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showing that a set S in not a rational interval watch

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    Hi,

    Let

    S := \{x \in \mathbb{Q} | x^2 \leq 2\}

    Show that S has the property that if x, y \in S and x \leq z \leq y then also z \in S. Show also that S is not a rational interval - that is, there does not exist a,b \in \mathbb{Q} such that

    S = \{ x \in \mathbb{Q} | a < x < b \}

    or S = \{ x \in \mathbb{Q} | a \leq x \leq b \}

    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 x, y \in S which implies that x^2 \leq 2 and y^2 \leq 2. As x \leq z \leq y then this implies that z^2 \leq 2 and therefore z \in S. Is this sufficient for the first part?

    For the second part I was thinking of a sort of proof by contradiction? Firstly, assume that \dfrac{m}{n} is the biggest fraction below \sqrt{2}. Now surely a number that is bigger that \dfrac{m}{n} but less than \sqrt{2} is

    \dfrac{1}{2}(\dfrac{m}{n} + \sqrt{2})

    Therefore for any rational less than \sqrt{2} I can generate a bigger one that's still less than \sqrt{2}, there is therefore no possible value of b in

    S = \{ x \in \mathbb{Q} | a < x < b \}

    that would work. Is this sufficient for the second part? Have I expressed it in the correct way?

    Thank you
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    (Original post by so it goes)
    For the first part I said:

    We know that x, y \in S which implies that x^2 \leq 2 and y^2 \leq 2. As x \leq z \leq y then this implies that z^2 \leq 2 and therefore z \in S. Is this sufficient for the first part?
    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).


    For the second part I was thinking of a sort of proof by contradiction? Firstly, assume that \dfrac{m}{n} is the biggest fraction below \sqrt{2}. Now surely a number that is bigger that \dfrac{m}{n} but less than \sqrt{2} is

    \dfrac{1}{2}(\dfrac{m}{n} + \sqrt{2})

    Therefore for any rational less than \sqrt{2} I can generate a bigger one that's still less than \sqrt{2}.
    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).
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    (Original post by so it goes)
    For the second part I was thinking of a sort of proof by contradiction? Firstly, assume that \dfrac{m}{n} is the biggest fraction below \sqrt{2}. Now surely a number that is bigger that \dfrac{m}{n} but less than \sqrt{2} is

    \dfrac{1}{2}(\dfrac{m}{n} + \sqrt{2})

    Therefore for any rational less than \sqrt{2} I can generate a bigger one that's still less than \sqrt{2}
    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 \{b \in \mathbb{Q} : b^2 &lt; 2\}, you can find c \in \mathbb{Q} &gt; b such that c^2 &lt; 2

    Since c will clearly only be a "bit" bigger than b, write:

    c=b+\epsilon with \epsilon \in \mathbb{Q}, 0 &lt; \epsilon &lt; 1

    so

    c^2 = (b+\epsilon)^2 = b^2+2b\epsilon + \epsilon^2 = b^2 +\epsilon(2b+\epsilon) &lt; b^2 + \epsilon(2b+1)

    Now what condition must hold on \epsilon?
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    (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 \{b \in \mathbb{Q} : b^2 &lt; 2\}, you can find c \in \mathbb{Q} &gt; b such that c^2 &lt; 2

    Since c will clearly only be a "bit" bigger than b, write:

    c=b+\epsilon with \epsilon \in \mathbb{Q}, 0 &lt; \epsilon &lt; 1

    so

    c^2 = (b+\epsilon)^2 = b^2+2b\epsilon + \epsilon^2 = b^2 +\epsilon(2b+\epsilon) &lt; b^2 + \epsilon(2b+1)

    Now what condition must hold on \epsilon?
    Ahh okay, thank you - I feel silly for missing a mistake that obvious. I'm just reading some lecture notes and trying the exercises but I still haven't got the hang of it. I'll see if I can make any headway with this now
    Sorry atsruser but for some reason I can't up-vote your comment.
    Thanks again.
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    (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).
    For z &lt; 0 I said:

    z&lt;0
    \Rightarrow x&lt;z&lt;0
    \Rightarrow 0 &lt; z^2 &lt; x^2 &lt; 2
    \Rightarrow z^2 &lt; 2

    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 \{b \in \mathbb{Q} : b^2 &lt; 2\}, you can find c \in \mathbb{Q} &gt; b such that c^2 &lt; 2

