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I'm a relative newcomer to these stackexchange websites, and this post will serve as my introduction to the Mathematics stackexchange site. After perusing some of the related questions, I found these to be the most relevant to my question: 1) Why can't you count real numbers this way?, 2) Why there is not the next real number?, 3) Proof that the real numbers are countable: Help with why this is wrong, and 4) Why does the Dedekind Cut work well enough to define the Reals? .

In related question #1), the concept of an "index" with regard to the un/countability of the set of real numbers was mentioned. I have encountered that term before in past discussions regarding the countability of transfinite ordinal sets (in other math forum websites), but I'm not sure what it means.

In any case, I understand the "real-number line" to be composed of the set of rational numbers (whose cardinality is $\aleph_0$) which have sets of "gaps" between them, each of which is filled in by a set of irrational numbers (whose cardinality is $\aleph_1$). My "understanding" is that there are no "gaps" between any two consecutive irrational numbers, and likewise, no "gaps" between a rational number and the irrational numbers that immediately precede and follow it. If all of this is true, then given some number $n$ (rational or irrational), why can't I write the next consecutive irrational number that immediately follows n as $\lim_{t\to\infty}n+10^{-t}$.

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  • $\begingroup$ I think the reason is quite simply that the limit you pose is actually $n$ since you can make it arbitrarily close to $n$ as you like. $\endgroup$ – Cameron Williams Oct 21 '14 at 23:59
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    $\begingroup$ Between any two distinct real numbers there's an infinite number of rationals and an infinite number of irrationals, therefore, the phrase My "understanding" is that there are no "gaps" between any two consecutive irrational numbers is quite false. $\endgroup$ – TZakrevskiy Oct 22 '14 at 0:08
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    $\begingroup$ Furthermore, there is no such thing as the irrational number that "precedes" or "follows" a rational number. The real numbers are simply not organized in a sequence the way the integers are. $\endgroup$ – MJD Oct 22 '14 at 0:10
  • $\begingroup$ "I think the reason is quite simply that the limit you pose is actually n" True! But that is precisely why I'm seeking some way to represent irrational numbers. I would like to find a way to represent them that enables us to differentiate them from one another mathematically. $\endgroup$ – jpbrooks-user153707 Oct 22 '14 at 0:11
  • $\begingroup$ My answer here addresses several of your confusions, including your mistaken understanding of the $\lim$ notation. $\endgroup$ – MJD Oct 22 '14 at 0:13
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You asked specifically

why can't I write the next consecutive irrational number that immediately follows n as $\lim_{t\to\infty}n+10^{-t}$.

This $\lim_{t\to \infty}$ notation is a mathematical term of art; it has a specific meaning, and it simply doesn't mean what you think it does. You have the idea that it means $n$ plus some tiny tiny number that is smaller than every other number. But it doesn't mean that, because there is no such thing as a tiny tiny number that is smaller than every other number. What you have written is actually a complicated way of writing the number $n$, nothing more.

I sympathize with your desire to invent a tiny tiny number that is smaller than every other number, because it sounds plausible, and almost everyone tries to do it at some point. But on closer investigation, the idea turns out to be incoherent, and this is why there is no notation for that tiny tiny number.

Consider this analogy: the phrase “the biggest purple hat in my closet” sounds reasonable, and there is no obvious reason why such a thing doesn't exist. But in fact there is no such thing as the biggest purple hat in my closet, although you have to know a bit about the contents of my closet to know this. (I don't have any purple hats.)

Similarly, there is no such thing as the number that is the next bigger real number after $n$, although you have to know a little bit about real numbers to realize this. For suppose there were such a number; let us call it $y$. Then let $\epsilon = y - n$. Since $y$ is bigger than $n$, the number $\epsilon$ is positive. But then $\frac\epsilon2$ is positive also, although smaller than $\epsilon$, so we have $0 < \frac\epsilon2 < \epsilon$. Then adding $n$ to each part of that inequality, we get $$n < n+\frac\epsilon2 < n+\epsilon = y$$ which shows that $n+\frac\epsilon2$ is between $n$ and $y$, and therefore that $y$ was not the next number bigger than $n$. But this argument works for any possible number $y$ that you could invent, so the whole idea of a number $y$ that is the next bigger number after $n$ has to be scrapped.

I'm not sure this answers your question, but your question is based on several basic (but very common) misunderstandings of how numbers work, so I hope this helps clear up some of it. But the question posed by the title of your question, “Why can't consecutive irrational numbers…” can be answered without even reading the whole thing; the answer is “because there is no such thing as consecutive irrational numbers.”

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  • $\begingroup$ Yes, I follow your reasoning. I see that the way irrational numbers are structured (as a set) is fundamentally different (and not simply quantitatively different) from the way the rational numbers are structured. $\endgroup$ – jpbrooks-user153707 Oct 22 '14 at 0:33
  • $\begingroup$ For these purposes, the rationals and the irrationals are not very different. Neither one is like the natural numbers, where there is a tiny tiny natural number smaller than every other natural number (we call this number “$1$”) and where there is a clear notion of "next bigger natural number". It might be instructive to see why, if you want $y$ to be the next bigger natural number after $n$, my argument above about the nonexistence of $y$ fails, as it must. $\endgroup$ – MJD Oct 22 '14 at 0:37
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    $\begingroup$ Well, your argument wouldn't apply in that case because the natural numbers are not closed with regard to division. (But it would still be valid where it did apply.) But, Yes, I do see your point. Your argument would also work for the rationals. $\endgroup$ – jpbrooks-user153707 Oct 22 '14 at 1:01
  • $\begingroup$ Yes, exactly so. There are no consecutive rationals. $\endgroup$ – MJD Oct 22 '14 at 1:07
  • $\begingroup$ Ok. So what all of this seems to suggest to me is that between any two rational numbers on the "real-number line", there are aleph-one irrational numbers that each have aleph-one rational and irrational numbers between them, and so on. Would that be correct? $\endgroup$ – jpbrooks-user153707 Oct 22 '14 at 1:15
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You can't have two 'adjacent' real numbers like that. Suppose $\alpha$ and $\beta$ were two such numbers, then $\frac{\alpha+\beta}{2}$ would be between them.

As already pointed out in the comments, $$\lim_{t\to\infty}n+10^{-t}=n.$$

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  • $\begingroup$ Well, now this raises another issue about cardinality that I'm considering as a new question $\endgroup$ – jpbrooks-user153707 Oct 22 '14 at 0:39
  • $\begingroup$ @jpbrooks-user153707: Ask away! $\endgroup$ – Charles Oct 22 '14 at 2:06

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