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I know that this kind of question has been answered before in many places, but I'd like a proof with a little spin on it. In his book "A course of pure mathematics", Hardy defines a real number $\alpha$ as a section composed by a lower class (a) (in which a is rational and a$\lt$$\alpha$) and an upper class (A) (in which A is rational and A$\ge$$\alpha$).

I'd like to prove that there are infinitely many rational numbers between any given two real numbers definied in this way by using this idea of sets.

I'm not sure if it's possible, but I believe so, because Hardy says "All these results are immediate consequences of our definitions" after a series of examples I managed to prove except for this one.

It might be useful to tell you that, according to thoses definitions, the relations of magnitude between two real numbers are defined in this way:

$\alpha$$\lt$$\beta$ if, and only if, (a)$\subset$(b) and (A)$\supset$(B)

$\alpha$$\gt$$\beta$ if, and only if, (a)$\supset$(b) and (A)$\subset$(B)

(where (a), (A) and (b), (B) are the lower and upper classes of $\alpha$ and $\beta$).

Evidently, if $\alpha$$\lt$$\beta$ and x is a rational number between $\alpha$ and $\beta$, then x$\in$(A)$\cap$(b). Showing that there are infinitely many elements in (A)$\cap$(b) would show that there are infinitely many rational numbers between $\alpha$ and $\beta$, but I'm stuck at this point and cannot find a way to develop the proof from this approach.

Both hints and solutions are welcome.

Greetings!

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  • $\begingroup$ See whether this helps you: math.stackexchange.com/questions/445514/… $\endgroup$ – Rohan Jan 28 '17 at 14:19
  • $\begingroup$ So, you want to prove that there are infinitely many rational between any two real number.right? $\endgroup$ – MatheMagic Jan 28 '17 at 14:19
  • $\begingroup$ Right. I have seen some proofs of it, but none using this idea of real numbers as sections of a line of rational numbers. $\endgroup$ – R. Maia Jan 28 '17 at 14:24
  • $\begingroup$ Thank you, Rohan, I'll take a look. $\endgroup$ – R. Maia Jan 28 '17 at 14:25
  • $\begingroup$ You MUST use a definition of $\Bbb R$ or a consequence of the definition because $\Bbb R $ can be extended to a larger ordered field $\Bbb R^*$ that has positive members that are smaller than any positive rational. If $x$ is one of them then $0<x/2<x$ but there are no rationals in the interval $(x/2,x)$ ..... $\Bbb R^*$ does not have the Archimedean property. $\endgroup$ – DanielWainfleet Oct 1 '18 at 10:57
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Let $a,b\in\mathbb R$, WLOG $a< b$ then $b-a>0$. Let us assume that $x=b-a$ and $y=5$ then there exist (by Archimedean property) $n\in\mathbb N$ s.t $$nx>y$$ $$n(b-a)>5$$ $$nb-na>5$$. Since difference between $nb$ and $na$ is greater than $5$,then there exists atleast one $m\in\mathbb Z$ s.t $$na<m<nb$$ $$a<\frac mn<b$$ $$a<r<b$$,where $r=\frac mn\in\mathbb Q$ i.e there exist one rational number $r$ between $a$ and $b$.Again $a<r$,there exist $r_1\in\mathbb Q$ s.t $$a<r_1<r$$.Proceeding this way we can generate as many as rational number as we wish.

Hope this will help!!!

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  • $\begingroup$ Let me know if you have any difficulties regarding this answer. $\endgroup$ – MatheMagic Jan 28 '17 at 15:03
  • $\begingroup$ Oh, thanks. Beautifully done. Unfortunately, I think there isn't a proof using the concepts of real numbers I mentioned. But it's strange, because before introducing these ideas Hardy had showed a proof very similar to yours, and then, after defining real numbers as sections and discussing some properties of them he invited the reader to prove again the same thing and claimed that the solution would be a consequence of the previous definitions. What do you think of that? $\endgroup$ – R. Maia Jan 29 '17 at 3:30
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This is really an immediate consequence of the definition of real numbers as given by Hardy. To clarify I repeat the definition here:

A real number $\alpha$ is a pair of sets $A, B$ such that

  • $A\cap B=\emptyset, A\cup B=\mathbb{Q}, A\neq \emptyset \neq B$.
  • if $x\in A, y\in B$ then $x<y$.
  • if $x\in A$ then there is a $y\in A$ with $x<y$. Note that repeated use of this property shows that there are infinitely many $y\in A$ with $x<y$.

If $\alpha=(A, B), \beta=(C, D) $ are two real numbers defined in this manner then we say that $\alpha$ is less than $\beta$ and write $\alpha<\beta$ if $A\subset C$.

Also if $r$ is rational number then there is a corresponding real number $\alpha_r$ defined as $(A_r, B_r) $ where $A_r=\{x\mid x\in\mathbb {Q}, x<r\} $ and $B_r=\{x\mid x\in\mathbb {Q}, x\geq r\} $.


Let's now suppose we are given two real numbers $\alpha=(A, B), \beta=(C, D) $ such that $\alpha <\beta$. By definition this means that $A\subset C$ and hence $S=C\setminus A\neq \emptyset$. It is easy to show that the set $S$ contains infinitely many rationals.

Since $S\neq \emptyset $ there is a member $x\in S$ and then $x\in C, x\notin A$. Since $\beta=(C, D) $ is a real number it follows that there are infinitely many members $y\in C$ with $y>x$. And none of these members $y$ are in $A$ because if $y\in A$ then $x<y$ implies $x\in A$ which is not the case. So all these members $y\in S=C\setminus A$.

And if $r$ is one such rational number lying in this set $S=C\setminus A$ then the corresponding real number $\alpha_r=(A_r, B_r) $ as defined earlier is such that $\alpha<\alpha_r<\beta$. It is in this sense that there are infinitely many rational numbers between any two given real numbers.

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The proof is called diagonal argument : Write each number with its decimal expansion in base and move along the diagonal to create a number that is not on the list. So there is contradiction and real numbers can't be listed.

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