    Since c will clearly only be a "bit" bigger than b, write:

    c=b+\epsilon with \epsilon \in \mathbb{Q}, 0 &lt; \epsilon &lt; 1

    so

    c^2 = (b+\epsilon)^2 = b^2+2b\epsilon + \epsilon^2 = b^2 +\epsilon(2b+\epsilon) &lt; b^2 + \epsilon(2b+1)

    Now what condition must hold on \epsilon?
    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

    c^2 &lt; b^2 + \epsilon (2b + 1) &lt; 2 + \epsilon (2b + 1)

    But I just can't see how that could help me. I think I might need another hint, thank you
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    (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

    c^2 &lt; b^2 + \epsilon (2b + 1) &lt; 2 + \epsilon (2b + 1)

    But I just can't see how that could help me. I think I might need another hint, thank you
    Well, you identified that the b \in \mathbb{Q} must be less than or equal to \sqrt{2} (though you haven't proved it), or equivalently that b^2 \le 2. This follows since if we had \sqrt{2} &lt; b \Rightarrow 2 &lt; b^2 then we could find u \in \mathbb{Q} with 2 &lt; u &lt; b^2 (why?) so with x^2 = u then x could not be in S.

    Now you want to show that for any such choice of b, you can find  x \in S with x &gt; b, contradicting the suggested hypothesis.

    I suggested that you consider c = b+\epsilon where 0 &lt; \epsilon &lt; 1 is some rather small rational quantity. We want to find out just how small it has to be so that not only b^2 but also c^2 &lt; 2.

    I showed that *c^2 &lt; b^2 + \epsilon(2b+1)*. You want, however, c^2 &lt; 2.

    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 \epsilon must satisfy. Find that condition in the form:

    \epsilon &lt; \text{something depending only on b}

    3. Now you've found the condition for \epsilon, write the argument in the correct order.

    [I've edited this about 8 times - I hope it now makes sense]
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    (Original post by atsruser)
    Well, you identified that the b \in \mathbb{Q} must be less than or equal to \sqrt{2} (though you haven't proved it), or equivalently that b^2 \le 2. This follows since if we had \sqrt{2} &lt; b \Rightarrow 2 &lt; b^2 then we could find u \in \mathbb{Q} with 2 &lt; u &lt; b^2 (why?) so with x^2 = u then x could not be in S.
    Can we find the u \in \mathbb{Q} where 2 &lt; u &lt; b^2 with the same idea I used before? By saying

    u = \dfrac{2+b^2}{2}

    So if

    x^2 = u

    \Rightarrow x = \dfrac{\sqrt{2+b^2}}{\sqrt{2}}

    which is not rational and so x is not in S? Is that correct? If so I'm really sorry but I don't really see what it's getting at because isn't my x^2 supposed to be less than or equal to 2 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 b, you can find  x \in S with x &gt; b, contradicting the suggested hypothesis.

    I suggested that you consider c = b+\epsilon where 0 &lt; \epsilon &lt; 1 is some rather small rational quantity. We want to find out just how small it has to be so that not only b^2 but also c^2 &lt; 2.

    I showed that *c^2 &lt; b^2 + \epsilon(2b+1)*. You want, however, c^2 &lt; 2.
    All of this makes perfect sense.

    (Original post by atsruser)

    1. What do you need to stick on the end of the starred inequality to ensure that?
    Could I just say:

    c^2 &lt; b^2 + \epsilon (2b+1) &lt; 2 ... (1)

    (Original post by atsruser)

    2. Once you have stuck that thing on, you have a condition that \epsilon must satisfy. Find that condition in the form:

    \epsilon &lt; \text{something depending only on b}
    So if I rearrange equation (1) I get:

    \epsilon (2b+1) &lt; 2 - b^2

    for -\frac{1}{2} &lt; b then \epsilon &lt; \dfrac{ 2 - b^2}{2b+1}

    but if b &lt; -\frac{1}{2} then \epsilon &gt; \dfrac{2 - b^2}{2b + 1}

    If we assume b &lt; -\frac{1}{2}:

    \Rightarrow c &gt; b + \dfrac{2 - b^2}{2b + 1}

    As we know that -\sqrt{2} &lt; b &lt; -\frac{1}{2} we know that 2 - b^2 is always positive. However, we'll be diving it by 2b +1 which is negative and so \dfrac{2 - b^2}{2b +1} is negative, meaning b + \dfrac{2 - b^2}{2b +1} is always less than b and so c &gt; "some number less than b". This doesn't seem helpful.

