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showing that a set S in not a rational interval Watch

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    (Original post by DFranklin)
    No. Why do you care whether b is in S?
    Oh no, I don't do I? I just need to show that for a 2 < b^2 there will be a smaller rational x^2 also greater than 2. I think the reason I was worried about whether it was in S or not was because I didn't really understand what I was trying to find when dealing with numbers outside the set S but I think I just had a "click" moment.
    Thank you again.
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    I hate to say it, but there seems to be a massive hole in the argument I'm trying to make you take, which I wrote down at the beginning and missed completely. I'll have to come back to this later though.

    --------------------------------

    OK. Here is a proper argument. It runs very much along the lines of the previous one but actually makes sense this time, I think.

    1. Consider S= \{ x \in \mathbb{Q} : x^2 \le 2 \} and P = \{ x \in \mathbb{Q} : -b \le x \le b \} .

    We claim that for all b \in \mathbb{Q},  S \neq P i.e. that there either exists x \in S with x \notin P or x \in P with x \notin S

    2. For all b \in \mathbb{Q}, we have either:

    a) b^2 < 2 or
    b) b^2 > 2

    3. We prove the result for case a):

    Consider x = b+\epsilon with 0 < \epsilon < \frac{2-b^2}{2b+1} < 1. Then we have:

    x^2 = (b+\epsilon)^2 <  b^2 + \epsilon(2b+1) < b^2 + \frac{2-b^2}{2b+1}(2b+1) = 2

    Hence x^2 < 2 \Rightarrow x \in S but x = b+\epsilon > b \Rightarrow x \notin P

    4. We prove the result for case b):

    To be done by so it goes, along the lines of her somewhat incorrect attempt above.

    so it goes: Your final line should say:

    Hence 2 < x^2 \Rightarrow x \notin S but x < b \Rightarrow x \in P
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    similar to the BMO maths problem in 2005 or 2006 i think or 2004.


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    (Original post by atsruser)
    I hate to say it, but there seems to be a massive hole in the argument I'm trying to make you take, which I wrote down at the beginning and missed completely. I'll have to come back to this later though.

    --------------------------------

    OK. Here is a proper argument. It runs very much along the lines of the previous one but actually makes sense this time, I think.

    1. Consider S= \{ x \in \mathbb{Q} : x^2 \le 2 \} and P = \{ x \in \mathbb{Q} : -b \le x \le b \} .

    We claim that for all b \in \mathbb{Q},  S \neq P i.e. that there either exists x \in S with x \notin P or x \in P with x \notin S

    2. For all b \in \mathbb{Q}, we have either:

    a) b^2 < 2 or
    b) b^2 > 2

    3. We prove the result for case a):

    Consider x = b+\epsilon with 0 < \epsilon < \frac{2-b^2}{2b+1} < 1. Then we have:

    x^2 = (b+\epsilon)^2 <  b^2 + \epsilon(2b+1) < b^2 + \frac{2-b^2}{2b+1}(2b+1) = 2

    Hence x^2 < 2 \Rightarrow x \in S but x = b+\epsilon > b \Rightarrow x \notin P

    4. We prove the result for case b):

    To be done by so it goes, along the lines of her somewhat incorrect attempt above.

    so it goes: Your final line should say:

    Hence 2 < x^2 \Rightarrow x \notin S but x < b \Rightarrow x \in P
    (Original post by DFranklin)
    ...
    case b)

    Consider x = b-\epsilon with  0 < \frac{2-b^2}{1-2b} < \epsilon < 1. Then we have:

    x^2 = (b-\epsilon)^2 <  b^2 + \epsilon(1-2b) > b^2 + \frac{2-b^2}{1-2b}(1-2b) = 2

    Hence x^2 < b^2 + \epsilon(1-2b) > 2 \Rightarrow 2 < x^2 ? I don't think I can say this, but if I can it implies that  x \notin S but x < b \Rightarrow x \in P

    I tried using the fact that 0 < \epsilon to show that 2<x^2 but all it gave me was that x^2 < b^2

    Perhaps I have my inequality for \epsilon wrong? I worked it out by saying:

    x^2 = (b-\epsilon)^2

    \Rightarrow 2 < b^2 - 2b\epsilon + \epsilon ^2

    \Rightarrow2 < b^2 + \epsilon (\epsilon )-2b but 0<\epsilon<1

    \Rightarrow2<b^2 + \epsilon (1-2b)

    \Rightarrow \epsilon > \dfrac{2-b^2}{1-2b}

    \Rightarrow 0 < \dfrac{2-b^2}{1-2b} < \epsilon < 1

    Thank you
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    (Original post by physicsmaths)
    similar to the BMO maths problem in 2005 or 2006 i think or 2004.


