Hi,
Let
Show that has the property that if and then also . Show also that is not a rational interval  that is, there does not exist such that
or any one of the other possibilities for an interval.
Now I think I've managed to do the first part of this question and perhaps the second but I'm unsure with both of them whether I have gone into enough detail and presented it in the right way.
For the first part I said:
We know that which implies that and . As then this implies that and therefore . Is this sufficient for the first part?
For the second part I was thinking of a sort of proof by contradiction? Firstly, assume that is the biggest fraction below . Now surely a number that is bigger that but less than is
Therefore for any rational less than I can generate a bigger one that's still less than , there is therefore no possible value of b in
that would work. Is this sufficient for the second part? Have I expressed it in the correct way?
Thank you

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 21072014 09:57

DFranklin
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 21072014 11:40
(Original post by so it goes)
For the first part I said:
We know that which implies that and . As then this implies that and therefore . Is this sufficient for the first part?
Now sometimes that is completely valid (sometimes you just need to point out the obvious), but I'm unconvinced here.
There's a certain amount of subtlety here, basically because x^2 is not monotonic, so x < y doesn't automatically imply x^2 < y^2. Everything you've said is true, but you're making the examiner do the work of deciding "yes, it really is true", and i don't think that's right here.
I would divide into two cases:
if 0 <= z, then 0 < z < y and so 0 < z^2 < y^2 < 2, so z^2 < 2.
If 0 > z, then .. (I'll leave this one for you).
Now I'm not sure how your course is approaching this; I'm used to the approach that essentially says "you can't assume irrational numbers exist (as things you can calculate with etc)". In which case the number you've written "isn't allowed" and you need a different approach.
If your course is allowing you to talk about irrational numbers, then you can argue / prove that if a and b are real numbers with a < b, then there is always a rational number between a and b. (Exactly how you do this depends on exactly how you define or think of irrational numbers, but you could, for instance, consider the decimal expansions of a, b and truncate at the first point where they differ to get rational numbers a*, b* with a<=a* <=b* <=b. Then (a*+b*)/2 will be between a and b).
The rest of your argument is fine (and the way you're supposed to do it). 
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 21072014 11:49
(Original post by so it goes)
For the second part I was thinking of a sort of proof by contradiction? Firstly, assume that is the biggest fraction below . Now surely a number that is bigger that but less than is
Therefore for any rational less than I can generate a bigger one that's still less than
The various cases make it a bit messy, I guess, but you want to show that for all , you can find such that
Since will clearly only be a "bit" bigger than , write:
with
so
Now what condition must hold on ? 
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 21072014 13:07
(Original post by DFranklin)
I'd say it's questionable. You should be very uneasy of your proof if you don't actually "do" anything in it. By which I mean, all you've done is taken what they've told you, and said it implies something and that means the result is true.
Now sometimes that is completely valid (sometimes you just need to point out the obvious), but I'm unconvinced here.
There's a certain amount of subtlety here, basically because x^2 is not monotonic, so x < y doesn't automatically imply x^2 < y^2. Everything you've said is true, but you're making the examiner do the work of deciding "yes, it really is true", and i don't think that's right here.
I would divide into two cases:
if 0 <= z, then 0 < z < y and so 0 < z^2 < y^2 < 2, so z^2 < 2.
If 0 > z, then .. (I'll leave this one for you).
It's not particularly clear that you can do this, because the number you have just provided is not rational, and therefore isn't a member of S.
Now I'm not sure how your course is approaching this; I'm used to the approach that essentially says "you can't assume irrational numbers exist (as things you can calculate with etc)". In which case the number you've written "isn't allowed" and you need a different approach.
If your course is allowing you to talk about irrational numbers, then you can argue / prove that if a and b are real numbers with a < b, then there is always a rational number between a and b. (Exactly how you do this depends on exactly how you define or think of irrational numbers, but you could, for instance, consider the decimal expansions of a, b and truncate at the first point where they differ to get rational numbers a*, b* with a<=a* <=b* <=b. Then (a*+b*)/2 will be between a and b).
The rest of your argument is fine (and the way you're supposed to do it).
(Original post by atsruser)
You haven't generated a rational, so this is no good.
The various cases make it a bit messy, I guess, but you want to show that for all , you can find such that
Since will clearly only be a "bit" bigger than , write:
with
so
Now what condition must hold on ?
Sorry atsruser but for some reason I can't upvote your comment.
Thanks again. 
