May 21st, 2017, 05:56 AM  #1 
Member Joined: Jan 2016 From: United Kingdom Posts: 35 Thanks: 0  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 
May 21st, 2017, 07:59 AM  #2 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,664 Thanks: 2644 Math Focus: Mainly analysis and algebra 
It's a good attempt, but I worry that it's circular where you say that $0 \lt p_n p \lt k_2k_1$ for some $n$. My approach would be to say that for $k_2k_1 \gt 0$, there exists $n \in \mathbb N$ such that $n(k_2k_1) \gt 1$ because $n(k_2k_1)$ is unbounded above. Then there exists $m \in \mathbb Z$ such that $$\begin{gather*} nk_1 \lt m \lt nk_2 \\ k_1 \lt \tfrac{m}{n} \lt k_2 \end{gather*}$$ Last edited by v8archie; May 21st, 2017 at 08:34 AM. 
May 21st, 2017, 11:17 PM  #3 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,664 Thanks: 2644 Math Focus: Mainly analysis and algebra 
Just to add some detail to the above:

May 22nd, 2017, 03:33 PM  #4 
Math Team Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233 
[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 
May 22nd, 2017, 03:55 PM  #5 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,664 Thanks: 2644 Math Focus: Mainly analysis and algebra 
Try $0.2989989998\ldots$ and $0.3010010001\ldots$. I think your rational is $0.25$ which isn't between the two irrationals.

May 22nd, 2017, 04:02 PM  #6 
Math Team Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233 
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. 
May 22nd, 2017, 04:28 PM  #7 
Math Team Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233 
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. 
May 22nd, 2017, 04:32 PM  #8 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,664 Thanks: 2644 Math Focus: Mainly analysis and algebra 
You might have more luck rounding the lower one up, but we might be able to arrange for them to fail still.

May 22nd, 2017, 04:39 PM  #9 
Math Team Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233 
That's a great numerical counter example , I don't think I can save my argument. kudos 
May 22nd, 2017, 05:17 PM  #10 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,664 Thanks: 2644 Math Focus: Mainly analysis and algebra 
Just truncating the upper one might work.


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