    If we assume -\frac{1}{2}&lt;b then c &lt; b + \dfrac{2 - b^2}{2b +1}. We know that -\dfrac{1}{2} &lt; b &lt; \sqrt{2} and so \dfrac{2 - b^2}{2b +1 } is always positive and so c &lt; "some number more than b". 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]
    Thank you very much! I really appreciate this I doubt I'd be able to figure this out with just the books.
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    (Original post by so it goes)
    ..
    The case b < -1/2 (or indeed b < 0) is a non-issue. 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 b \in S 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" \sqrt{2}, but since it's rational, we're mentally thinking the only thing that can work is if b is very close to \sqrt{2}.

    So the key argument is what happens when b is close to \sqrt{2}. 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].
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    I think I'd approach it like this:

    1. We have S=\{ x \in \mathbb{Q} : x^2 &lt; 2 \}

    Since f(x) = x^2 is an even function, we can say by symmetry that if S were a rational interval then it must look like S= \{ x \in \mathbb{Q} : -b \le x \le b \} i.e. whereas the question suggested an interval of the form a \le x \le b, we don't, in fact, need to think about this any more.

    2. Now we can concentrate on \{x \in \mathbb{Q} : 0 &lt; x \le b \} 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 b is \sqrt{2}. We must have one of:

    b &lt; \sqrt{2} \Rightarrow b^2 &lt; 2
    b = \sqrt{2} \Rightarrow b^2 = 2
    b &gt; \sqrt{2} \Rightarrow b^2 &gt; 2

    But b=\sqrt{2} \notin \mathbb{Q}, we can rule this out as a candidate. We now have to rule out the other two candidate ranges too.

    4. For the case b^2 &lt; 2, 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 b with b^2 &lt; 2, we can find c &gt; b with c \in \mathbb{Q}, c^2 &lt; 2 i.e. now matter how close b^2 is to 2, then we can always fit another square rational in between it and 2.

    We are thinking only about the range 0 &lt; x &lt; b, remember. So consider 0 &lt; c = b+\epsilon with 0 &lt; \epsilon &lt; 1. Then

    c^2 = (b+\epsilon)^2 = b^2 + 2b\epsilon + \epsilon^2 = b^2 +\epsilon(2b+\epsilon) &lt;  b^2 +\epsilon(2b+1)

    Now if \epsilon &lt; \frac{2-b^2}{2b+1} then c^2 &lt;  b^2 + \epsilon(2b+1) &lt;  b^2 +\frac{2-b^2}{2b+1}(2b+1) = 2

    So with that value for \epsilon we can ensure that b &lt; c and c^2 &lt; 2 and c \in \mathbb{Q} (since 0 &lt; c = b+\epsilon).

    Note that it looked like I pulled the value for \epsilon out of a hat, but I showed the justification in an earlier post.

    So no rational b with b^2 &lt; 2 will allow us to define the interval.

    5. We now must rule out the case with b^2 &gt; 2. 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 b^2 &gt; 2 then you can find b^2 &gt; c^2 &gt; 2 i.e. that the hypothesised interval will admit points that do not meet the definition of S.

    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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    (Original post by DFranklin)
    The case b < -1/2 (or indeed b < 0) is a non-issue. 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 b \in S 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" \sqrt{2}, but since it's rational, we're mentally thinking the only thing that can work is if b is very close to \sqrt{2}.

    So the key argument is what happens when b is close to \sqrt{2}. 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].
    Ahh, okay thank you. I don't know why I'm finding it so tricky to wrap my head around this question

    Thanks again!
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    (Original post by atsruser)
    I think I'd approach it like this:

    1. We have S=\{ x \in \mathbb{Q} : x^2 &lt; 2 \}

    Since f(x) = x^2 is an even function, we can say by symmetry that if S were a rational interval then it must look like S= \{ x \in \mathbb{Q} : -b \le x \le b \} i.e. whereas the question suggested an interval of the form a \le x \le b, we don't, in fact, need to think about this any more.

    2. Now we can concentrate on \{x \in \mathbb{Q} : 0 &lt; x \le b \} 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 b is \sqrt{2}. We must have one of:

    b &lt; \sqrt{2} \Rightarrow b^2 &lt; 2
    b = \sqrt{2} \Rightarrow b^2 = 2
    b &gt; \sqrt{2} \Rightarrow b^2 &gt; 2

    But b=\sqrt{2} \notin \mathbb{Q}, we can rule this out as a candidate. We now have to rule out the other two candidate ranges too.

    4. For the case b^2 &lt; 2, 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 b with b^2 &lt; 2, we can find c &gt; b with c \in \mathbb{Q}, c^2 &lt; 2 i.e. now matter how close b^2 is to 2, then we can always fit another square rational in between it and 2.