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 21072014 15:42
(Original post by DFranklin)
I'd say it's questionable. You should be very uneasy of your proof if you don't actually "do" anything in it. By which I mean, all you've done is taken what they've told you, and said it implies something and that means the result is true.
Now sometimes that is completely valid (sometimes you just need to point out the obvious), but I'm unconvinced here.
There's a certain amount of subtlety here, basically because x^2 is not monotonic, so x < y doesn't automatically imply x^2 < y^2. Everything you've said is true, but you're making the examiner do the work of deciding "yes, it really is true", and i don't think that's right here.
I would divide into two cases:
if 0 <= z, then 0 < z < y and so 0 < z^2 < y^2 < 2, so z^2 < 2.
If 0 > z, then .. (I'll leave this one for you).
It's not particularly clear that you can do this, because the number you have just provided is not rational, and therefore isn't a member of S.
Now I'm not sure how your course is approaching this; I'm used to the approach that essentially says "you can't assume irrational numbers exist (as things you can calculate with etc)". In which case the number you've written "isn't allowed" and you need a different approach.
If your course is allowing you to talk about irrational numbers, then you can argue / prove that if a and b are real numbers with a < b, then there is always a rational number between a and b. (Exactly how you do this depends on exactly how you define or think of irrational numbers, but you could, for instance, consider the decimal expansions of a, b and truncate at the first point where they differ to get rational numbers a*, b* with a<=a* <=b* <=b. Then (a*+b*)/2 will be between a and b).
The rest of your argument is fine (and the way you're supposed to do it).
Is that okay for the first part?
(Original post by atsruser)
You haven't generated a rational, so this is no good.
The various cases make it a bit messy, I guess, but you want to show that for all , you can find such that
Since will clearly only be a "bit" bigger than , write:
with
so
Now what condition must hold on ?
But I just can't see how that could help me. I think I might need another hint, thank you 
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 21072014 16:47
(Original post by so it goes)
For the second part I'm really not too sure. What you've written makes sense but I'm not too sure where I should go with it. I tried
But I just can't see how that could help me. I think I might need another hint, thank you
Now you want to show that for any such choice of , you can find with , contradicting the suggested hypothesis.
I suggested that you consider where is some rather small rational quantity. We want to find out just how small it has to be so that not only but also .
I showed that **. You want, however, .
1. What do you need to stick on the end of the starred inequality to ensure that?
2. Once you have stuck that thing on, you have a condition that must satisfy. Find that condition in the form:
3. Now you've found the condition for , write the argument in the correct order.
[I've edited this about 8 times  I hope it now makes sense] 
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 22072014 11:35
(Original post by atsruser)
Well, you identified that the must be less than or equal to (though you haven't proved it), or equivalently that . This follows since if we had then we could find with (why?) so with then could not be in S.
So if
which is not rational and so is not in ? Is that correct? If so I'm really sorry but I don't really see what it's getting at because isn't my supposed to be less than or equal to but here it's greater than, so I'm guessing a proof by contradiction? I feel I'm misunderstanding this part.
(Original post by atsruser)
Now you want to show that for any such choice of , you can find with , contradicting the suggested hypothesis.
I suggested that you consider where is some rather small rational quantity. We want to find out just how small it has to be so that not only but also .
I showed that **. You want, however, .
(Original post by atsruser)
1. What do you need to stick on the end of the starred inequality to ensure that?
(Original post by atsruser)
2. Once you have stuck that thing on, you have a condition that must satisfy. Find that condition in the form:
for then
but if then
If we assume :
As we know that we know that is always positive. However, we'll be diving it by which is negative and so is negative, meaning is always less than and so "some number less than ". This doesn't seem helpful.
If we assume then . We know that and so is always positive and so "some number more than ". This also doesn't seem helpful.
I must be missing something but I can't spot what.
(Original post by atsruser)
[I've edited this about 8 times  I hope it now makes sense] 
DFranklin
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 22072014 12:41
(Original post by so it goes)
..
What this is "really about" is saying that S doesn't have a maximal element. That is, it's not possible to find with b >= s for all s in S. I know I said "don't assume irrational numbers exist", but since you do know about irrationals, it's fine to use them as a guide about whats going on. So that said, we know that ideally, b "should be" , but since it's rational, we're mentally thinking the only thing that can work is if b is very close to .
So the key argument is what happens when b is close to . Now what often happens at this point is that the argument will need to use the fact that "b is close to \sqrt{2}" (so you might rely on the fact that b is positive, or that b is < 2, for example). You should still "close the gaps" by showing "we don't have to worry about what happens when b < 0 etc", but that's not really the key point here].Last edited by DFranklin; 22072014 at 12:42. 
