Who is right here? Are the amount of numbers between both 0 to 1 and 0 to 2 the same?

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In summary, the cardinalities of the intervals [0,1] and [0,2] are the same, meaning there are an equal number of numbers in each interval. However, there are different types of infinities, with uncountable infinities being larger than countable infinities. This can be seen in the comparison of the number of real numbers between 0 and 1 and the number of natural numbers, as well as the power set of the reals.
  • #1
tahayassen
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leroyjenkens said:
I used to have a problem with infinity. I kept using it like it was a number.
For example, I couldn't understand that the amount of numbers between both 0 and 1 and 0 and 2 were both the same.

tahayassen said:
Aren't some infinities larger than other infinities?

Jack21222 said:
In a way, yes, that's what Cantor demonstrated in the late 1800s.

If you took the number of numbers in between 0-1 and divided it by the number of numbers between 0-2, you should get 1/2.

Let x be the number of numbers between 0-1. There are an equal number of numbers between 0-1 and between 1-2, so the number of numbers between 0-2 is x + x, or 2x. So you have x/2x, and even if x is infinity, they cancel (they're the same infinity).

I'm sure mathematicians will murder me for doing it that way, since I probably did all kinds of things wrong, but I think that's the general idea.

leroyjenkens said:
The problem is you're using infinity as if it's a number. You added infinity with infinity. That makes no sense if infinity isn't a number.

Jack21222 said:
It makes plenty of sense. For every number in the 0-1 set, there is a corresponding number in the 1-2 set. In my example, x is not necessarily infinity, it's the number of numbers in between 0-1.

The concept of infinities cancelling out, and one infinity being "bigger" than the other, is used ALL THE TIME in calculus when dealing with limits. For example, consider (2^x)/(x!) As x goes to infinity, the top and bottom are both infinity. However, the bottom infinity is "larger" so the limit as it goes to infinity is zero.

EricVT said:
For x equal to infinity, both the numerator and denominator are infinitely large, but their ratio is not zero.

For x approaching infinity -- but still finite -- the numerator and denominator also have finite values and their ratio is close to zero, but not zero.

Taking the limits of functions like this is not the same as dividing infinity by infinity.

This is a discussion we had in another part of the forum and I'm wondering who is correct. The discussion is becoming increasingly confusing and annoyingly (regardless of the posts in between the discussion), no one with a "Science Advisor", "Homework Helper", or "PF Mentor" title is stepping into end the argument, so I would appreciate it if you would end the argument.
 
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  • #2


The cardinalities of the intervals [0 1] and [0 2] are the same. In fact, there are as many numbers in the interval [0 1] as there are real numbers.

For example, consider (2^x)/(x!) As x goes to infinity, the top and bottom are both infinity. However, the bottom infinity is "larger" so the limit as it goes to infinity is zero.

This is a pretty poor explanation of what's happening. Taking limits in calculus is a process; we examine what happens as a quantity increases without bound. The limit of that function as x goes to infinity is zero because x! increases faster as x increases, not because it is a "larger quantity". "Infinity" in this context has a completely different meaning than the "infinity" used to describe the size of sets.
 
  • #3


there is an infinite, but countably infinite amount of rational numbers between 0 and 1. this is also the case for the interval from 0 to 2.

but there is an uncountably infinite amount of real numbers between 0 and 1. this is also the case for the interval from 0 to 2.
 
  • #4


Jack212222 is wrong. Cantor's theory is nothing like what he describes. He seems to be confusing things with limits.

And yes, the cardinality of [0,1] is exactly the same as the cardinality of [0,2]. So both sets have the same size.

EricVT is wrong too, since he assumes that [itex]\frac{+\infty}{+\infty}[/itex] is defined when it is not. So the ratio doesn't even make sense.
 
  • #5


As for your question:

Aren't some infinities larger than other infinities?

Yes, this is true. But the example of [0,1] and [0,2] is not a good example since both infinities are the same here.

However, we can look at the sets [itex]\mathbb{N}[/itex] and [itex]\mathbb{R}[/itex]. Those are infinite sets, but the latter set is much larger than the former.

See the following FAQ post: https://www.physicsforums.com/showthread.php?t=507003 [Broken] (also check out the sequels whose link is at the bottom of the thread).
 