    We are thinking only about the range 0 &lt; x &lt; b, remember. So consider 0 &lt; c = b+\epsilon with 0 &lt; \epsilon &lt; 1. Then

    c^2 = (b+\epsilon)^2 = b^2 + 2b\epsilon + \epsilon^2 = b^2 +\epsilon(2b+\epsilon) &lt;  b^2 +\epsilon(2b+1)

    Now if \epsilon &lt; \frac{2-b^2}{2b+1} then c^2 &lt;  b^2 + \epsilon(2b+1) &lt;  b^2 +\frac{2-b^2}{2b+1}(2b+1) = 2

    So with that value for \epsilon we can ensure that b &lt; c and c^2 &lt; 2 and c \in \mathbb{Q} (since 0 &lt; c = b+\epsilon).

    Note that it looked like I pulled the value for \epsilon out of a hat, but I showed the justification in an earlier post.

    So no rational b with b^2 &lt; 2 will allow us to define the interval.
    Ahh that's clever, thank you. I think I understand up to this point.

    (Original post by atsruser)

    5. We now must rule out the case with b^2 &gt; 2. 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 b^2 &gt; 2 then you can find b^2 &gt; c^2 &gt; 2 i.e. that the hypothesised interval will admit points that do not meet the definition of S.
    What I don't understand is this point. Why do I need to worry about b^2 &gt; 2 when isn't it defined in the question that b^2 must be less than 2 to be in the set S? What am I missing?

    Thank you again, and sorry I'm struggling so much with this.
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    (Original post by DFranklin)
    ...

    (Original post by atsruser)
    ...
    Would it be possible for me to approach the problem like this?

    If b^2 &lt; 2 let \mu = 2 - b^2

    Now can't I find some number less than \mu by expressing it as \dfrac{1}{n} which I can make as small as I like by increasing n sufficiently?

    I will then have found b^2 &lt; b^2 + \dfrac{1}{n} &lt; 2

    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 re-written \epsilon as \dfrac{1}{n}?

    Thank you
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    (Original post by so it goes)
    Why do I need to worry about b^2 &gt; 2 when isn't it defined in the question that b^2 must be less than 2 to be in the set S? What am I missing?
    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.

    (Original post by so it goes)
    Would it be possible for me to approach the problem like this?

    If b^2 &lt; 2 let \mu = 2 - b^2

    Now can't I find some number less than \mu by expressing it as \dfrac{1}{n} which I can make as small as I like by increasing n sufficiently?

    I will then have found b^2 &lt; b^2 + \dfrac{1}{n} &lt; 2
    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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    (Original post by so it goes)
    What I don't understand is this point. Why do I need to worry about b^2 &gt; 2 when isn't it defined in the question that b^2 must be less than 2 to be in the set S? What am I missing?
    We want to show that no b \in \mathbb{Q} exists such that the set S can be written in the form S=\{ x \in \mathbb{Q} : -b \le x \le b \}

    We have already dismissed the possibilities b^2 \le 2. The only remaining possibility is that b^2 &gt; 2. We want to make the argument that no matter how close b^2 may be to 2, it still isn't close enough to stop us finding a rational x \in \mathbb{Q} : x &lt; b, x^2 &gt; 2.

    If we can do that, then by the first inequality, we have x &lt; b \Rightarrow x \in S, but by the second x^2 &gt; 2 \Rightarrow x \notin S which is a contradiction, which tells us that our hypothesis that such a b exists is false.

    At this point, you may say that it is "obvious" that if b^2 &gt; 2 then we can find such an x, 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 b^2 &lt; 2 then we can find x &gt; b, x^2 &lt; 2.

    Note that you may claim that since both b^2 \in \mathbb{Q}, 2 \in \mathbb{Q}, then you can certainly find a rational u : 2 &lt; u &lt; b^2. But in fact that's not good enough: we need to find a rational x^2 : 2 &lt; x^2 &lt; b^2 and for that I think we need to prove it.

    Thank you again, and sorry I'm struggling so much with this.
    There's no need to apologise. This is incredibly tricky stuff and these kinds of arguments require a totally different set of skills to the ones that you learn at A level.
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    (Original post by so it goes)
    I will then have found b^2 &lt; b^2 + \dfrac{1}{n} &lt; 2
    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 b &lt; x with x \in \mathbb{Q} and x^2 &lt; 2.

    This is not the same as finding b^2 &lt; u &lt; 2 with u \in \mathbb{Q} since this does not imply that x = \sqrt{u} is in \mathbb{Q}.

