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I am wondering if the following is a valid proof of how there is always a rational number between 2 distinct reals:

First I show that there is a real number between any 2 distinct real numbers which has a finite representation. Given any 2 distinct real numbers, first simplify any repeating 9's by eliminating them and adding 1 to a number accordingly. Then, if they were to be indexed left to right (e.g. in $3.24$, assign position 1 to $3$, position 2 to $2$, etc.), there exists a first such index for which the number at this position is different between these 2 real numbers. The one with the number being smaller here is the smaller real number. Then, if to the smaller number we augment the first non-9 digit by 1, we have obtained a real number smaller than the bigger real number out of the 2 with which we started, but bigger than the smaller one out of the 2 with which we started- so this is a real number between the 2 arbitrary distinct real numbers we started with. Then, this can be made into a rational number by having the numerator be the entire real number but omitting the decimal point, and having the denominator be $10^d$ where $d$ is the number of digits after the decimal point in the real number between the 2 real numbers with which we started.

I am unsure of this because in proofs of this such as this, much more complicated arguments are used, leading me to believe there is a flaw in my rather simple argument.

Princess Mia
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  • You lost me at "..if they were to be indexed left to right, there exists a first such index.." – AlvinL Sep 21 '23 at 05:46
  • @AlvinL is this not a valid statement? – Princess Mia Sep 21 '23 at 05:47
  • It's valid, AlvinL just doesn't understand you mean a smallest 'n' where n is the negative of the power of ten of the place value. – Nij Sep 21 '23 at 05:52
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    @Shmuel when you write "If we added the number $1$ or anything up to $9$ at the end of the smaller real number," Since when do real numbers have to have ends? – M W Sep 21 '23 at 05:52
  • So you take two real numbers $a.a_1a_2a_3\ldots$ and $b.b_1b_2b_3\ldots$. If $a<b$, then $b$ is a rational between the two. So we can assume them to be between $0$ and $1$. Since the numbers are distinct, there exists a decimal place at which they differ .. – AlvinL Sep 21 '23 at 05:57
  • @MW I see, would this work if instead of saying that I affix it to the end of the number, I just augment the first existing number not equal to $9$ after this index, or, in the case of all of the other numbers being $9$, augment this first index at which the numbers differ? – Princess Mia Sep 21 '23 at 06:00
  • @Shmuel No, because if the smaller number ends in all 9s, and you augment the first index where they differ, you may overshoot - you need to adress all 9s a different way (hint: every real number has a representation that does not end in a sequence of infinitely many 9s). This gets to a larger issue that it isn't clear the extent to which you have a precise definition of what a real number is, and how to decide if two are equal. – M W Sep 21 '23 at 06:02
  • @MW So in the case of us having a real number with 9's repeating infinitely, then we rewrite this such that the number the repeating 9's is augmented, then we like normal compare these numbers finding a first such index? Also, is it the case that the real number we find between these 2 distinct real numbers always has a finite representation, such that we could represent it as a rational number in the manner I have described? – Princess Mia Sep 21 '23 at 06:09
  • @Shmuel Still doesn't work, now if its .49999... and .5002 then you still overshoot. There is a key fact that you will have to use, though, which is that every number that ends in all 9s has an equivalent representation that is not ending in all 9s. that's a nontrivial fact that requires careful definition of what it means for two decimal expansions to be equivalent. Finally, it is true we can always find a terminating decimal between two real numbers, yes. – M W Sep 21 '23 at 06:18
  • @MW but when we first replace $.49999$ by $.5$, then the first such index at which they differ is when $.5002$ has 2- so then we augment the number after this, producing $.50001$, which is in between them, right? – Princess Mia Sep 21 '23 at 06:21
  • @Shmuel How do you know one number isn't .4999999... and the other number is .5 exactly? – M W Sep 21 '23 at 06:22
  • That works but how do you guarantee such a procedure for ALL weird cases? – AlvinL Sep 21 '23 at 06:22
  • @MW would that not contradict the fact that these 2 real numbers are distinct? – Princess Mia Sep 21 '23 at 06:23
  • @Shmuel, yes, it would, but that's a nontrivial fact that needs to be established, which is exactly what I'm getting at. Your general idea can be made to work once you establish as a fact that .499999...=.5 exactly. then once you've established that, just first choose representations that never terminate in all 9s, and augment the first non-9 digit after the differing digit, as you proposed earlier. But you need to establish that fact first – M W Sep 21 '23 at 06:30
  • Isn't it simpler to point out that $(a+b)/2$ is a real number that is strictly between $a$ and $b$? – Arturo Magidin Sep 21 '23 at 06:32
  • @ArturoMagidin that's the sensible thing to do, but somehow I fell down a rabbit hole of trying to critique a method using decimal expansions. [Edit - he wants a rational number actually] – M W Sep 21 '23 at 06:35
  • @ArturoMagidin is this average assured to have a finite representation in the reals, or in a form which we can easily represent as a rational number? – Princess Mia Sep 21 '23 at 06:37
  • If both $a,b$ are rational that's fine, otherwise we'd eventually need some density-esque argument to conjure a rational with desired property. – AlvinL Sep 21 '23 at 06:38
  • You said you wanted to first show that there is a real between any two distinct reals, and then you engage in all sorts of very complicated and obtuse contorsions using decimal expansions. If all you want as a first step is to prove that if $a$ and $b$ are distinct real numbers there will be a real number strictly between them, then $(a+b)/2$ does the trick. Why you think this is useful to prove the existence of a rational between them I do not know, nor do I understand what makes you think that what you are trying to do is less "complicated" than using the Archimedean Property. – Arturo Magidin Sep 21 '23 at 06:41
  • @ArturoMagidin in fairness, though original question was phrased a little awkwardly, he wanted to find a real "with finite representation" between the other reals, which he would then argue was in fact rational. So his goal was to get a rational in between two reals all along as I understand it. – M W Sep 21 '23 at 06:44
  • (If you are dead set on using decimal expansions, it is also rather mysterious why you do not seem to use the fact that a decimal expansion corresponds to a rational number if and only if it is eventually peripdic...) – Arturo Magidin Sep 21 '23 at 06:45
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    @MW In fairness, the government doesn't like it when I read minds without a warrant, so I go by what they said/wrote. – Arturo Magidin Sep 21 '23 at 06:46
  • You also seem to have forgotten about the existence of negative numbers, which makes claims like that "the one with the number being smaller is the smaller real number" possibly false. Is $-.5$ smaller than $-.6$? – Arturo Magidin Sep 21 '23 at 06:49
  • The thing you are trying to prove requires a handle to rationals in real number system. Either you need to use Archimedean property (in axiomatic approach) or use some construction of real numbers. – Paramanand Singh Sep 22 '23 at 03:47

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