Real Numbers are a Subset of the Rationals

Mar 2015
1,720
126
New Jersey
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.
 

romsek

Math Team
Sep 2015
2,958
1,672
USA
$\pi \in \mathbb{R}$

$\pi \not \in \mathbb{Q}$

$\therefore \mathbb{R} \not \subset \mathbb{Q}$
 
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Mar 2015
1,720
126
New Jersey
$\pi \in \mathbb{R}$
$\pi \not \in \mathbb{Q}$
$\therefore \mathbb{R} \not \subset \mathbb{Q}$
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 2016
1,310
551
USA
$\pi \in \mathbb{R}$

$\pi \not \in \mathbb{Q}$

$\therefore \mathbb{R} \not \subset \mathbb{Q}$
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:

romsek

Math Team
Sep 2015
2,958
1,672
USA
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.
 

v8archie

Math Team
Dec 2013
7,710
2,679
Colombia
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.