# Cube of irrational Dedekind cuts

Let $A$ be a set that represents the Dedekind cut for the real number $\sqrt[3]{3}$. Let $B$ be the set representing the Dedekind cut for the real number 3. Prove that $A*A*A = B$, where $*$ is the Dedekind multiplication operator.

I have no idea how to prove this, but this is what I got so far. Multiplication for two positive Dedekind cuts/sets is defined in the following way ($(x,y)$ is a tuple in the cartesian join $X \times Y$ of those sets): $$X*Y = \{q \in \mathbb{Q}: (q \leq 0) \lor (\exists (x,y) \in X\times Y: q = xy) \}.$$ For brevity, I'm omitting both the $q\in \mathbb{Q}$ and $q \leq 0$ part in the above expression so that it becomes $$X*Y = \{\exists (x,y) \in X\times Y: q = xy\}.$$ It is implied that all sets contain all negative rationals. Using this notation, my sets are $$A = \{q^3 < 3\}\quad\mathrm{and}\quad B=\{q < 3\}.$$ So $$A*A*A = \{\exists (x,y,z)\in A*A*A : q = xyz)\} = \{\exists (x,y,z)\in \{q^3 < 3\}^3 : q = xyz)\}$$ It is clear that we can extract $x,y,z$ in the above and the set is equal to $$\{\exists x,y,z \in\mathbb{Q}: x^3,y^3,z^3 < 3 \land q = xyz\}$$ Cubing both sides of the equality filter $$\{\exists x,y,z \in\mathbb{Q}: x^3,y^3,z^3 < 3 \land q^3 = x^3y^3z^3\}$$ Since $x^3,y^3,z^3 < 3$ we can add the following condition to the set generating expression: $$\{q^3 < 27 \land \exists x,y,z \in\mathbb{Q}: x^3,y^3,z^3 < 3 \land q^3 = x^3y^3z^3\}$$ The above can be written as the intersection of two sets: $$\{q^3 < 27\} \cap \{\exists x,y,z \in\mathbb{Q}: x^3,y^3,z^3 < 3 \land q^3 = x^3y^3z^3\}$$ We see that the left set is equal to $B$: $$B \cap \{\exists x,y,z \in\mathbb{Q}: x^3,y^3,z^3 < 3 \land q^3 = x^3y^3z^3\}$$ If we can now show that $B = \{\exists x,y,z \in\mathbb{Q}: x^3,y^3,z^3 < 3 \land q^3 = x^3y^3z^3\}$ the proof is complete. Clearly, $a \in \{\exists x,y,z \in\mathbb{Q}: x^3,y^3,z^3 < 3 \land q^3 = x^3y^3z^3\} \implies a \in B$. So the last part is showing $$a \in B \implies a \in \{\exists x,y,z \in\mathbb{Q}: x^3,y^3,z^3 < 3 \land q^3 = x^3y^3z^3\}$$ That is the same as $$a^3 < 27 \implies \exists x,y,z \in\mathbb{Q}: x^3,y^3,z^3 < 3 \land a^3 = x^3y^3z^3$$ $a^3 < 27 \implies a < 3$, so what if I just choose $x, y, z$ so that $x^3 = y^3 = z^3 = a$? $$a^3 < 27 \implies a,a,a < 3 \land a^3 = a\cdot a \cdot a$$ Proof complete!?

• Why do you think you need to use the least upper bound property? – xarles Aug 22 '18 at 21:03
• @xarles Because of the book hint :) My question is a rewrite of exercise 1.14 of Chapman Pugh's Real Mathematical Analysis. – Björn Lindqvist Aug 22 '18 at 21:09
• Your definition of multiplication doesn't look quite right. I think you need to add that $x$ and $y$ are both positive. – Barry Cipra Aug 22 '18 at 21:11
• The only point it is not clear is the one after "Then it follows:", since it is not the same for three numbers to be less than 3, that their product is less than $3^3$. One implies the other, so you only get $A*A*A\subset B$. – xarles Aug 22 '18 at 21:17

You have proved (if I follow your notation) that $A*A*A\subseteq B$, but you seem to be missing an argument that every rational number $<3$ can be written as a product $xyz$ where $x^3, y^3, z^3$ are all less than $3$.
Doing so might for example involve using that $q\mapsto q^3$ is uniformly continuous between $1$ and $2$.
• Iiuc, the missing piece is proving $\forall q < 3\, \exists x^3, y^3, z^3 < 3 : q = xyz$? I wonder is my proof on the right track or should I start over? – Björn Lindqvist Aug 23 '18 at 15:23