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    Oh cool, thanks
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    (Original post by so it goes)
    case b)

    Consider x = b-\epsilon with  0 < \frac{2-b^2}{1-2b} < \epsilon < 1. Then we have:

    x^2 = (b-\epsilon)^2 <  b^2 + \epsilon(1-2b) > b^2 + \frac{2-b^2}{1-2b}(1-2b) = 2
    Note that in this case, you want to show 2 < x^2, so your inequality stumbles at the first step, when you write x^2 = (b-\epsilon)^2 <  . You then try to turn things around by introducing an inequality going the other way, but that's doomed to fail, I'm afraid.

    Look at the following:

    x^2 = (b-\epsilon)^2 = b^2 -2b\epsilon +\epsilon^2 > \text{?} > 2

    We want b^2 -2b\epsilon +\epsilon^2 to be bigger (i.e more positive) than "something". That means that it must be that "something" + "something > 0". If you look at that expression, then you should be able to identify "something > 0" rather easily. (Hint: what is the range of f: x \mapsto x^2?)
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    (Original post by atsruser)
    Note that in this case, you want to show 2 < x^2, so your inequality stumbles at the first step, when you write x^2 = (b-\epsilon)^2 <  . You then try to turn things around by introducing an inequality going the other way, but that's doomed to fail, I'm afraid.

    Look at the following:

    x^2 = (b-\epsilon)^2 = b^2 -2b\epsilon +\epsilon^2 > \text{?} > 2

    We want b^2 -2b\epsilon +\epsilon^2 to be bigger (i.e more positive) than "something". That means that it must be that "something" + "something > 0". If you look at that expression, then you should be able to identify "something > 0" rather easily. (Hint: what is the range of f: x \mapsto x^2?)
    This is what I've been able to come up with. However I don't think it really works as at one point I say "for 0<b<\sqrt{2}" however for part b) I'm working with b^2 > 2 so this doesn't make much sense.

    b^2 - 2b\epsilon + \epsilon ^2 > b^2 - 2b\epsilon > b^2 - \dfrac{2b(2-b^2)}{1-2b} as \epsilon > \dfrac{2-b^2}{1-2b}

    \Rightarrow b^2 - 2b\epsilon + \epsilon ^2 > \dfrac{b^2 - 4b}{1-2b}

    for 0<b<\sqrt{2} we know \dfrac{b^2 - 4b}{1-2b} > 2

    Therefore b^2 - 2b\epsilon + \epsilon ^2 > 2

    Therefore x^2 > 2

    I've also tried saying b^2 - 2b\epsilon + \epsilon ^2 > -2b\epsilon + \epsilon ^2 and b^2 - 2b\epsilon + \epsilon ^2 > -2b\epsilon but I couldn't get either of them to work. Should I have another look at one of them?

    Thank you
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    (Original post by so it goes)
    This is what I've been able to come up with. However I don't think it really works as at one point I say "for 0<b<\sqrt{2}" however for part b) I'm working with b^2 > 2 so this doesn't make much sense.
    You seem to be losing track of what you are actually trying to show. (It's very easy to do this, sadly). You have made some progress, but then you go off on a tangent.

    You wrote:

    b^2 - 2b\epsilon + \epsilon ^2 > b^2 - 2b\epsilon > b^2 - \dfrac{2b(2-b^2)}{1-2b} as \epsilon > \dfrac{2-b^2}{1-2b}

    However, you seem to have some preconceived idea of a condition on \epsilon already. However, it is *precisely that condition* that we are trying to find. We don't yet know how small \epsilon must be. We want to say:

    if \epsilon < \text{some expression only in b} then x^2 > 2

    So let's take what you wrote and modify it a bit:

    x^2 = b^2 - 2b\epsilon + \epsilon ^2 > b^2 - 2b\epsilon > \text{?}

    I have:

    a) added x^2 at the beginning to remind you that this is the quantity for which you want to find an inequality - all of the b, \epsilon stuff is merely a sideshow in finding this inequality and

    b) stuck ? at the end.