    You can approach it this way, but you would have to show that you can find b^2 &lt; p^2/q^2 &lt; 2 with p,q \in \mathbb{Q} since then x = \sqrt{p^2/q^2} = p/q \in \mathbb{Q}.
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    (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.
    Ahh, okay. I think I understand these points now. Thank you
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    (Original post by atsruser)
    We want to show that no b \in \mathbb{Q} exists such that the set S can be written in the form S=\{ x \in \mathbb{Q} : -b \le x \le b \}

    We have already dismissed the possibilities b^2 \le 2. The only remaining possibility is that b^2 &gt; 2. We want to make the argument that no matter how close b^2 may be to 2, it still isn't close enough to stop us finding a rational x \in \mathbb{Q} : x &lt; b, x^2 &gt; 2.

    If we can do that, then by the first inequality, we have x &lt; b \Rightarrow x \in S, but by the second x^2 &gt; 2 \Rightarrow x \notin S which is a contradiction, which tells us that our hypothesis that such a b exists is false.
    Okay, so if I'm understanding this correctly would it look something like this?

    2 &lt; x^2 &lt; b^2 or \sqrt{2} &lt; x &lt; b

    let x = b - \epsilon where \epsilon \in \mathbb{Q} and 0&lt;\epsilon &lt; 1

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

    \Rightarrow 2 &lt; b^2 - 2b\epsilon + \epsilon ^2

    \Rightarrow 2 &lt; b^2 + \epsilon (\epsilon - 2b) but 0 &lt; \epsilon &lt;1

    \Rightarrow 2 &lt; b^2 + \epsilon (1 -2b)

    \Right \epsilon &gt; \dfrac{2 - b^2}{1 - 2b}

    \Rightarrow x &gt; b - \dfrac{2 - b^2}{1 - 2b}

    \Rightarrow x &gt; \dfrac{(b-2)(b+1)}{2b - 1}

    \Rightarrow x^2 &gt; \dfrac{(b-2)^2 (b+1)^2}{(2b - 1)^2}

    Now, if we assume b \in S then x \in S as x &lt; b and x is rational.

    However, x^2 &gt; 2 therefore x \not\in S

    We therefore have a contradiction, meaning our original assumption was wrong, therefore b \not\in S if b^2 &gt;2

    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 b &lt; x with x \in \mathbb{Q} and x^2 &lt; 2.

    This is not the same as finding b^2 &lt; u &lt; 2 with u \in \mathbb{Q} since this does not imply that x = \sqrt{u} is in \mathbb{Q}.

    You can approach it this way, but you would have to show that you can find b^2 &lt; p^2/q^2 &lt; 2 with p,q \in \mathbb{Q} since then x = \sqrt{p^2/q^2} = p/q \in \mathbb{Q}.
    Yes that makes sense. Hopefully I've understood this as well as I think and havn't made this mistake in my above work.

    Thank you.
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    (Original post by so it goes)
    Have I understood you correctly?
    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 b \in S 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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    (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 b \in S 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?]
    Ahh okay, so instead of saying this at the end:

    (Original post by so it goes)
    \Rightarrow x &gt; \dfrac{(b-2)(b+1)}{2b - 1}

    \Rightarrow x^2 &gt; \dfrac{(b-2)^2 (b+1)^2}{(2b - 1)^2}

    Now, if we assume b \in S then x \in S as x &lt; b and x is rational.

    However, x^2 &gt; 2 therefore x \not\in S

    We therefore have a contradiction, meaning our original assumption was wrong, therefore b \not\in S if b^2 &gt;2
    Should it be more like this?

    \Rightarrow x &gt; \dfrac{(b-2)(b+1)}{2b - 1}

    \Rightarrow x^2 &gt; \dfrac{(b-2)^2 (b+1)^2}{(2b - 1)^2}

    x^2 &gt; 2 therefore x \not\in S

    as 2&lt;x^2&lt;b^2 this implies that b \not\in S

    Does that work?

    Thank you so much for this!
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    (Original post by so it goes)
    Ahh okay, so instead of saying this at the end:



    Should it be more like this?

    \Rightarrow x &gt; \dfrac{(b-2)(b+1)}{2b - 1}

    \Rightarrow x^2 &gt; \dfrac{(b-2)^2 (b+1)^2}{(2b - 1)^2}

    x^2 &gt; 2 therefore x \not\in S

    as 2&lt;x^2&lt;b^2 this implies that b \not\in S
    No. Why do you care whether b is in S?
 
 
 
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