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Proof there are as many numbers between 0 and 1 as 1 and infinity?


Realintruder
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If a/b is greater than 1 than b/a must greater than 0 but less than 1.<br><br>

 

For every number greater than 1 there is a multiplicative inverse greater than 0 but less than 1. <br><br>

 

Examples 3/4-4/3, 7/8-8/7, 9/10-10/9, 1/3-3/1, 3/5-5/3 <br><br>

 

But is there a proof for this?<br><br>

Edited by Realintruder
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The reasoning you just used is essentially the proof. By pairing each real number in [math](1, \infty)[/math] with exactly one number in [math](0,1)[/math] (taking into account, of course, the equivalence between certain numbers in the sets, e.g. 6/4 = 3/2), you've constructed a bijection (specifically, the mapping f(x) = 1/x) between the two intervals, which in turn means they have the same cardinality, i.e. the same number of elements.

In fact, when dealing with the real numbers (or even just the rationals), there are as many elements in any non-empty, non-degenerate interval as there are in any other non-empty, non-degenerate interval.

 

Edit: To be clear, pairing each real number in [math](1, \infty)[/math] with exactly one number in [math](0,1)[/math] forms an injection, but it's easy to show the mapping in question is also surjective, thus it is a bijection.

Edited by John
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John

By pairing each real number in 9f96c55e3f4517afc70e5c5be4307d83-1.png with exactly one number in b6dbc33006b907f2db1855810abfce98-1.png (taking into account, of course, the equivalence between certain numbers in the sets, e.g. 6/4 = 3/2), you've constructed a bijection (specifically, the mapping f(x) = 1/x) between the two intervals, which in turn means they have the same cardinality, i.e. the same number of elements.

 

 

 

John Cuthber

That proves that there are at least as many numbers between 0 and 1 as there are between 1 and infinity.

 

I did wonder at the OP's use of fractions since it implies his underlying set is the set of rational numbers, not the reals.

 

Which is intended needs to be made clear before constructing a proof.

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That proves that there are at least as many numbers between 0 and 1 as there are between 1 and infinity.

There are , I think, actually rather more

http://en.wikipedia.org/wiki/Aleph_number

No, the cardinality of (0, 1) is the same as the cardinality of (1, infinity). The fact that there exists a bijection between the two intervals is proof.

 

 

I did wonder at the OP's use of fractions since it implies his underlying set is the set of rational numbers, not the reals.

 

Which is intended needs to be made clear before constructing a proof.

The same reasoning applies with the rationals.

 

Edit: To add more detail, the function in question is f : (0, 1) -> (1, infinity) defined by f(x) = 1/x. The construction of the rationals and the reals ensures that in either set, every non-zero element has a multiplicative inverse. Furthermore, since every x will be less than 1 and positive, every y = 1/x will be greater than 1 and positive. Thus, our function is valid on (0, 1) in either the rationals or the reals, i.e. every x in (0, 1) will be mapped to some y = 1/x in (1, infinity).

 

Consider two distinct elements y1 = 1/x1 and y2 = 1/x2 such that y1 = y2. This means 1/x1 = 1/x2, therefore x1 = x2. Thus the function is injective.

 

Now take any y in (1, infinity). If y = 1/x, then we have x = 1/y. Since every non-zero element of the reals (or rationals) has a multiplicative inverse, this x certainly exists. Furthermore, since y > 1 and the positive reals (or rationals) are closed under multiplication, 1/y must be less than 1 and positive, i.e. 1/y = x must be in (0, 1). Thus the function is surjective.

 

Since the function is injective and surjective, it is bijective by definition. And since the existence of a bijection between two sets implies the two sets are equinumerous, we have that (0, 1) and (1, infinity) are equinumerous.

Edited by John
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It's not clear from the OP whether he is talking about the number of numbers between one and infinity or the number of rationals in that range.

