Real Numbers: 10^{\aleph} Possibilities?

[cragar]

If the set of natural numbers is $\aleph$
and when we write a real number we have 10 choices for each position 0-9
so can we say that there are $10^{\aleph}$ real numbers ?

 

[micromass]

Hi cragar!

If the set of natural numbers is $\aleph$

You probably mean $\aleph_0$, right?

and when we write a real number we have 10 choices for each position 0-9
so can we say that there are $10^{\aleph}$ real numbers ?

Yes, that is correct, the real numbers have cardinality $10^{\aleph_0}$. Also note that

$$2^{\aleph_0}=10^{\aleph_0}=\aleph_0^{\aleph_0}$$

But, I also must give you a warning. Saying that you have "10 choices for each position 0-9" is not exactly true, there are technical details. For example 1.00000000... and 0.9999999... are the same numbers, so some choice yield the same number. Also, choice like ...9999999.999999... are not allowed: we must only have a finite number of 1-9 in front of the dot.

These technical matters can be fixed however.

 

[cragar]

Hi cragar!
You probably mean $\aleph_0$, right?
Yes, that is correct, the real numbers have cardinality $10^{\aleph_0}$. Also note that

$$2^{\aleph_0}=10^{\aleph_0}=\aleph_0^{\aleph_0}$$

how is this true $$2^{\aleph_0}=10^{\aleph_0}=\aleph_0^{\aleph_0}$$

 

[micromass]

how is this true $$2^{\aleph_0}=10^{\aleph_0}=\aleph_0^{\aleph_0}$$

Well, to give an intuitive explanation. You showed that the real numbers have cardinality $10^{\aleph_0}$, but you used decimal representation here. We can also use binary representation. In that way, you have numbers of the form 111.0101101 for example. So you have to choose 0 or 1 a countable number of times. So by the same reasoning, the real numbers have cardinality $2^{\aleph_0}$.
When using hexadecimal, you'll obtain $16^{\aleph_0}$ as cardinality of the reals. So

$$2^{\aleph_0}=3^{\aleph_0}=...=10^{\aleph_0}=...$$

 

[cragar]

I seen the proof where the set has 2^n subsets . like for example if i have a sub set {3,2} this would mean I would put a one in the 3rd position and a 2 in the second position and zeros in the rest. but i thought this was a proof where we couldn't repeat numbers. We didn't start with a multiset. So are you saying the reals are all of the subsets of the naturals.

 

[micromass]

I seen the proof where the set has 2^n subsets . like for example if i have a sub set {3,2} this would mean I would put a one in the 3rd position and a 2 in the second position and zeros in the rest. but i thought this was a proof where we couldn't repeat numbers. We didn't start with a multiset. So are you saying the reals are all of the subsets of the naturals.

Well, the reals aren't the set of all subsets of the naturals, but they certainly have as much elements!

 

[cragar]

when you use binary for your list count, what do you mean by your 0 or 1 .

 

[micromass]

when you use binary for your list count, what do you mean by your 0 or 1 .

Just use the binary system: for example 10=2, 11=3, 100=4, 0.1=1/2, etc.