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Let [itex]x[/itex] and [itex]y[/itex] be real numbers such that [itex]y<x[/itex], using the Dedekind cut construction of reals prove that there is always a rational [itex]q[/itex] such that [itex]y<q<x [/itex]

What I've done

Since I can associate a cut to every real number, let [itex]x^∗[/itex] be the cut associated to [itex]x[/itex] and [itex]y^∗[/itex] the one associated with [itex]y[/itex].

Since [itex]y<x \implies y^* \subsetneq x^*[/itex] then [itex]\exists q \in \Bbb Q[/itex] such that [itex]q \in x^*[/itex] and [itex]q\not\in y^*[/itex]. Next I associate a cut [itex]q^∗[/itex] to [itex]q[/itex]. Now how can I deduce from there that [itex]q^* \subsetneq x^*[/itex] and [itex]y^* \subsetneq q^*[/itex], thus proving [itex]y<q<x[/itex]

Any help will be appreciated,

M.

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# Proof of rational density using Dedekind cuts

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