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Firstly the title may not explain my question well but i cant think of a logical title for my question as its been a long time since i did math.

i know that the set of rational numbers is countably infinite (have done the proof before its a pain and way to long to write so im going to take this one as a fact)

Now i also know a proof to prove that between every rational number there exists an irrational number. ( i swear i used to know at least 3 ways to prove this from algebra to calculus)

i also used to know that the irrational numbers are unaccountably infinite. it seems to me that this means there exists irrational numbers such that there is no rational number between them.

1)is there a way to prove that this pair exists?

2)is there anyway to express 2 such irrational numbers

Edit: so the answer im receiving are causing me more confusion if my assumption isn't true and there is a rational between any two irrational numbers then it seems like they should have the same cardinality.

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  • $\begingroup$ pick a name for it that make sense an ill rename it. $\endgroup$ – Faust Oct 31 '15 at 1:43
  • $\begingroup$ Did the proof of the bijection between the rationals and the naturals not fit in the margin of the notebook? A good title for this would be "Density of the Rationals?" $\endgroup$ – Prince M Oct 31 '15 at 1:57
  • $\begingroup$ That's "uncountably infinite" & not "unaccountably infinite" (which, I have to say, is pretty funny). $\endgroup$ – BrianO Oct 31 '15 at 2:10
  • $\begingroup$ On behalf of new user @user1280378: For confusion in your edited part: Edit: so the answer im receiving are causing me more confusion if my assumption isn't true and there is a rational between any two irrational numbers then it seems like they should have the same cardinality Between any two irrational numbers also have infinitely many of irrational numbers. I think you misunderstood the 'any two irrational numbers'. It is not two irrationals "next" each other, actually there is no such "next" in $\mathbb{R}$ (in contrast, there is "next" in $\mathbb{N}$) $\endgroup$ – dantopa Jan 7 at 10:05
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Between every two real numbers (in particular between two irrational numbers), there are rational numbers (in fact infinitely many of them).

Depending on which definition of real numbers you are using, this is more or less obvious. For example, if you define real numbers as equivalence classes of rational Cauchy sequences, it is clear that the rational numbers are dense in $\Bbb{R}$.

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  • $\begingroup$ I understand where your coming from and i agree that the rationals are dense in $ \mathbb {R} $ but if theirs a rational between every irrational then why are the irrationals uncountable whilst the rationals are? $\endgroup$ – Faust Oct 31 '15 at 1:59
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    $\begingroup$ Because infinity is funny that way. $\endgroup$ – Brian Tung Oct 31 '15 at 2:01
  • $\begingroup$ More seriously, there is no "next real number" or "next rational number." So our intuition about stringing numbers together in a line fails at that level. $\endgroup$ – Brian Tung Oct 31 '15 at 2:02
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    $\begingroup$ Because the definition of uncountable has to do with existence of a bijection to the naturals, not how the reals are ordered. $\endgroup$ – Prince M Oct 31 '15 at 2:03
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    $\begingroup$ I see how it can feel funny that between every irrationals there is infinitely many rationals, even though there are a lot less rationals. But if you wanted to make that argument rigorous, it would be something like: There is function $f:\Bbb{R}\times \Bbb{R}\to \Bbb{Q}$ that assign to a pair of real numbers, a rational sitting between them. The problem is that there is no way to prove that this map is injective, and actually, I suspect that using sime kind of diagonal argument, we could show directly that it cannot be injective. $\endgroup$ – Nitrogen Oct 31 '15 at 2:04
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Imagine that such two numbers exists and denote them $a$ and $b$, $a < b$. Then definitely $\epsilon = b-a>0$. But it is clear, that $\forall a$ irrational and $\forall \delta >0$ you can find $q \in \mathbb Q$ such that $|a-q|<\delta, a<q$. But if you pick $\epsilon=\delta$ you just find rational number between $a$ and $b$.

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That there is a rational between any two irrationals comes from the definition of the real number system;the reals can be extended to larger ordered fields that have "infinitesimals", members that are positive but less than any positive rational.If x is one of these "infinitesimals" then there is no rational between x/2 and x, but x is not in the real-number system. The reals are usually defined by Dedekind cuts, or by equivalence classes of Cauchy sequences of rationals.

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