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 May 26th, 2017, 07:34 PM #21 Member   Joined: May 2017 From: USA Posts: 31 Thanks: 0 Talking about real numbers and natural numbers. To me, as a nonscientist, there is only one real number and its a 1. Any other numbers, that you can imagine are the natural ( maid up numbers). Like: 2 - two ones 10 - ten ones 500 - five hundred ones 10000 - ten thousand ones As I mentioned earlier, any activities that are happened in the interval 0 to 1 automatically happening in any intervals you can imagine as: 1 to 2 2 to 3 117 to 118 10001 to 10002 and so on.
May 26th, 2017, 07:42 PM   #22
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 Originally Posted by Microlab Talking about real numbers and natural numbers. To me, as a nonscientist, there is only one real number and its a 1. Any other numbers, that you can imagine are the natural ( maid up numbers). Like: 2 - two ones 10 - ten ones 500 - five hundred ones 10000 - ten thousand ones As I mentioned earlier, any activities that are happened in the interval 0 to 1 automatically happening in any intervals you can imagine as: 1 to 2 2 to 3 117 to 118 10001 to 10002 and so on.
What about numbers less than 1?

Edit: If you work in binary notation, I suppose you could express all the real numbers as strings of 0's and/or 1's. I'm not sure what your point is though. Can you help elaborate?

Last edited by AplanisTophet; May 26th, 2017 at 07:45 PM.

May 26th, 2017, 07:50 PM   #23
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 Originally Posted by Microlab Thanks and I still believe that they are real. There is no abstraction. If I have one full apple and I know I can make 0.5 of it, that means 0.1234512345... part of apple is real too, because its a part of one full apple, even knowing that I will not be able to see or touch it. From this view of point, one full apple is a combination of infinite infinity particles of it.
But how can you know? How are you going to measure that it is, physically, exactly half?

May 26th, 2017, 08:01 PM   #24
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 Originally Posted by Microlab As I mentioned earlier, any activities that are happened in the interval 0 to 1 automatically happening in any intervals you can imagine as: 1 to 2 2 to 3 117 to 118 10001 to 10002 and so on.
There are an infinite number of reals on the interval [0, 0.5], so at first glance maybe it's plausible to suspect we can add a certain number 'b' to go from one to the next: 0 + b + b + b + b + b + ... = 0.5

This fails what is known as the Archimedean property though, and the reals are supposed to have this property.

Personally, I am very perplexed by this fact. Maschke might appreciate my attempt at a little humor here, but I like numbers that fail to exhibit the Archimedean property, meaning numbers such as x or y where:

x + x + x + x + ... = 1

y + y + y + y + ... = $\pi \cong$ 3.14159265359...

Lol though, maybe not, as Maschke can be a stubborn ol' goat at times.

 May 26th, 2017, 08:04 PM #25 Member   Joined: May 2017 From: USA Posts: 31 Thanks: 0 This is what I have being working on all this time! How many numbers are there, that are less than 1 and more than 0? Thanks! Last edited by Microlab; May 26th, 2017 at 08:14 PM.
 May 26th, 2017, 08:24 PM #26 Member   Joined: May 2017 From: USA Posts: 31 Thanks: 0 You was right Joppy! I guess, I am high. But I still can't stop thinking. Thanks again!
May 26th, 2017, 09:29 PM   #27
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 Originally Posted by Microlab This is what I have being working on all this time! How many numbers are there, that are less than 1 and more than 0? Thanks!
The cardinality (or 'size') of the set of reals on the interval (0, 1) is, again, $2^{\aleph_0}$. This is pronounced "two to the aleph null."

Consider the natural numbers $\mathbb{N}$ = {1, 2, 3, 4, 5, ... }. The cardinality of the natural numbers is $\aleph_0$, where $\aleph_0 < 2^{\aleph_0}$.

The powerset of $\mathbb{N}$, denoted $P( \,\mathbb{N}) \,$, is the set containing all possible subsets of $\mathbb{N}$:

$P( \,\mathbb{N}) \,$ = { $\emptyset$, {1}, {2}, {3}, ..., {1,2}, {1,3}, {1,4}, ..., {evens}, {odds}, $\mathbb{N}$ }

It turns out that the cardinality of $P( \,\mathbb{N}) \,$ is equal to the cardinality of the set of reals on the interval (0, 1), but what does that mean?