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 22072014 16:34
I think I'd approach it like this:
1. We have
Since is an even function, we can say by symmetry that if were a rational interval then it must look like i.e. whereas the question suggested an interval of the form , we don't, in fact, need to think about this any more.
2. Now we can concentrate on and when we've sorted out this part of the set, we can fix up the arguments for the negative values (which should be trivial by symmetry).
3. We now use our intuition to reason that the only candidate for is . We must have one of:
But , we can rule this out as a candidate. We now have to rule out the other two candidate ranges too.
4. For the case , we can proceed as I suggested. I'll now fill in the details as I think you were getting a little lost.
We want to show that for any with , we can find with i.e. now matter how close is to 2, then we can always fit another square rational in between it and 2.
We are thinking only about the range , remember. So consider with . Then
Now if then
So with that value for we can ensure that and and (since ).
Note that it looked like I pulled the value for out of a hat, but I showed the justification in an earlier post.
So no rational with will allow us to define the interval.
5. We now must rule out the case with . I handwaved this a bit in an earlier post (with at least one logical error), but I think that you can make a very similar argument to that which I've given above: that if then you can find i.e. that the hypothesised interval will admit points that do not meet the definition of .
I'll leave the details to you as an exercise.
6. You also need to tidy up the "negative values can be dealt with by symmetry" argument, I guess, and check that I'm right about that. 
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 23072014 06:47
(Original post by DFranklin)
The case b < 1/2 (or indeed b < 0) is a nonissue. If b < 0, then just take c = 0 to get c > b but c^2 < 2.
What this is "really about" is saying that S doesn't have a maximal element. That is, it's not possible to find with b >= s for all s in S. I know I said "don't assume irrational numbers exist", but since you do know about irrationals, it's fine to use them as a guide about whats going on. So that said, we know that ideally, b "should be" , but since it's rational, we're mentally thinking the only thing that can work is if b is very close to .
So the key argument is what happens when b is close to . Now what often happens at this point is that the argument will need to use the fact that "b is close to \sqrt{2}" (so you might rely on the fact that b is positive, or that b is < 2, for example). You should still "close the gaps" by showing "we don't have to worry about what happens when b < 0 etc", but that's not really the key point here].
Thanks again! 
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 23072014 07:57
(Original post by atsruser)
I think I'd approach it like this:
1. We have
Since is an even function, we can say by symmetry that if were a rational interval then it must look like i.e. whereas the question suggested an interval of the form , we don't, in fact, need to think about this any more.
2. Now we can concentrate on and when we've sorted out this part of the set, we can fix up the arguments for the negative values (which should be trivial by symmetry).
3. We now use our intuition to reason that the only candidate for is . We must have one of:
But , we can rule this out as a candidate. We now have to rule out the other two candidate ranges too.
4. For the case , we can proceed as I suggested. I'll now fill in the details as I think you were getting a little lost.
We want to show that for any with , we can find with i.e. now matter how close is to 2, then we can always fit another square rational in between it and 2.
We are thinking only about the range , remember. So consider with . Then
Now if then
So with that value for we can ensure that and and (since ).
Note that it looked like I pulled the value for out of a hat, but I showed the justification in an earlier post.
So no rational with will allow us to define the interval.
(Original post by atsruser)
5. We now must rule out the case with . I handwaved this a bit in an earlier post (with at least one logical error), but I think that you can make a very similar argument to that which I've given above: that if then you can find i.e. that the hypothesised interval will admit points that do not meet the definition of .
Thank you again, and sorry I'm struggling so much with this. 
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 23072014 08:07
(Original post by DFranklin)
...
(Original post by atsruser)
...
If let
Now can't I find some number less than by expressing it as which I can make as small as I like by increasing sufficiently?
I will then have found
I'm pretty sure this isn't rigorous enough to be a proper proof but is it a possible way of solving this? Or will this end up being the same as the one before, except I've rewritten as ?
Thank you 
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 23072014 12:41
(Original post by so it goes)
Why do I need to worry about when isn't it defined in the question that must be less than to be in the set ? What am I missing?
If I have T = {x : 0 < x < 1}, then I can't find values a, b in T s.t. T = (a, b) (or [a, b], etc).
But I can still write T as the interval (0, 1), even though 0 and 1 aren't members of T.
So you need to show something similar can't happen with S.