 

There are more numbers (including transendentals) in the range 0 ->1 than there are integers in the range 1-> infinity or rationals in the range 0->1

Edited by John Cuthber
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It's not clear from the OP whether he is talking about the number of numbers between one and infinity or the number of rationals in that range.

 

There are more numbers (including transendentals) in the range 0 ->1 than there are integers in the range 1-> infinity or rationals in the range 0->1

 

 

Yes I think we are all agreed on this.

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Dunno, the wording of the first line of the OP seems to indicate that, whether he means rationals or reals, he intends for both intervals to be subsets of the same set. But perhaps he'll respond and say otherwise.

 

 

Yes so long as you do not try to make injections from the reals to the rationals.

 

Certainly.

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The reasoning you just used is essentially the proof. By pairing each real number in [math](1, \infty)[/math] with exactly one number in [math](0,1)[/math] (taking into account, of course, the equivalence between certain numbers in the sets, e.g. 6/4 = 3/2), you've constructed a bijection (specifically, the mapping f(x) = 1/x) between the two intervals, which in turn means they have the same cardinality, i.e. the same number of elements.

 

In fact, when dealing with the real numbers (or even just the rationals), there are as many elements in any non-empty, non-degenerate interval as there are in any other non-empty, non-degenerate interval.

 

Edit: To be clear, pairing each real number in [math](1, \infty)[/math] with exactly one number in [math](0,1)[/math] forms an injection, but it's easy to show the mapping in question is also surjective, thus it is a bijection.

 

Thank You for the well written response. I think the question has been answered just fine despite all the needless replies.

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  • 4 years later...
  • 1 year later...

Wouldn’t the answer depend on whether or not 0 and 1 are included in the statement of each number?

 

Say for example they are included, then 0 to 1 will always have two more numbers than 1 to infinity. Those numbers being 0 and 1. 
 

Every number that can be made between 0 and 1 (barring the two themselves) can be made in replication essentially speaking by removing {0.} from the front of it: so 0.1 would become one and so on. Admittedly this is perhaps over simplifying the process but I believe the general concept of it still works.

However, if 0&1 are included then those would become the only numbers 1-infinity cannot replicate, as 1 would be matching the smallest number starting trailing only zeros until ending in 1. Leaving only zero and one left without a match because one to infinity does not end in a finite number.

 

In the case where they aren’t included in either 0-1 still has more numbers but only by one, that being what I described as one’s normal match up above.

 

[After thought]: for numbers that 0.0 then something i.e. 0.01 it’s probably best to consider the 1- infinity equivalent of natural numbers as putting the zero/zeros at the end of the equation after removing the decimal. So 0.01 would be 10, 0.0054 would be 5400 and so on.


Sorry for the poor wording of it. Any idea of how to word it better or issues with the concept?
 

 

Though admittedly, the layout of what’s included in each part makes to most sense to me set up as follows:

[0,1) and [1, infinity)

Even still [0,1) would still have one more number than [1, infinity) that being 0. The only way I see them being equal is the set up as follows: (0,1) and [1, infinity)

Edited by Wyatt
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1 hour ago, Wyatt said:

Say for example they are included, then 0 to 1 will always have two more numbers than 1 to infinity. Those numbers being 0 and 1. 

1/x is a bijection from (0,1) to (1, infinity). 

There's a bijection from [0,1) to (0,1). How do we do this? The rationals in (0,1) are countable so they can be enumerated as [math]r_1, r_2, r_3, \dots[/math]. 

Now you define a bijection [math]f: [0,1) \to (0,1)[/math] as follows:

If [math]x[/math] is irrational, then [math]f(x) = x[/math]. That is, [math]f[/math] leaves all the irrationals unchanged.

[math]f(0) = r_1[/math]. 

And [math]f(r_n) = r_{n+1}[/math]. In other words [math]f(r_1) = r_2, f(r_2) = r_3[/math], and so forth.

Now you can see that every rational in (0,1) is in the range of [math]f[/math], and [math]f[/math] is reversible; that is, it's both injective and surjective. So it's a bijection.