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  • #6


Aren't some infinities larger than other infinities?
You also have to be careful how you define "larger". Micromass is completely correct about cardinality and that will be appropriate if 'larger' means "has more numbers in the set" as was originally asked. But one can also argue that the interval [0, 2] is twice as long, and so twice as large, as the interval [0, 1]. It depends upon what you are comparing.
 
  • #7


Thanks everyone. The FAQ was very helpful. I also found this helpful:
 
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  • #8


Two sets have the same cardinaility if there exists a 1-1 correspondence between them.
The function f(x) = 2x establishes such a 1-1 correspondence between the rationals in
[0,1] and [0,2]. It also establishes a 1-1 correspondence between the reals in [0,1] and [0,2].
 
  • #9


Some infinities are larger than others. Namely, uncountable infinities are larger than countable infinities and that is all there is to it.

The number of real numbers between 0 and 1 is uncountably infinite. The number of real numbers between 0 and 2 is uncountably infinite. The number of real numbers period is uncountably infinite. These all describe the same cardinality.

The natural numbers are countably infinite, the real numbers are uncountably infinite. Therefore the cardinality of natural numbers is smaller than that of the real numbers.

There is no differentiation between any two uncountably infinite, or two countably infinite sets, that I am aware of, and such an idea doesn't really make sense to me.
 
  • #10


1MileCrash said:
There is no differentiation between any two uncountably infinite, or two countably infinite sets, that I am aware of, and such an idea doesn't really make sense to me.

Sure there is. There are many types of uncountably infinite sets. For example, the set [itex]\mathcal{P}(\mathbb{R})[/itex] (power set of the reals) has a strictly larger cardinality than [itex]\mathbb{R}[/itex]. and [itex]\mathcal{P}(\mathcal{P}(\mathbb{R}))[/itex] is even larger! This process continues indefinitely.
 
  • #11


1MileCrash said:
There is no differentiation between any two uncountably infinite, or two countably infinite sets, that I am aware of, and such an idea doesn't really make sense to me.

Ah, thanks for pointing that out.

Edit: I take my thank you back.
 
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  • #12


amount of numbers
^ This is the crux of the problem.

Talking about the "amount of numbers" in a set or which set is "bigger" (or "smaller" or "equal in size" or the notion of "size" at all) is an imprecise claim.

When you're comparing two sets, there are many equally legitimate ways to do it. You can for example...

Designate a count to each set, where the count is either the number of elements (if that number is finite) or infinity. Under this definition, the reals have the same count as the natural numbers because both are infinite.

Designate a cardinality to each set. We say two sets have the same cardinality if there is a a bijection between them. We order the cardinalities with a relation, saying that A <= B when there is a bijection between A and some subset of B. In this case, card N < card R. However, card [0, 1] = card [0, 2]

The notion of subset, too, gives us a way to compare sets. People take offense to the idea that [0, 1] is in some sense "the same" as [0, 2] because they are conflating two different senses of "same". One sense where [0, 1] and [0, 2] are clearly different is via a subset relation. Since [0, 1] is a subset of [0, 2], we can claim that [0, 1] is "less than" [0, 2] in this sense. However, this notion is different than the notions of cardinality or count above because it is a partial order (the other two are total orders). This means that you can't compare certain pairs of sets this way: [0, 1] has no relation to [2, 3].

There's one last common one that only works for special kinds of classes (including the real numbers), and that is measure. The sets [0, 1] has a certain property that, when written down on a number line, it gives us a certain length: 1 - 0 = 1. Similarly, the set [2, 5] also has a length: 5 - 2 = 3. This leads us to the notion of a measure. However, this is a very restricted notion, as it applies only to sets endowed with a measure space structure. And some sets, such as {0} in R or Q in R, have a measure of 0, despite having an infinite count.

I post this because I feel too many people quickly jump on the idea that cardinality is the most important way to define a "size" of a set. I don't believe this is the case. In every day problem solving, count and the subset relation are just as important (if not more).
 
  • #13


micromass said:
Sure there is. There are many types of uncountably infinite sets. For example, the set [itex]\mathcal{P}(\mathbb{R})[/itex] (power set of the reals) has a strictly larger cardinality than [itex]\mathbb{R}[/itex]. and [itex]\mathcal{P}(\mathcal{P}(\mathbb{R}))[/itex] is even larger! This process continues indefinitely.