    Questions:

    1. What should I have written in place of ? to get our desired inequality
    2. Does this allow you to find a condition on \epsilon that makes x^2 > 2? If so, how?

    One other point: it may help if you sketched out the required locations of \sqrt{2}, x, b on a number line, and then thought about the inequality that implies on \epsilon - do we want \epsilon bigger or smaller than something i.e. are we looking for \epsilon > \text{?} or \epsilon < \text{?} (For example, since x = b - \epsilon, and b is, say a bit bigger than \sqrt{2}, then if we make \epsilon quite big, then we could make x \approx 0. Would that be any good for our inequality?)
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    (Original post by atsruser)
    One other point: it may help if you sketched out the required locations of \sqrt{2}, x, b on a number line, and then thought about the inequality that implies on \epsilon - do we want \epsilon bigger or smaller than something i.e. are we looking for \epsilon > \text{?} or \epsilon < \text{?} (For example, since x = b - \epsilon, and b is, say a bit bigger than \sqrt{2}, then if we make \epsilon quite big, then we could make x \approx 0. Would that be any good for our inequality?)
    I don't think this would be good as we want to find an x between \sqrt{2} and b so I'm thinking we should keep \epsilon < b - \sqrt{2}?

    (Original post by atsruser)
    You seem to be losing track of what you are actually trying to show. (It's very easy to do this, sadly). You have made some progress, but then you go off on a tangent.

    You wrote:

    b^2 - 2b\epsilon + \epsilon ^2 > b^2 - 2b\epsilon > b^2 - \dfrac{2b(2-b^2)}{1-2b} as \epsilon > \dfrac{2-b^2}{1-2b}

    However, you seem to have some preconceived idea of a condition on \epsilon already. However, it is *precisely that condition* that we are trying to find. We don't yet know how small \epsilon must be. We want to say:

    if \epsilon < \text{some expression only in b} then x^2 > 2

    So let's take what you wrote and modify it a bit:

    x^2 = b^2 - 2b\epsilon + \epsilon ^2 > b^2 - 2b\epsilon > \text{?}

    I have:

    a) added x^2 at the beginning to remind you that this is the quantity for which you want to find an inequality - all of the b, \epsilon stuff is merely a sideshow in finding this inequality and

    b) stuck ? at the end.

    Questions:

    1. What should I have written in place of ? to get our desired inequality
    2. Does this allow you to find a condition on \epsilon that makes x^2 > 2? If so, how?
    If I say:

    x^2 = b^2 -2b \epsilon + \epsilon ^2 > b^2 -2b\epsilon > 2

    If \epsilon > \dfrac{b^2 -2}{2b} \Rightarrow b^2 - 2b\epsilon > 2

    \Rightarrow \epsilon > \dfrac{b^2 -2}{2b}

    If \epsilon > \dfrac{b^2 - 2}{2b} \Rightarrow x^2 > b^2 - (b^2 - 2) + \dfrac{(b^2 - 2)^2}{4b^2} \Rightarrow x^2 > \dfrac{b^4 + 4b^2 + 4}{4b^2} > 2 from graph

    \Rightarrow 2 < x^2 < b^2

    Does that look okay?
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    (Original post by so it goes)
    I don't think this would be good as we want to find an x between \sqrt{2} and b so I'm thinking we should keep \epsilon < b - \sqrt{2}?
    The precise details are unimportant here, but the point is the we need to keep \epsilon pretty small, so x = b-\epsilon fits between \sqrt{2} and b.

    Remember that we're thinking about the b's that are very, very close to \sqrt{2} and the closer they are, the smaller we will have to make \epsilon. So:

    1. \epsilon must be less than something.
    2. That something must depend on (i.e. be a function of) b

    If I say:

    x^2 = b^2 -2b \epsilon + \epsilon ^2 > b^2 -2b\epsilon > 2
    This is good. But then you have the implications (and inequality) going the wrong way. Instead of

    If \epsilon > \dfrac{b^2 -2}{2b} \Rightarrow b^2 - 2b\epsilon > 2
    You want x^2 > b^2 -2b\epsilon > 2 \Rightarrow b^2 > 2+ 2b\epsilon \Rightarrow  \dfrac{b^2 -2}{2b} > \epsilon

    And now you're done. We have imposed the requirement that x^2 > 2 and found that it implies that \dfrac{b^2 -2}{2b} > \epsilon .

