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I have studied the definition of Dedekind cut and from what I understand $\{r ∈ \Bbb{Q}:r^3<2\}$ is Dedekind cut. I can explain this by making a cut on the real line and making two subsets whose union forms $\Bbb{R}$ and satisfy the conditions for Dedekind cut. However, I don't know how to show this with a formal prove. This post: Dedekind cut of an irrational number talks a bit about this but does not give any formal prove.

I need help in providing a formal prove for $\{r∈Q:r^3<2\}$.

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You are not supposed to find two sets whose union is $\mathbb{R} $ so your statement about making a cut on the real line does not seem relevant here. You should first write down the definition of Dedekind cut and verify each of the properties for the given set $A=\{r\in\mathbb{Q},r^3<2\} $. Most books list the following properties for Dedekind cut $A$:

  1. $\emptyset \neq A\neq\mathbb {Q} $ and $A\subset\mathbb {Q} $.
  2. If $r, s$ are rational numbers such that $r>s$ and $r\in A$ then $s\in A$ ie if some rational number lies in $A$ then all smaller rationals also lie in $A$.
  3. $A$ does not possess a maximum element ie if $r\in A$ then there is an element $s\in A$ with $r<s$.

Now you need to verify the above properties for your given set $A$. The first property is obvious because $1\in A, 2\notin A$. For second property let $r\in A$ and $r>s$ then we can see that $s^3<r^3<2$ so that $s\in A$.

It is the third property which requires some effort to verify. You have to show that if $r\in A$ then there is another $s\in A$ with $r<s$. Since $r\in A$ we have $r^3<2$ and then $d=2-r^3>0$. We will use $r, d$ to find a number $s\in A$ with $r<s$. Since we want $r<s$ we can first try writing $s=r+h$ where $h>0$ and we need to find $h$ suitably so that $s=r+h\in A$ ie $(r+h) ^3<2$ ie $$r^3+3r^2h+3rh^2+h^3<2$$ or $$h(3r^2+3rh+h^2)<2-r^3=d$$ Next let's choose $h<1$ so that $$3r^2h+3rh+h^2<3r^2+3r+1$$ ie $$h(3r^2+3rh+h^2)<h(3r^2+3r+1)$$ If in addition to $h<1$ we also ensure that $h<d/(3r^2+3r+1)$ then we have the desired inequality $(r+h) ^3<2$. Thus we just need to take $h$ less than $\min(1,d/(3r^2+3r+1))$ and then $s=r+h\in A$ and thus $A$ has no maximum element.

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Let $A=\{r\in\mathbb{Q}:r^3<2\}$ and $B=\mathbb{Q}\setminus A$. Then $A$ and $B$ form a partition for $\mathbb{Q}$. To show this is a cut we need to show that for every $a\in A$ and $b\in B$, $a<b$.

  • Observe that if $q<0$ with $q\in\mathbb{Q}$, then $q\in A$ since $q^3<0$. Therefore, $A$ contains all the negative numbers, so all numbers in $B$ must be nonnegative. It follows that if $a<0$, then for all $b\in B$, $a<b$.

  • Observe that $(m^3-n^3)=(m-n)(m^2+mn+n^2)$. Suppose $a\in A$ and $b\in B$ are both nonnegative. We know that $a^3<b^3$. Moreover, $(b^3-a^3)=(b-a)(b^2+ab+a^2)$. The LHS is positive by assumption and since $a$ and $b$ are both nonnegative, $b^2+ab+a^2$ is nonnegative. Therefore, $b-a$ is also positive, or $b>a$.

  • All that is left is to show that $A$ has no greatest element. This can be done by observing that $\sqrt[3]{2}$ is irrational, so for any $a\in A$, we can find $a+\varepsilon$ that is also in $A$.

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  • $\begingroup$ Did you miss out something in the first bullet? Your sentence ends abruptly. $\endgroup$ – The.Shark Feb 28 '18 at 3:44
  • $\begingroup$ I've edited - a misplaced dollar sign. $\endgroup$ – Michael Burr Feb 28 '18 at 15:14

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