Just as a curiosity, what's the connection between the discrete/countable infinite sets (sequences) [itex]\mathcal{P}^{n}(\mathbb{R})[/itex] and [itex] \aleph_{n} [/itex] ?
 
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  • #14
micromass said:
Sure there is. There are many types of uncountably infinite sets. For example, the set [itex]\mathcal{P}(\mathbb{R})[/itex] (power set of the reals) has a strictly larger cardinality than [itex]\mathbb{R}[/itex]. and [itex]\mathcal{P}(\mathcal{P}(\mathbb{R}))[/itex] is even larger! This process continues indefinitely.

Ah.. I failed to think of deriving a set from an uncountable set in such a way that the cardinality must be greater. It doesn't make much sense at all to say that a set and it's power set have the same cardinality. That is easy to see.

Thank you.

Is this limited to the idea of power sets? It feels like this could only occur when you "build" a "higher order" uncountable set from a previously uncountable set.

Also- the power set of the naturals would certainly be of higher cardinality than the naturals and thus not at a one-to-one correspondence with the naturals and therefore uncountable, right?

Does that make P(N) have the same cardinality as R, and P(P(N)) have the same cardinality as P(R)?

Or am I oversimplifying the idea?
 
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  • #15


P^n (N) (=P(P(P...P(N)))) for arbitrary n has the same cardinality with N.
 
  • #16
dextercioby said:
P^n (N) (=P(P(P...P(N)))) for arbitrary n has the same cardinality with N.

I don't understand how that could be the case.
 
  • #17


Who's P(N) ? Write an explicit formula using the 3 symbols: ..., { and }. Then calculate its cardinality.
 
  • #18
dextercioby said:
Who's P(N) ? Write an explicit formula using the 3 symbols: ..., { and }. Then calculate its cardinality.

Forgive me if I'm not understanding but,

The number of elements of P(N) is the number of possible subsets of natural numbers that can be formed.

That is surely uncountable, as I can combine any number of any natural numbers I want to form some subset. If I call one subset, the non-proper subset, the entire set of natural numbers, one subset, what's another subset? The subset of 1 million of any natural numbers? 2 million any natural numbers? Considering all singleton sets of each natural number is already a countably infinite number of subsets, and there are uncountably many subsets besides those, it doesn't work in my brain to call P(N) a set of countable cardinality.

EDIT:

I found this

http://www.earlham.edu/~peters/writing/infapp.htm#thm3

Which I guess is a much better and more formal version of what I'm thinking.
 
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  • #19
1MileCrash said:
Forgive me if I'm not understanding but,

The number of elements of P(N) is the number of possible subsets of natural numbers that can be formed.

That is surely uncountable, as I can combine any number of any natural numbers I want to form some subset. If I call one subset, the non-proper subset, the entire set of natural numbers, one subset, what's another subset? The subset of 1 million of any natural numbers? 2 million any natural numbers? Considering all singleton sets of each natural number is already a countably infinite number of subsets, and there are uncountably many subsets besides those, it doesn't work in my brain to call P(N) a set of countable cardinality.

EDIT:

I found this

http://www.earlham.edu/~peters/writing/infapp.htm#thm3

Which I guess is a much better and more formal version of what I'm thinking.

You're right, P(N) is uncountable. An easy way to see this is to identify each element of P(A) with an infinite binary string (easy to show that this is a bijection) and note that the set of all such strings is uncountable.
 
  • #20


Okay, I'm not sure how to construct these infinite binary strings but I do understand why P(N) is of uncountable cardinality. I have a few more questions.

I have been reading about "beth numbers" and this seems to be the name for the "order of infinity."

So here are my questions:

Is the beth number the sole indicator of cardinality of uncountable sets? If so, this would imply that P(N) and R have the exact same cardinality, right?
Is the power set concept the only source of constructing a "higher order" infinity?

Are there any sets that we naturally consider that aren't derived from taking the power set of some other set with a beth number greater than two?

I do realize that this question is pretty much identical to my previous one, but I guess what I'm asking is if there is a set with a beth number greater than or equal to 2 that we could describe in some other way than a power set of some other set?

I'm sorry for kind of derailing the topic but I find this really cool, as a grad, what focuses on this type of thing? Is it just set theory?
 