Again, if we can have a bijection from one set onto another (that is, a function that results in a 1-to-1 correspondence between the elements of two sets), then we say those sets have equal cardinality.

No set can have cardinality equal to its powerset. Try this first with a finite set:

Q = {1, 2}
$P$(Q) = { $\emptyset$, {1}, {2}, {1, 2} } (all subsets of {1,2})

Try to define a function that maps each element of Q 1-to-1 onto $P$(Q), and it should be obvious that you cannot do it.

Let function $f$ be bijective from Q onto $P$(Q). Say f(1) = {1,2} and f(2) = {1}. Obviously, we missed two elements of $P$(Q): $\emptyset$ and {2}. Let's do something 'sneaky' though. Let me define a set $A$ as follows:

Let $A$ = { n in Q : n is not in f(n) } = {2} in $P$(Q) (ie, this is because 1 was in f(1) = {1, 2} but 2 was not in f(2) = {1}). Notice that our function $f$ did not map an element of Q to the set $A$ = {2} in $P$(Q).

No matter what set you have, this sneaky trick can be used to show that, if you try to biject a set onto its powerset, that you cannot do it.

For example, let a function $g$ be bijective (or surjective) from $\mathbb{N}$ onto $P( \,\mathbb{N}) \,$. I can define my 'sneaky' set $A$:

Let $A$ = { n in $\mathbb{N}$ : n is not in g(n) }

We know that $A$ is in $P( \,\mathbb{N}) \,$ because $A$ is either a set of natural numbers or $A$ = $\emptyset$, therefore, if function $g$ is bijective, there must be an element of $\mathbb{N}$, call it $k$, such that g($k$) = $A$. But, we also know that it's not possible for an element $k$ to exist such that g($k$) = $A$. Why? Just as with the finite set example above, we get rubbish:

If $k$ is in $A$, then $k$ is not in g($k$) = $A$ (a contradiction)
If $k$ is not in $A$, then $k$ is in g($k$) = $A$ (also a contradiction)

Thus, there is no bijective (or surjective) function $g$ from $\mathbb{N}$ onto its powerset $P( \,\mathbb{N}) \,$ because there is no element $k$ in $\mathbb{N}$ such that g($k$) = $A$ in $P$($\mathbb{N})$. We now know that the cardinality of $P( \,\mathbb{N}) \,$ is greater than the cardinality of $\mathbb{N}$.

Wow, right!? So, we have different 'sizes' of infinity, or more specifically, infinite sets with differing cardinalities.

Now, to answer your question, it is possible to have a bijection between the reals on the interval (0, 1) and $P( \,\mathbb{N}) \,$, so the cardinality of the set of reals on the interval (0, 1) is equal to the cardinality of $P( \,\mathbb{N}) \,$ = $2^{\aleph_0}$.

So now, what about the cardinality of the powerset of the powerset of $\mathbb{N}$, meaning $P$($P$($\mathbb{N}$))? Next, $P$($P$($P$($\mathbb{N}$)))? So, we have an infinite number of different cardinalities. Cool, right?!

Last edited by AplanisTophet; May 26th, 2017 at 09:37 PM.

 May 27th, 2017, 06:56 AM #28 Member   Joined: May 2017 From: USA Posts: 31 Thanks: 0 Absolutely gorgeous! There is only one aspect. If I have way smaller interval compare to larger interval, like: ( 0 to. 0.001 ) compare to ( 0 to 1 ), I am running to " spooky action ". The set of infinite numbers, in the smaller interval equals to set of infinite numbers in the larger interval. The infinity is constant no matter where you look for it. Thanks Last edited by Microlab; May 27th, 2017 at 06:58 AM.
 May 27th, 2017, 07:28 AM #29 Member   Joined: May 2017 From: USA Posts: 31 Thanks: 0 So, how big the universe is? Is it to large? Big? Or small enough that some one exploring it through the microscope?
 May 27th, 2017, 08:30 AM #30 Member   Joined: May 2017 From: USA Posts: 31 Thanks: 0 Thank you guys, thanks for all your suggestions. You guys are the open minds! I am out.

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