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 May 18th, 2018, 07:27 AM #1 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 125 Real Numbers are a Subset of the Rationals The following table, which is countable (google), includes all pos rationals and reals. The reals are a subset of the rationals if you define: Real: m/100000........., all m Note: Left column and Top row are endless, \begin{matrix} &1 & 2 & 3 & 4 & . & . &. \\ 1 &\frac{1}{1} & \frac{1}{2} & \frac{1}{3} &\frac{1}{4} & . & . &. \\ 2 & \frac{2}{1} & \frac{2}{2} & \frac{2}{3} & \frac{2}{4} & .&. &. \\ 3 & \frac{3}{1}& \frac{3}{2} & \frac{3}{3} &\frac{3}{4} &. & . &. \\ 4 & \frac{4}{1} &\frac{4}{2} &\frac{4}{3} & \frac{4}{4} &. &. &. \\ . &. &. & . & . & . & . & .\\ .&. &. & . & . & . & . &. \\ . &. & . & . & .& . & . & . \end{matrix}\\ 0 assumed included. May 18th, 2018, 08:40 AM #2 Senior Member   Joined: Sep 2015 From: USA Posts: 2,463 Thanks: 1340 $\pi \in \mathbb{R}$ $\pi \not \in \mathbb{Q}$ $\therefore \mathbb{R} \not \subset \mathbb{Q}$ Thanks from zylo May 18th, 2018, 09:11 AM   #3
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Quote:
 Originally Posted by romsek $\pi \in \mathbb{R}$ $\pi \not \in \mathbb{Q}$ $\therefore \mathbb{R} \not \subset \mathbb{Q}$
True

$\displaystyle \pi$ $\displaystyle \equiv$ 3 "+ " 1415926....../1000000.......

That's awkward and requires refinement: my original version was:

The real numbers in [0,1] are a subset of the rational numbers in [0,1]. From table.

Real numbers in [0,1] = m/1000......... , all m

In that case, a real number is N.r which is still countable: N is a subset of the rationals and so is r. May 18th, 2018, 09:29 AM   #4
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Quote:
 Originally Posted by romsek $\pi \in \mathbb{R}$ $\pi \not \in \mathbb{Q}$ $\therefore \mathbb{R} \not \subset \mathbb{Q}$
Romsek

I doubt zylo will accept your proof. He simply assumes again that an infinite set has the same cardinality as any other infinite set.

There is no doubt that he has described an infinite set that includes all rationals. In fact, it includes every possible decimal representation of every rational number. He still has not shown how he matches each real to one of these representations. Obviously, he will have no difficulty matching a unique rational to one of those representations. But he neglects to show that the irrationals match up with the remaining representations. Instead he implicitly relies on his assumption.

Last edited by JeffM1; May 18th, 2018 at 09:32 AM. May 18th, 2018, 09:33 AM   #5
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Quote:
 Originally Posted by zylo True $\displaystyle \pi$ $\displaystyle \equiv$ 3 "+ " 1415926....../1000000....... That's awkward and requires refinement: my original version was: The real numbers in [0,1] are a subset of the rational numbers in [0,1]. From table. Real numbers in [0,1] = m/1000......... , all m In that case, a real number is N.r which is still countable: N is a subset of the rationals and so is r.
You don't seem to be able to discern between an actual and a limiting value. Archie has tried to pound this difference into your head apparently to no avail.

It's my opinion you just want attention. Have fun. May 18th, 2018, 09:41 AM #6 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,663 Thanks: 2643 Math Focus: Mainly analysis and algebra I recommend closing this thread and stopping Zylo from creating more like it. I can't see any value at all in rehashing this nonsense again. Tags numbers, rationals, real, subset Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post phillip1882 Number Theory 7 September 10th, 2017 02:29 PM santosingh Advanced Statistics 1 May 16th, 2012 04:58 PM elim Probability and Statistics 0 July 21st, 2010 02:03 PM uberbandgeek6 Number Theory 1 February 4th, 2010 06:27 AM boxerdog246 Real Analysis 3 October 6th, 2008 10:35 AM

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