Graphically, we've mapped the rationals like this:

[math]0 \to r_1 \to r_2 \to r_3 \to r_4 \to \dots[/math], in effect sliding 0 into the enumerated rationals in (0,1), which do NOT include 0. This trick amounts to embedding Hilbert's hotel in the unit interval. Initially the hotel is full, with guest 1 in room 1, guest 2 in room 2, etc. Now an "extra guest" named 0 shows up. We move guest 1 to room 2, guest 2 to room 3, and so forth, leaving room 1 empty. Then we move the extra guest 0 into room 1. In effect we've made room for the extra point at 0 by sliding everyone else up one room.

To biject [0,1] to (0,1) we just do the same trick twice. Map 0 to [math]r_1[/math], map 1 to [math]r_2[/math], map [math]r_n[/math] to [math]r_{n+2}[/math], and leave the irrationals alone.

Then to map [0,1) or [0,1] to (1, infinity), just compose the bijections. First map [0,1) or [0,1] to (0,1), then map (0,1) to (1, infinity).

There are some other clever solutions in these Stackexchange threads.

https://math.stackexchange.com/questions/28568/bijection-between-an-open-and-a-closed-interval

https://math.stackexchange.com/questions/213391/how-to-construct-a-bijection-from-0-1-to-0-1

The checked answer by Asaf Karagila in the first link gives a completely explicit solution by sliding along 1/2, 1/4, 1/8, etc., without the need for a mysterious enumeration.

That is, to map from [0,1) to (0,1) you map 0 to 1/2, 1/2 to 1/4, etc., and leave everything else alone. To map [0,1] to (0,1) you map 0 to 1/2, 1 to 1/4, 1/2 to 1/8, 1/4 to 1/16, etc. I like that one, clean and simple.

Edited by wtf
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23 minutes ago, Wyatt said:

I don’t know, it just feels wrong.

Ugh, I’m not sure how to word.

Here is a very clear demonstration of how working with infinite sets / sequences is non intuitive.

The top line is just a list zero plus the positive integers.

The matching line underneath is a list of the squares of the line above.

You can clearly see that because every integer has a square, there is one to one correspondence between the upper and lower set / sequence.

That is every number in the upper sequence corresponds to one and only one number in the lower sequence uniquely.

So both sets have the same count of members.

Yet the upper sequence contains every number in the lower sequence plus as many more as you choose.


[math]\begin{array}{*{20}{c}}
   0 \hfill & 1 \hfill & 2 \hfill & 3 \hfill & 4 \hfill  \\
   0 \hfill & 1 \hfill & 4 \hfill & 9 \hfill & {16} \hfill  \\
\end{array}[/math]

So the upper set has numbers, not in the bottom set, yet it has the same count of numbers.

 

Edited by studiot
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7 hours ago, Wyatt said:

I don’t know, it just feels wrong.

Ugh, I’m not sure how to word.

Walk through it step by step. Convince yourself that the mapping is:

* Injective: Different inputs go to different outputs; and

* Surjective: Everything in the target set gets hit.

By definition, the mapping is a bijection. Stepping through the details will make the idea clear. 

To see exactly how it works, I suggest using the method I showed at the end (instead of my solution). Map 0 to 1/2, map 1/2 to 1/4, 1/4 to 1/8, etc. What happens is that 0 gets absorbed and all the powers of 2 get slid down one space, mapping [0,1) to (0,1).

Edited by wtf
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10 hours ago, Wyatt said:

I don’t know, it just feels wrong.

Ugh, I’m not sure how to word.

I found a picture of the mapping from (0,1] to (0,1). Instead of using the inverse powers of 2, it uses the sequence 1/2, 1/3, 1/4, 1/5, ...

See how this works? The 1 at the right end of (0,1] gets mapped to 1/2. Then 1/2 gets mapped to 1/3; 1/3 gets mapped to 1/4, and so forth. The mapping is reversible so this is a bijection. 

 

download.png.85c96fdddf34b1917e849bd976c6a81c.png

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