Diagnol Argument #2

**ThePerfectHacker** · February 3rd 2006, 10:34 AM

If you do not like Cantor's Diagnol Argument, possibly because the same real number might not be unique, or the fact that rationals have repeating decimals. Then consider, a another variation of the Diagnol Argument.
Define $\mathbb{I}=\{0<x<1|x\in\mathbb{R}, x \not \in \mathbb{Q}\}$ . Thus this set is the set of all irrationals. Next, by the theory of continued fractions, every irrational number can be expressed as a unique infinite continued fraction. By that you assume that the irrationals are countable thus you can list them for example
$[0;2,4,5,1,...$
$[0;8,2,1,1,...$
$[0;5,2,7,1,...$
$[0;1,2,3,4,...$
.....
Now employ the diagnol argument and you demonstrated that $\mathbb{I}$ is uncountable.
Switch from number theory to set theory:
we have that,
$|\mathbb{I}|>\aleph_0\mbox{ and } |\mathbb{Q}|=\aleph_0$ Thus,
$\mathbb{I}\cup\mathbb{Q}>\aleph_0$ .
Thus,
$\{0<x<1|x\in\mathbb{R}\}$ is uncountable.
Q.E.D.

**CaptainBlack** · February 6th 2006, 12:54 AM

Originally Posted by ThePerfectHacker

If you do not like Cantor's Diagnol Argument, possibly because the same real number might not be unique, or the fact that rationals have repeating decimals. Then consider, a another variation of the Diagnol Argument.

Slight problem here, If you object to the diagonal argument because of
"the fact that rationals have repeating decimals", don't you still have that
problem here, as at least the real quadratic irrational have periodic
continued fraction expansions?

RonL

**ThePerfectHacker** · February 6th 2006, 02:13 PM

Originally Posted by CaptainBlack

Slight problem here, If you object to the diagonal argument because of
"the fact that rationals have repeating decimals", don't you still have that
problem here, as at least the real quadratic irrational have periodic
continued fraction expansions?

RonL

I noticed that too but I paid no attention of that

Perhaps,
the sets of quadradic solutions is countable!!!

Another question, is the set of cardinal numbers itself countable?

One more, how do we know that there exists such a thing as $\aleph_1$ ? Perhaps, the cardinals have a property of the rationals between any
$\aleph_n<\aleph_m$
There exists one such as,
$\aleph_n<\aleph_p<\aleph_m$ Thus there is no such thing as $\aleph_1$ ? This definitely looks like the countinuum hypothesis.

**CaptainBlack** · February 6th 2006, 02:32 PM

Originally Posted by ThePerfectHacker

I noticed that too but I paid no attention of that

Perhaps,
the sets of quadradic solutions is countable!!!

The set of algebraic numbers is countable so yes

RonL

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