  • #21


dextercioby said:
Just as a curiosity, what's the connection between the discrete/countable infinite sets (sequences) [itex]\mathcal{P}^{n}(\mathbb{R})[/itex] and [itex] \aleph_{n} [/itex] ?

The only thing we can say for sure is that [itex]|\mathcal{P}^n(\mathbb{R})|\geq \aleph_{n+1}[/itex]. Equality is not true in general. If you assume the GCH (generalized continuum hypothesis) as an axiom, then equality is true.

1MileCrash said:
Ah.. I failed to think of deriving a set from an uncountable set in such a way that the cardinality must be greater. It doesn't make much sense at all to say that a set and it's power set have the same cardinality. That is easy to see.

Thank you.

Is this limited to the idea of power sets? It feels like this could only occur when you "build" a "higher order" uncountable set from a previously uncountable set.

Also- the power set of the naturals would certainly be of higher cardinality than the naturals and thus not at a one-to-one correspondence with the naturals and therefore uncountable, right?

Does that make P(N) have the same cardinality as R, and P(P(N)) have the same cardinality as P(R)?

Or am I oversimplifying the idea?

It is certainly true that [itex]\mathcal{P}(\mathbb{N})[/itex] has the same cardinality as [itex]\mathbb{R}[/itex]. But this requires an explicit proof.
Note, that it might happen that there exists a set A such that [itex]|\mathbb{N}|\leq |A|\leq |\mathbb{R}|[/itex]. So it is certainly not the case that [itex]\mathbb{R}[/itex] is the smallest uncountable set. This is only true under the continuum hypothesis.

dextercioby said:
P^n (N) (=P(P(P...P(N)))) for arbitrary n has the same cardinality with N.

No this is false, they all have strictly larger cardinality as [itex]\mathbb{N}[/itex] as long as n>0.

1MileCrash said:
Okay, I'm not sure how to construct these infinite binary strings but I do understand why P(N) is of uncountable cardinality. I have a few more questions.

I have been reading about "beth numbers" and this seems to be the name for the "order of infinity."

The notion you want are actually aleph numbers and not beth numbers. Beth numbers might miss some orders of infinity. Only under the GCH, it is true that the beth numbers capture all orders of infinity.

So here are my questions:

Is the beth number the sole indicator of cardinality of uncountable sets? If so, this would imply that P(N) and R have the exact same cardinality, right?

Again: it is the aleph numbers that indicates the cardinality of uncountable sets. The beth numbers are intimately related to power sets.
Incidentally, it is true that [itex]|\mathbb{R}|=|\mathbb{P}(\mathbb{N})|[/itex]. But this does not follow from beth numbers.

Is the power set concept the only source of constructing a "higher order" infinity?

The ZFC axioms (axioms for set theory) give some ways to construct new sets. But it is indeed only the power set axiom that allows you to construct new orders of infinity. Note that this does not mean that the power sets are the only orders of infinity.

For example, [itex]\mathcal{P}(\mathbb{N})[/itex] are a higher order of infinity than [itex]\mathbb{N}[/itex]. But there might be a subset A of [itex]\mathcal{P}(\mathbb{N}[/itex] such that
[tex]|\mathbb{N}|<|A|<|\mathcal{P}(\mathbb{N})|[/tex]

So the power set might immediately give us many new orders of infinity instead of just one. Only under the GCH, we get one new order of infinity.

Are there any sets that we naturally consider that aren't derived from taking the power set of some other set with a beth number greater than two?

This question cannot be answered in ZFC framework and requires a new axiom. The GCH axiom says that all orders of infinity correspond to power sets. However, it is also consistent with ZFC that there are sets with a beth number greater than two that aren't derived from taking the power set (however, they will be "subsets" of some power set though)

I'm sorry for kind of derailing the topic but I find this really cool, as a grad, what focuses on this type of thing? Is it just set theory?

You need to study set theory for this. A very good book is "Introduction to Set theory" by Hrbacek and Jech.
 
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  • #22


Thank you

Unfortunately I just searched my university's course offerings and the only mention of set theory is for classes I've already taken, but were introductory and only "applied" basic set theory to proofs and such. Looks like this will be an independent study.
 
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  • #23


micromass said:
Only under the GCH, it is true that the beth numbers capture all orders of infinity.

