username53983805
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#1
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Because there are infinitely many rationals on (0,1) I thought that the nicest way of proving S contains all rationals on (0,1) was to show that the sum of the elements in S diverges, as if it converged there must only exist finitely many elements in S, however I feel like my argument is not solid enough. Anyways here it is

Consider S to contain finitely many elements, such that the sum of these elements converged to a limit, L. We can note that \frac{1}{x+1}+\frac{x}{x+1}=1, so we can pair elements in S such that their sums equal 1. Noting that x,\frac{1}{x+1},\frac{x}{x+1} are all distinct if x is rational, given an element in S we can create two new elements in S such that their sum is equal to 1. So now the sum converges to L+1. Given that this process can be done infinitely many times, we note that the sum converges to L+1+1+1+... i.e. the sum diverges. Therefore S must contain infinitely many elements, and so we conclude S contains all rationals on the interval (0,1).

If it is too weak, what is the problem with it?
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davros
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(Original post by username53983805)
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Because there are infinitely many rationals on (0,1) I thought that the nicest way of proving S contains all rationals on (0,1) was to show that the sum of the elements in S diverges, as if it converged there must only exist finitely many elements in S, however I feel like my argument is not solid enough. Anyways here it is

...

If it is too weak, what is the problem with it?
If S consisted of only the unit fractions 1/2, 1/3, 1/4, 1/5 etc then their sum would diverge (the harmonic series) but S certainly doesn't contain all the rationals in (0, 1) - what about 2/5, 3/7, 4/19 etc?
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username53983805
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(Original post by davros)
If S consisted of only the unit fractions 1/2, 1/3, 1/4, 1/5 etc then their sum would diverge (the harmonic series) but S certainly doesn't contain all the rationals in (0, 1) - what about 2/5, 3/7, 4/19 etc?
But could I not just pair these three fractions with 3/5, 4/7, and 15/19 respectively (as these must also be elements of S) and then consider each of the pair sums (1)?

Now I look over it I feel like I'm saying S has some sort of infinite set property in the sense that we can keep creating these new fraction pairs in S infinitely many times, and so this implies the sum diverges, which means S contains all rational numbers in (0,1), which does not make much sense as I'm basically just assuming S is an infinite set when I use the property
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ghostwalker
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(Original post by username53983805)
But could I not just pair these three fractions with 3/5, 4/7, and 15/19 respectively (as these must also be elements of S) and then consider each of the pair sums (1)?
Why must 3/5, 4/7, 15/19 be elements of S?

There's nothing obvious in the definition of S that tells you they are elements of it.

The elements of S are defined inductively. 1/2 is in S, and then anything you can generate by the use of the second rule.

E.g. 1/( (1/2) +1) = 2/3 is in S, etc.

Also, even if you can show that S contains an infinite number of elements, this does not imply that it contains all rationals in the interval (0,1).
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username53983805
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(Original post by ghostwalker)
Why must 3/5, 4/7, 15/19 be elements of S?

There's nothing obvious in the definition of S that tells you they are elements of it.

The elements of S are defined inductively. 1/2 is in S, and then anything you can generate by the use of the second rule.

E.g. 1/( (1/2) +1) = 2/3 is in S, etc.

Also, even if you can show that S contains an infinite number of elements, this does not imply that it contains all rationals in the interval (0,1).
Oh yeah it doesn't tell you that

Okay so I should try an alternative approach

Why is this the case?
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(Original post by username53983805)
Oh yeah it doesn't tell you that

Okay so I should try an alternative approach

Why is this the case?
What do you mean? Why is what the case?

Edit: I see you've hightlighted a line that wasn't showing, and DFranklin's addressed the point below.
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DFranklin
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(Original post by username53983805)
Oh yeah it doesn't tell you that

Okay so I should try an alternative approach

Why is this the case?
You've been given an explicit example of a set S that is (a) an infinite set of rationals from (0,1), (b) S is not all the rationals in (0,1). This is sufficient to tell you that S infinite does not imply S contains all the rational in (0,1). That's the whole point of a (counter) example.
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(Original post by DFranklin)
You've been given an explicit example of a set S that is (a) an infinite set of rationals from (0,1), (b) S is not all the rationals in (0,1). This is sufficient to tell you that S infinite does not imply S contains all the rational in (0,1). That's the whole point of a (counter) example.
I'm sort of confused by this
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#9
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(Original post by username53983805)
I'm sort of confused by this
You're saying "If S is infinite, S must contain all the rationals in (0,1)." (*)
You've been given the example of the set S = {1/n : n = 2, 3, 4, ....}. S is infinite, but S does not contain all the rationals in (0,1).
Therefore (*) is not true.

If you don't understand, or don't agree, point out the exact point that you have a problem with.
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username53983805
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#10
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(Original post by DFranklin)
You're saying "If S is infinite, S must contain all the rationals in (0,1)." (*)
You've been given the example of the set S = {1/n : n = 2, 3, 4, ....}. S is infinite, but S does not contain all the rationals in (0,1).
Therefore (*) is not true.

If you don't understand, or don't agree, point out the exact point that you have a problem with.
Oh yes now I see
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DFranklin
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Suggestion (I've spoilered it, but the spoilers are fairly mild. It's much more "how you might approach this" than "do this, this and that").

Spoiler:
Show

(a) start off from x = 1/2 and use the rule to make some new rationals. Start to get a feel for what happens.
(b) suppose x = p/q, what do the two new rationals you can make with the rule look like?
(c) Suppose instead you want to *make* y = a/b from the rule. What rational x = p/q would you need?

From there you should be able to set out some kind of proof I think.
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