    So as long as we make \epsilon at least that small, for any given value of b, we have found x with 2 < x^2 \Rightarrow x \notin S and x < b \Rightarrow x \in P, which is what we wanted.

    \Rightarrow \epsilon > \dfrac{b^2 -2}{2b}

    If \epsilon > \dfrac{b^2 - 2}{2b} \Rightarrow x^2 > b^2 - (b^2 - 2) + \dfrac{(b^2 - 2)^2}{4b^2} \Rightarrow x^2 > \dfrac{b^4 + 4b^2 + 4}{4b^2} > 2 from graph

    \Rightarrow 2 < x^2 < b^2

    Does that look okay?
    This is unnecessary and wrong since your \epsilon inequality is going the wrong way (and I'm not sure about your final b^4 inequality - why is that greater than 2?). Once we have found the condition on \epsilon, we are guaranteed that 2 < x^2 - we don't need to do any more work to show it.
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    (Original post by atsruser)
    The precise details are unimportant here, but the point is the we need to keep \epsilon pretty small, so x = b-\epsilon fits between \sqrt{2} and b.

    Remember that we're thinking about the b's that are very, very close to \sqrt{2} and the closer they are, the smaller we will have to make \epsilon. So:

    1. \epsilon must be less than something.
    2. That something must depend on (i.e. be a function of) b

    This is good. But then you have the implications (and inequality) going the wrong way. Instead of

    You want x^2 > b^2 -2b\epsilon > 2 \Rightarrow b^2 > 2+ 2b\epsilon \Rightarrow  \dfrac{b^2 -2}{2b} > \epsilon

    And now you're done. We have imposed the requirement that x^2 > 2 and found that it implies that \dfrac{b^2 -2}{2b} > \epsilon .

    So as long as we make \epsilon at least that small, for any given value of b, we have found x with 2 < x^2 \Rightarrow x \notin S and x < b \Rightarrow x \in P, which is what we wanted.
    Thank you so much! This question makes so much more sense now. You'be been super patient and helpful and I really appreciate it

    If I'm remembering this correctly, because x^2 is symmetrical does proving it for the positives imply it for the negatives?

    (Original post by atsruser)
    This is unnecessary and wrong since your \epsilon inequality is going the wrong way (and I'm not sure about your final b^4 inequality - why is that greater than 2?).
    I tried plotting y = \dfrac{x^4 +4x^2 +4}{4x^2} and found that it was symmetrical about the y-axis with minimum turning points at (\pm \sqrt{2}, 2)

    (Original post by atsruser)
    Once we have found the condition on \epsilon, we are guaranteed that 2 < x^2 - we don't need to do any more work to show it.
    Oh, of course, thank you. I think I was losing track of what I was trying to show again.

    Thank you!
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    (Original post by so it goes)
    If I'm remembering this correctly, because x^2 is symmetrical does proving it for the positives imply it for the negatives?
    Well, I made a kind of handwaving argument to that effect, but maybe you should try to justify it properly. Also the original question raised the possibility of strict/non-strict inequalities at either end of the interval e.g.  a < x \le b and so on. I have given no thought to that (and indeed can't be bothered to ). I think DFranklin made a comment about this earlier somewhere.

    I tried plotting y = \dfrac{x^4 +4x^2 +4}{4x^2} and found that it was symmetrical about the y-axis with minimum turning points at (\pm \sqrt{2}, 2)
    That doesn't really cut the mustard, but as you don't need that inequality for this question, it doesn't matter too much. Maybe you ought to practise a bit of inequality manipulation though. It's very useful in analysis.

    Do you fully understand the way that I suggested proving this i.e. can you see the logic in creating two sets S and P and showing that they can never contain the same elements?
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    (Original post by atsruser)
    Well, I made a kind of handwaving argument to that effect, but maybe you should try to justify it properly. Also the original question raised the possibility of strict/non-strict inequalities at either end of the interval e.g.  a < x \le b and so on. I have given no thought to that (and indeed can't be bothered to ). I think DFranklin made a comment about this earlier somewhere.
    Ahh, okay. Thanks, I'll try and figure those out

    (Original post by atsruser)
    Do you fully understand the way that I suggested proving this i.e. can you see the logic in creating two sets S and P and showing that they can never contain the same elements?
    I definitely see the logic behind it - it wouldn't have occurred to me to do that (hopefully it will in the future though) but I can see what it's doing.

    Thank you again
 
 
 
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