Statement:

Let $$X=$$ $$\{(a,b) \in \mathbb{R} \setminus \{0\} \times \mathbb{R}\setminus \{0\}:a<b\}$$ There exists a function $f:X \rightarrow \mathbb{R} \setminus \{0\}$ such that for all $(a,b) \in X$, $|f(a,b)-b| \le \frac{3}{4}|a-b|$ and $|f(a,b)-a| \le \frac{3}{4}|a-b|$.

I proved this statement constructively without using the axiom of countable choice, but for doing that, I assumed real numbers as regular sequences of rational numbers (according to Bishop).

The idea of the proof is that you construct a rational number $q$ such that $|q-b| < \frac{3}{4}|a-b|$ and $|q-a| < \frac{3}{4}|a-b|$. If $q=0$ then put $f(a,b)=\frac{\frac{a+b}{2}+b}{2}$ and if $\neg(q=0)$ then put $f(a,b)=q$.

Now I wonder how can we prove this statement by assuming real numbers as Dedekind cuts of rational numbers? i.e if we consider real numbers as Dedekind cuts, can we prove this statement constructively without using the axiom of countable choice?

The proof that I mentioned doesn't go well if reals are Dedekind reals, because we cannot easily construct the desired rational number, if we restrict ourselves to not use the axiom of countable choice.

Can anyone help me with this? Thank you.

Statement:

Let $$X=$$ $$\{(a,b) \in \mathbb{R} \setminus \{0\} \times \mathbb{R}\setminus \{0\}:a<b\}$$ There exists a function $f:X \rightarrow \mathbb{R} \setminus \{0\}$ such that for all $(a,b) \in X$, $|f(a,b)-b| \le \frac{3}{4}|a-b|$ and $|f(a,b)-a| \le \frac{3}{4}|a-b|$.

I proved this statement constructively without using the axiom of countable choice, but for doing that, I assumed real numbers are regular sequences of rational numbers (according to Bishop).

The idea of the proof is that you construct a rational number $q$ such that $|q-b| < \frac{3}{4}|a-b|$ and $|q-a| < \frac{3}{4}|a-b|$. If $q=0$ then put $f(a,b)=\frac{\frac{a+b}{2}+b}{2}$ and if $\neg(q=0)$ then put $f(a,b)=q$.

Now I wonder how can we prove this statement by considering real numbers as Dedekind cuts of rational numbers? i.e if we consider real numbers as Dedekind cuts, can we prove this statement constructively without using the axiom of countable choice?

The proof that I mentioned doesn't go well if reals are Dedekind reals, because we cannot easily construct the desired rational number, if we restrict ourselves to not use the axiom of countable choice.

Can anyone help me with this? Thank you.

Statement:

Let $$X=\{(a,b) \in \mathbb{R} \setminus \{0\} \times \mathbb{R}\setminus \{0\}:a<b\}$$ There exists a function $f:X \rightarrow \mathbb{R} \setminus \{0\}$ such that for all $(a,b) \in X$, $|f(a,b)-b| \le \frac{3}{4}|a-b|$ and $|f(a,b)-a| \le \frac{3}{4}|a-b|$.

I proved this statement constructively without using the axiom of countable choice, but for doing that, I assumed real numbers as regular sequences of rational numbers (according to Bishop).

The idea of the proof is that you construct a rational number $q$ such that $|q-b| < \frac{3}{4}|a-b|$ and $|q-a| < \frac{3}{4}|a-b|$. If $q=0$ then put $f(a,b)=\frac{\frac{a+b}{2}+b}{2}$ and if $\neg(q=0)$ then put $f(a,b)=q$.

Now I wonder how can we prove this statement by assuming real numbers as Dedekind cuts of rational numbers? i.e if we consider real numbers as Dedekind cuts, can we prove this statement constructively without using the axiom of countable choice?

The proof that I mentioned doesn't go well if reals are Dedekind reals, because we cannot easily construct the desired rational number, if we restrict ourselves to not use the axiom of countable choice.

Can anyone help me with this? Thank you.

Statement:

Let $$X=$$ $$\{(a,b) \in \mathbb{R} \setminus \{0\} \times \mathbb{R}\setminus \{0\}:a<b\}$$ There exists a function $f:X \rightarrow \mathbb{R} \setminus \{0\}$ such that for all $(a,b) \in X$, $|f(a,b)-b| \le \frac{3}{4}|a-b|$ and $|f(a,b)-a| \le \frac{3}{4}|a-b|$.

I proved this statement constructively without using the axiom of countable choice, but for doing that, I assumed real numbers as regular sequences of rational numbers (according to Bishop).

The idea of the proof is that you construct a rational number $q$ such that $|q-b| < \frac{3}{4}|a-b|$ and $|q-a| < \frac{3}{4}|a-b|$. If $q=0$ then put $f(a,b)=\frac{\frac{a+b}{2}+b}{2}$ and if $\neg(q=0)$ then put $f(a,b)=q$.

Now I wonder how can we prove this statement by assuming real numbers as Dedekind cuts of rational numbers? i.e if we consider real numbers as Dedekind cuts, can we prove this statement constructively without using the axiom of countable choice?

The proof that I mentioned doesn't go well if reals are Dedekind reals, because we cannot easily construct the desired rational number, if we restrict ourselves to not use the axiom of countable choice.

Can anyone help me with this? Thank you.