(Original post by so it goes)
Would it be possible for me to approach the problem like this?
If let
Now can't I find some number less than by expressing it as which I can make as small as I like by increasing sufficiently?
I will then have found 
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 23072014 15:48
(Original post by so it goes)
What I don't understand is this point. Why do I need to worry about when isn't it defined in the question that must be less than to be in the set ? What am I missing?
We have already dismissed the possibilities . The only remaining possibility is that . We want to make the argument that no matter how close may be to 2, it still isn't close enough to stop us finding a rational .
If we can do that, then by the first inequality, we have , but by the second which is a contradiction, which tells us that our hypothesis that such a exists is false.
At this point, you may say that it is "obvious" that if then we can find such an , but would you be happy making that argument to a scary maths professor in an Oxford tutorial? Bear in mind that in maths, if you say that something is "obvious", what you really mean is that the proof is so trivial that it can be stated with no further thought. In this case, I don't think that the proof is obvious; in fact, I think it's of precisely the same difficulty as the statement that says that if then we can find .
Note that you may claim that since both , then you can certainly find a rational . But in fact that's not good enough: we need to find a rational and for that I think we need to prove it.
Thank you again, and sorry I'm struggling so much with this.Last edited by atsruser; 23072014 at 18:38. 
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 23072014 16:00
You want to find with and .
This is not the same as finding with since this does not imply that is in .
You can approach it this way, but you would have to show that you can find with since then . 
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 24072014 09:57
(Original post by DFranklin)
So, forget all the sqrt(2) stuff, and think about how we describe intervals.
If I have T = {x : 0 < x < 1}, then I can't find values a, b in T s.t. T = (a, b) (or [a, b], etc).
But I can still write T as the interval (0, 1), even though 0 and 1 aren't members of T.
So you need to show something similar can't happen with S.
You're showing you can add something small to b^2 and keep b^2 less than 2, but you need to show you can add something small to b and keep b^2 less that 2. 
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 24072014 10:40
(Original post by atsruser)
We want to show that no exists such that the set can be written in the form
We have already dismissed the possibilities . The only remaining possibility is that . We want to make the argument that no matter how close may be to 2, it still isn't close enough to stop us finding a rational .
If we can do that, then by the first inequality, we have , but by the second which is a contradiction, which tells us that our hypothesis that such a exists is false.
or
let where and
but
Now, if we assume then as and is rational.
However, therefore
We therefore have a contradiction, meaning our original assumption was wrong, therefore if
Have I understood you correctly?
(Original post by atsruser)
As Dfranklin pointed out, this is no good. You're making the same error that I made earlier (though I didn't notice it at the time).
You want to find with and .
This is not the same as finding with since this does not imply that is in .
You can approach it this way, but you would have to show that you can find with since then .
Thank you. 
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 24072014 11:46
(Original post by so it goes)
Have I understood you correctly?
"S can be written in the form {x : b <= x <= b} for some rational b > 2".
And then at the end you've said "if we assume then ... contradiction".
So you've made a second assumption and got a contradiction. Now in this case, it's actually obvious that the second assumption follows from the first (since we're claiming S is of the form {x: b <= x <= b} is obvious that b is in S).
But you actually need to prove it for all types of interval (e.g. {x : b < x < b}, {x : b <= x < b} etc), at which point you're making two contradictory assumptions and everything falls down.
[It's actually trivial to fix this: why make the second assumption at all?] 
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 24072014 13:37
(Original post by DFranklin)
IYou're close, but your argument gets quite confused towards the end,. The problem I have with what is that the "real" assumption (i.e. the thing we're going to show isn't true) is:
"S can be written in the form {x : b <= x <= b} for some rational b > 2".
And then at the end you've said "if we assume then ... contradiction".
So you've made a second assumption and got a contradiction. Now in this case, it's actually obvious that the second assumption follows from the first (since we're claiming S is of the form {x: b <= x <= b} is obvious that b is in S).
But you actually need to prove it for all types of interval (e.g. {x : b < x < b}, {x : b <= x < b} etc), at which point you're making two contradictory assumptions and everything falls down.
[It's actually trivial to fix this: why make the second assumption at all?]
(Original post by so it goes)
Now, if we assume then as and is rational.
However, therefore
We therefore have a contradiction, meaning our original assumption was wrong, therefore if
therefore
as this implies that
Does that work?
Thank you so much for this! 
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 24072014 13:51
(Original post by so it goes)
Ahh okay, so instead of saying this at the end:
Should it be more like this?
therefore
as this implies that
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