\alpha$ be an irrational number (which isn’t a Liouville number and which isn’t a solution to $\mu(\alpha)=\alpha$), then $\mu(\mu(\dots(\mu(\alpha))\dots))$ always falls to $1$ after a finite number of iterations.

Conjecture 2. Let $\alpha$ be an irrational number, then there exist infinitely many solutions to the equation $\mu(\alpha)=\alpha$ with $\alpha \in \mathopen]2, +\infty\mathclose[$.

Also, depending on 2:

Conjecture 2.1. There exist no finite interval contained in $]2, +\infty[$ such that there exist infinitely many solutions to the equation in this interval.

Then

Conjecture 3. Let $\alpha$ be an irrational number (which isn’t a Liouville number) such that $\mu(\alpha) \neq 2$, then $\mu(\alpha)$ is irrational.

Your comments are welcome!