Err, IIRC, GCH is only powerful enough to show weak limit cardinals are equivalent to strong limit cardinals. That doesn't imply ZF+GCH doesn't have inaccessible cardinals at all. Do you have a source?
 
  • #24


pwsnafu said:
Err, IIRC, GCH is only powerful enough to show weak limit cardinals are equivalent to strong limit cardinals. That doesn't imply ZF+GCH doesn't have inaccessible cardinals at all. Do you have a source?

Even innaccessible cardinals must have an aleph-number. So under GCH, all cardinals have a beth-number. Do you think these statements are wrong?
 
  • #25


micromass said:
Even innaccessible cardinals must have an aleph-number. So under GCH, all cardinals have a beth-number. Do you think these statements are wrong?

No you are right. I was only thinking about power sets and forgetting the supremum operation.
 
  • #26


micromass said:
[...]
It is certainly true that [itex]\mathcal{P}(\mathbb{N})[/itex] has the same cardinality as [itex]\mathbb{R}[/itex]. But this requires an explicit proof.[...]

Do you have a reference for this proof ?

Where does my naïve proof fail ?

P(N) - the power set of N = the set of all subsets of N.

[itex] \mathbb{N} \subset \mathcal{P}(\mathbb{N}) [/itex] and any proper subset of N is countable and so is their number

So P(N) is the disjoint reunion of N and the set of all proper subsets of N, hence countable, hence has the same cardinality with N.
 
  • #27


dextercioby said:
Do you have a reference for this proof ?

Where does my naïve proof fail ?

P(N) - the power set of N = the set of all subsets of N.

[itex] \mathbb{N} \subset \mathcal{P}(\mathbb{N}) [/itex] and any proper subset of N is countable and so is their number

So P(N) is the disjoint reunion of N and the set of all proper subsets of N, hence countable, hence has the same cardinality with N.

Identify each subset of N with an infinite binary string, where the n'th digit is a 1 if n is in the subset, and 0 otherwise. This is easily shown to be a bijection between P(N) and the set of all infinite binary strings. Cantor's diagonalization argument shows that the set of all such strings is uncountable, therefore P(N) is uncountable.

The union of countably many countable sets is countable, but that's no guarantee that the set of all such sets is countable.
 
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  • #28


dextercioby said:
Do you have a reference for this proof ?

Where does my naïve proof fail ?

P(N) - the power set of N = the set of all subsets of N.

[itex] \mathbb{N} \subset \mathcal{P}(\mathbb{N}) [/itex] and any proper subset of N is countable and so is their number

So P(N) is the disjoint reunion of N and the set of all proper subsets of N, hence countable, hence has the same cardinality with N.

First of all, what do you mean with a "proper" subset in this case?? Do you mean subsets of [itex]\mathbb{N}[/itex] that are not singletons?? In that case, there are uncountably many proper subsets of [itex]\mathbb{N}[/itex].

Here is a cool proof that [itex]\mathcal{P}(\mathbb{N})[/itex] is uncountable:

Assume we have a bijection
[tex]f:\mathbb{N}\rightarrow \mathcal{P}(\mathbb{N})[/tex]
Define the following subset of [itex]\mathbb{N}[/itex]:
[tex]A=\{x\in \mathbb{N}~\vert~x\notin f(x)\}[/tex]
Since f is surjective, there exists an [itex]a\in \mathbb{N}[/itex] such that [itex]f(a)=A[/itex].
There are two possibilities: either [itex]a\in A[/itex] or [itex]a\notin A[/itex].
But if [itex]a\in A=f(a)[/itex], then [itex]a\notin f(a)[/itex] by definition of [itex]A[/itex].
And if [itex]a\notin A=f(a)[/itex], then by definition of [itex]A[/itex], we would have [itex]a\in A[/itex].
So both possibilities lead to a contradiction. So the original statement that a bijection exists, must be a contradiction.

The proof that [itex]\mathcal{P}(\mathbb{N})[/itex] has the same cardinality as [itex]\mathbb{R}[/itex] is a bit harder. But here is a way to construct a bijection between [itex]\mathcal{P}(\mathbb{N})[/itex] and [0,1)(I'll leave the details to the reader and I'll try to find a reference).

