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I know that between any two reals, there is an irrational number.

See: Proving that there exists an irrational number in between any given real numbers

Now let a, b $\in$ $R$ such that a < b. And let M be the set of irrationals between a and b.

I want to show that M is uncountable. To do this, I think I need to show that there does not exist a bijection from M to the natural numbers.

  1. Can someone give me a hint about how to start to show this?

  2. What's the best way to approach "non-existence" proofs in general?

Thanks

Community
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  • Have you considered simply using a bijection from $[a,b]$ to $\mathbb R$? Like, say, $\tan x$? – abiessu Sep 30 '13 at 21:09
  • If you subtract a countable set from an uncountable one, what remains is uncountable. Any interval $(a,b)$ is uncountable, and the rationals are countable. – Andrés E. Caicedo Sep 30 '13 at 22:11

3 Answers3

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Instead you could show that there is a bijection between this set and interval (0,1)

Alex
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  • Is this the standard way to show a set is uncountable? –  Sep 30 '13 at 21:17
  • As far as I know: yes. As you have probably noticed when trying to solve this problem, to show "non-existence" directly can be tricky. – Alex Sep 30 '13 at 21:31
  • @Alex: What bijection between the set of irrational numbers in the interval $(a, b)$ and the set of all numbers in the interval $(a, b)$ are you proposing the OP should use? – Rob Arthan Sep 30 '13 at 21:40
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    I didn't say that the proof should one-step procedure. The bijection could be between the interval (a,b) and (0,1), and then additional argument about countability of rationals on both of the intervals. – Alex Sep 30 '13 at 21:45
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    @Alex: your comment reduces the problem to the special case when $a = 0$ and $b = 1$, without actually solving the problem. – Rob Arthan Sep 30 '13 at 21:57
  • Not sure I understand what you mean. First, the main purpose of my answer here was to tell OP to try another approach to the problem: instead of showing that there is no bijection to naturals, show that there is a bijection to a set that is known to be uncountable. It seems like kbball understood it. Second, bijection between $(a,b)$ and $(0, 1)$ is there to see that we can consider $(0,1)$ without loss of generality. – Alex Sep 30 '13 at 22:07
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    I don't understand how you plan to complete the proof in a way such that your suggestion results in a simplification. The obvious bijection between $(a,b)$ and $(0,1)$ does not map rationals to rationals (unless $a$ and $b$ happen to be rationals.) – Trevor Wilson Sep 30 '13 at 22:17
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First, show that there is a rational number between any two reals.

This can be done using the Archimedean axiom (or theorem depending on your axioms) that, for any two positive reals $x$ and $y$, there is an integer $n$ such that $nx > y$.

Then, using the same reasoning, show that there is another rational between the two reals.

Let these rationals be $r$ and $s$.

Finally, construct a linear mapping from $[0, 1]$ to $[r, s]$. This mapping maps rationals to rationals and irrationals to irrationals.

Since there an uncountable number of irrationals in $[0, 1]$, there are an uncountable number of irrationals in $[r, s]$.

Another proof, this time by contradiction, can be modeled on Cantor's original proof that the reals are uncountable, but I'll leave that as an exercise.

marty cohen
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I assume that you know that the interval $(a, b)$ is uncountable and that the set of rational numbers in that interval is countable. So the problem reduces to the following:

If $Y \subseteq X$ are sets such that $X$ is uncountable and $Y$ is countable, then $Z = X \mathrel{\backslash} Y$ is uncountable.

To see this, note that $Z$ is either (a) finite, (b) countably infinite or (c) uncountable. But the union of a finite set (or a countably infinite set) and a countable set is countable so (a) and (b) can't hold. So (c) holds, which was what we wanted.

Rob Arthan
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  • A bit harder is to prove that in fact $(a,b)\setminus \mathbb Q$ has the same size as $(a,b)$. (This does not require the axiom of choice.) – Andrés E. Caicedo Sep 30 '13 at 22:25
  • @Andres Caicedo: indeed, let's hope the OP enjoys the Schroeder-Bernstein theorem when he or she gets to it. – Rob Arthan Sep 30 '13 at 22:30