Counting problem

Originally posted by Acolyte
...
Believe it or not, this shows that there are exactly as many rational numbers as natural
numbers (though there are easier ways of showing this.)...

I might have misunderstood this but are you saying
that the set of natural numbers ( that is 0,1,2,3,4,5...? ) and the set of rational numbers (expressed a/b where a and b are natural numbers ? ) have the same size?

This time I feel confident (as I did with 0.99..!=1 :-) ).
This must be plain wrong based on,
the set of a/1 where a=[any natural number] is strictly the same size as the set of naturals. Therefore any rational number that is not a natural number makes the set larger..
Or????

The natural numbers (0,1,2,3,4,...)
The integers (...,-3,-2,-1,0,1,2,3,...)
The rationals (all numbers of the form a/b with a and b integers)
The reals (basically everything without i)
The complex numbers (everything)

If you look on page two, to my post there, it will show you how exactly the set of rationals is as big as the natural numbers.

For every natural number there is a rational combined with it, and vise versa. Thus they are just as big.

Originally posted by TheMaster37
The complex numbers (everything)

What about quaternions? 😛

Originally posted by piderman
What about quaternions? 😛

Quarternions are equivalent to Complex Numbers, so basically; yeah.

Even so, smartypants, the class of object we were discussing is the class of numbers. Wich you, I, and everyone else in this thread knows. I even told you what i meant, so this post of yours has absolutely no use.

Originally posted by royalchicken
How about this:

Let x(0) = 1, and let x(2n+1) = x(n) and x(2n+2) = x(n) + x(n+1). Then the nth positive rational number is x(n-1)/x(n).

It certainly looks good, but:

1. I don't see 0
2. I don't see negatives.

These aren't hard to implement:

x(0) = 0
x(0+1) = 1
x(4n+1+1) = x(n)
x(4n+2+1) = -x(4n+1)
x(4n+3+1) = x(n) + x(n+1)
x(4n+4+1) = -x(4n+3)

Haven't checked this, but the idea is clear, I hope.

Originally posted by royalchicken
How about this:

Let x(0) = 1, and let x(2n+1) = x(n) and x(2n+2) = x(n) + x(n+1). Then the nth positive rational number is x(n-1)/x(n).

That's cool! do you have a proof that this includes every rational number?

Originally posted by iamatiger
That's cool! do you have a proof that this includes every rational number?

Indeed. The basic ideas are scattered about this thread, and while I haven't looked carefully, piderman's post seems to improve it significantly.

Originally posted by Acolyte
I found a bijection...

You wouldn't happen to have that lying around, would you?

Originally posted by TheMaster37
The natural numbers (0,1,2,3,4,...)
The integers (...,-3,-2,-1,0,1,2,3,...)
The rationals (all numbers of the form a/b with a and b integers)
The reals (basically everything without i)
The complex numbers (everything)

If you look on page two, to my post there, it will show you how exactly the set of rationals is as big as the natural numbers.

F ...[text shortened]... natural number there is a rational combined with it, and vise versa. Thus they are just as big.

I'm not very knowledgeble in this :-)

But isn't there diffrent size infinites? Just because you can number the rationals it doesn't follow that there are as many rationals as there are naturals. Thats just messing with the measuring. Like saying all circles are the same size since you could plot an infinite number of points on their edge... ( a very nice metafor of how I see this. Think about it :-) )

Please explain what is wrong with the following statements:

How can the set rational numbers of the form a/1 (set A) be smaller than the set of natural numbers?

Further, the set i/1 where i is a negative integer (set B) also has the same size (?) (minus one. 0/1).

Since the intersection A B is empty, the union of the two sets should be double the size(?).

The set of rational numbers contains this union (?).
There are rationals not contained in A or B(?).

Originally posted by royalchicken
You wouldn't happen to have that lying around, would you?

I don't, but I can try to reconstruct it. Going from N to Q:

1 -> 0

For other numbers:
n even -> n/2
n odd -> -(n-1)/2
and then factorise the result into primes.
Now we change the power x of each prime in this factorisation as follows:
x even -> x/2
x odd -> -(x+1)/2

An example: 97 -> -48 -> -2^4*3 -> -2^2*3^-1 = -4/3

Does this work?

Originally posted by chasparos
I'm not very knowledgeble in this :-)

But isn't there diffrent size infinites? Just because you can number the rationals it doesn't follow that there are as many rationals as there are naturals. Thats just messing with the measuring. Like saying all circles are the same size since you could plot an infinite number of points on their edge... ( a very ...[text shortened]... f rational numbers contains this union (?).
There are rationals not contained in A or B(?).

"Just because you can number the rationals it doesn't follow that there are as many rationals as there are naturals."

Er, doch. It quite explicitly shows that there are as many rationals as naturals. If I have some apples and label them 1,2,3,4,5 (with each apple getting exactly one label), does it show that I have five apples?

"How can the set rational numbers of the form a/1 (set A) be smaller than the set of natural numbers?"

It isn't.

"Further, the set i/1 where i is a negative integer (set B) also has the same size"

Yep.

"Since the intersection A B is empty, the union of the two sets should be double the size(?)."

That's right. But 2*|N| = |N|: consider the number of even and odd numbers, for example.

"There are rationals not contained in A or B(?)."

Of course. Nevertheless, it is possible to count all the rationals, as should be illustrated in this thread.

Originally posted by Acolyte
I don't, but I can try to reconstruct it. Going from N to Q:

1 -> 0

For other numbers:
n even -> n/2
n odd -> -(n-1)/2
and then factorise the result into primes.
Now we change the power x of each prime in this factorisation as follows:
x even -> x/2
x odd -> -(x+1)/2

An example: 97 -> -48 -> -2^4*3 -> -2^2*3^-1 = -4/3

Does this work?

Wouldn't this always give us a fraction with an odd denominator - missing things like 1/64 and 3/2 out?

As acolyte said, two sets have the same size if you can combine one number of the first set to a number of the second set. If you don't leave out any number of one of the two sets you have what is called a 'bijection'. If such a bijection exists between two sets, they have the same size.

As you saw, i created a bijection between the natural numbers and the rationals, and thus they have the same size.

A bijection between the natural numbers and integers is this:

0 0
1 1
2 -1
3 2
4 -2
5 3

and so on.

Thus there are as many integers as naturals 🙂

There are two kinds of infinity;

countable and overcountable. Countable is that you can number all the elements of the set (so there are countable infinite integers and rationals)

Overcountable is if you can't number the elements. The set of reals is overcountable.

Is assure you, this is true 🙂

Originally posted by iamatiger
Wouldn't this always give us a fraction with an odd denominator - missing things like 1/64 and 3/2 out?

Nope: for example, 3/2 = 3*2^-1 <- 3^2*2 <- 3^2*2^2 = 36

Originally posted by Acolyte
I don't, but I can try to reconstruct it. Going from N to Q:

1 -> 0

For other numbers:
n even -> n/2
n odd -> -(n-1)/2
and then factorise the result into primes.
Now we change the power x of each prime in this factorisation as follows:
x even -> x/2
x odd -> -(x+1)/2

An example: 97 -> -48 -> -2^4*3 -> -2^2*3^-1 = -4/3

Does this work?

Yes 😀!

I was going to write a proof, first that each natural number yields exactly one rational (trivial) and then that the reverse is true (a little harder).

It's more interesting to look at each little bit and see how it fits together, because your function is remarkably constructive--intelligent design I'd say.