Take a subset [itex]A\subseteq \mathbb{N}[/itex]. Define [itex]x_n[/itex] as 0 if [itex]n\notin A[/itex] and define it as 1 if [itex]n\in A[/itex]. Then let
[tex]0.x_1x_2x_3x_4...[/tex]
be a number in [0,1] written in binary notation.
For example, the set [itex]\{1,4,6\}[/itex] corresponds to [itex]0.1001010000000...[/itex]. The even numbers correspond to [itex]0.010101010101...[/itex]. And the primes correspond to [itex]0.0110101000101...[/itex].
This is clearly a surjection. A technical detail is that this does not define an injection. The reason is that one number can have multiple binary representations. For example: [itex]0.01=0.00111111111111111...[/itex]. So the sets [itex]\{2\}[/itex] and [itex]\{3,4,5,6,7,...\}[/itex] are sent to the same number in [0,1]. This issue can be solved by remarking that such issue only arises with finite sets in [itex]\mathbb{N}[/itex] and that the collection of finite subsets of [itex]\mathbb{N}[/itex] is countable.
 
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  • #29
micromass said:
First of all, what do you mean with a "proper" subset in this case?? Do you mean subsets of [itex]\mathbb{N}[/itex] that are not singletons??
From both his context and the standard definition, he is talking about any subset of N that is not N itself.
 
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  • #30
dextercioby said:
Do you have a reference for this proof ?

Where does my naïve proof fail ?

P(N) - the power set of N = the set of all subsets of N.

[itex] \mathbb{N} \subset \mathcal{P}(\mathbb{N}) [/itex] and any proper subset of N is countable and so is their number

So P(N) is the disjoint reunion of N and the set of all proper subsets of N, hence countable, hence has the same cardinality with N.

Your proof fails in that, you are right that any proper subset of N is countable, but that is a nonissue because we are not concerned with the number of elements of the subsets, we are concerned with the number of subsets.

You do say that "they are countable and so is their number" but the latter part seems to be an assumption.

There are countably infinite different sizes a subset of N can be, and countably infinite choices for elements in that subset of a specified size. Therefore you could never "count" to the next "set size" choice because you are already at the "limit of countability" when considering how many choices of elements their are in that specific choice of set size.

When attempting to count you would be "stuck" on a set size of 1 (or w/e you chose) due to the countably infinite choices of the elements. This is exactly the same idea as being "stuck" when trying to count the real numbers, you are "stuck" because there are countably infinite decimal places to consider. So as a whole, the possibilities cannot be counted in either case.

But that's my weird way.
 
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  • #31


I thank you all for your kind replies.
 

1. What is the difference between the amount of numbers between 0 to 1 and 0 to 2?

The difference between the amount of numbers between 0 to 1 and 0 to 2 is that the range of numbers between 0 to 1 is smaller than the range of numbers between 0 to 2. This is because there are more numbers between 0 and 2, including all the numbers between 1 and 2, which are not included in the range of numbers between 0 and 1.

2. Is the amount of numbers between 0 to 1 and 0 to 2 the same?

No, the amount of numbers between 0 to 1 and 0 to 2 is not the same. As mentioned before, the range of numbers between 0 to 2 is larger than the range of numbers between 0 to 1. This means that there are more numbers between 0 and 2 compared to 0 and 1.

3. How do you determine the amount of numbers between 0 to 1 and 0 to 2?

The amount of numbers between 0 to 1 and 0 to 2 can be determined by counting the numbers within each range. For example, between 0 and 1, there are an infinite number of numbers, while between 0 and 2, there are twice as many numbers as between 0 and 1. Therefore, the amount of numbers between 0 to 2 is greater than the amount between 0 to 1.

4. Why is it important to understand the amount of numbers between 0 to 1 and 0 to 2?

Understanding the amount of numbers between 0 to 1 and 0 to 2 is important for various mathematical and scientific calculations. It helps in determining the probability of certain events, calculating averages and percentages, and understanding the concept of infinity. It is also important in fields such as physics, where precise measurements and calculations are necessary.

5. Can the amount of numbers between 0 to 1 and 0 to 2 change?

No, the amount of numbers between 0 to 1 and 0 to 2 cannot change. These ranges are fixed and will always have the same amount of numbers between them. However, the numbers within these ranges can change, as they are infinite and can be added or subtracted. But the overall amount of numbers between these ranges will remain the same.

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