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May 21st, 2017, 05:56 AM   #1
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Proof That a Rational lies between two Irrationals

Hi all,

I've attempted to prove that between any two irrational numbers there must exist a rational number. Please observe the image attached. Is this a sound argument?

Regards,

M
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May 21st, 2017, 07:59 AM   #2
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It's a good attempt, but I worry that it's circular where you say that $0 \lt p_n -p \lt k_2-k_1$ for some $n$.

My approach would be to say that for $k_2-k_1 \gt 0$, there exists $n \in \mathbb N$ such that $n(k_2-k_1) \gt 1$ because $n(k_2-k_1)$ is unbounded above. Then there exists $m \in \mathbb Z$ such that $$\begin{gather*} nk_1 \lt m \lt nk_2 \\ k_1 \lt \tfrac{m}{n} \lt k_2 \end{gather*}$$
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Last edited by v8archie; May 21st, 2017 at 08:34 AM.
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May 21st, 2017, 11:17 PM   #3
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Just to add some detail to the above:

Given that $n(k_2-k_1) \gt 1$, if $nk_1 > 0$, there are a finite number of non-negative integers less than $nk_1$ so we can find the greatest of them $a$. Thus we have
$$a \lt nk_1 \lt a+1 \lt nk_1+1 \lt nk_1 + (nk_2-nk_1)=nk_2$$
Writing $m=a+1$, we see that $m \in \mathbb Z$ and
$$nk_1 \lt m \lt nk_2$$

On the other hand, if $nk_1 \lt 0 0$, there are a finite number of non-negative integers less than $-nk_1 \gt 0$ and we can thus find the greatest of them $b$. Then, writing $a=-b-1$, we get
$$a \lt nk_1 \lt a+1 \lt nk_1+1 \lt nk_1 + (nk_2-nk_1)=nk_2$$
as before.
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May 22nd, 2017, 03:33 PM   #4
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[EDIT] The argument below is false. Thank you v8archie


If 2 irrationals are different then they must differ in at least 1 decimal position in their decimal representation , right?

If you accept that all real numbers have a decimal representation then you can pick any two irrationals , truncate them both AFTER the first decimal position that they differ (from left to right), add their truncated values and divide by 2 to get a rational between them.

Example: Find a rational $ \ \ \mathbb{ Q} \ \ $ such that

$ \sqrt{26.5} < \mathbb{Q} < \sqrt{27}$

$ \sqrt{26.5} \approx 5.1478 $

$ \sqrt{27} \approx 5.196 $

truncate at the 4th decimal position , add and divide by 2

$ \frac{5.14 + 5.19}{2} = 5.165 $

$ \sqrt{26.5} < 5.165 < \sqrt{27} $

This procedure works for any 2 irrationals but is not a rigorous proof.



[ENDEDIT]

Last edited by agentredlum; May 22nd, 2017 at 04:17 PM. Reason: false argument
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May 22nd, 2017, 03:55 PM   #5
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Try $0.2989989998\ldots$ and $0.3010010001\ldots$. I think your rational is $0.25$ which isn't between the two irrationals.
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May 22nd, 2017, 04:02 PM   #6
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You have to truncate AFTER the first decimal position that they differ

Calculate again

Oh Shoot , you're right

Last edited by agentredlum; May 22nd, 2017 at 04:13 PM.
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May 22nd, 2017, 04:28 PM   #7
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v8archie , I need your help

Will it work if we truncate 2 places AFTER the position that they differ?

So for your counter example we use 0.29 and 0.30

I see , still no good

Last edited by agentredlum; May 22nd, 2017 at 04:34 PM.
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May 22nd, 2017, 04:32 PM   #8
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You might have more luck rounding the lower one up, but we might be able to arrange for them to fail still.
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May 22nd, 2017, 04:39 PM   #9
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That's a great numerical counter example , I don't think I can save my argument. kudos

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May 22nd, 2017, 05:17 PM   #10
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Just truncating the upper one might work.
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