Show that S has the property that if x,y∈S and x≤z≤y then also z∈S. Show also that S is not a rational interval - that is, there does not exist a,b∈Q such that
S={x∈Q∣a<x<b}
or S={x∈Q∣a≤x≤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∈S which implies that x2≤2 and y2≤2. As x≤z≤y then this implies that z2≤2 and therefore z∈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 nm is the biggest fraction below 2. Now surely a number that is bigger that nm but less than 2 is
21(nm+2)
Therefore for any rational less than 2 I can generate a bigger one that's still less than 2, there is therefore no possible value of b in
S={x∈Q∣a<x<b}
that would work. Is this sufficient for the second part? Have I expressed it in the correct way?
We know that x,y∈S which implies that x2≤2 and y2≤2. As x≤z≤y then this implies that z2≤2 and therefore z∈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 nm is the biggest fraction below 2. Now surely a number that is bigger that nm but less than 2 is
21(nm+2)
Therefore for any rational less than 2 I can generate a bigger one that's still less than 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).
For the second part I was thinking of a sort of proof by contradiction? Firstly, assume that nm is the biggest fraction below 2. Now surely a number that is bigger that nm but less than 2 is
21(nm+2)
Therefore for any rational less than 2 I can generate a bigger one that's still less than 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∈Q:b2<2}, you can find c∈Q>b such that c2<2
Since c will clearly only be a "bit" bigger than b, write:
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).
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∈Q:b2<2}, you can find c∈Q>b such that c2<2
Since c will clearly only be a "bit" bigger than b, write:
c=b+ϵ with ϵ∈Q,0<ϵ<1
so
c2=(b+ϵ)2=b2+2bϵ+ϵ2=b2+ϵ(2b+ϵ)<b2+ϵ(2b+1)
Now what condition must hold on ϵ?
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.
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 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
c2<b2+ϵ(2b+1)<2+ϵ(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∈Q must be less than or equal to 2 (though you haven't proved it), or equivalently that b2≤2. This follows since if we had 2<b⇒2<b2 then we could find u∈Q with 2<u<b2 (why?) so with x2=u then x could not be in S.
Now you want to show that for any such choice of b, you can find x∈S with x>b, contradicting the suggested hypothesis.
I suggested that you consider c=b+ϵ where 0<ϵ<1 is some rather small rational quantity. We want to find out just how small it has to be so that not only b2 but also c2<2.
I showed that *c2<b2+ϵ(2b+1)*. You want, however, c2<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 ϵ must satisfy. Find that condition in the form:
ϵ<something depending only on b
3. Now you've found the condition for ϵ, write the argument in the correct order.
[I've edited this about 8 times - I hope it now makes sense]
Well, you identified that the b∈Q must be less than or equal to 2 (though you haven't proved it), or equivalently that b2≤2. This follows since if we had 2<b⇒2<b2 then we could find u∈Q with 2<u<b2 (why?) so with x2=u then x could not be in S.
Can we find the u∈Q where 2<u<b2 with the same idea I used before? By saying
u=22+b2
So if
x2=u
⇒x=22+b2
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 x2 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.
Now you want to show that for any such choice of b, you can find x∈S with x>b, contradicting the suggested hypothesis.
I suggested that you consider c=b+ϵ where 0<ϵ<1 is some rather small rational quantity. We want to find out just how small it has to be so that not only b2 but also c2<2.
I showed that *c2<b2+ϵ(2b+1)*. You want, however, c2<2.
2. Once you have stuck that thing on, you have a condition that ϵ must satisfy. Find that condition in the form:
ϵ<something depending only on b
So if I rearrange equation (1) I get:
ϵ(2b+1)<2−b2
for −21<b then ϵ<2b+12−b2
but if b<−21 then ϵ>2b+12−b2
If we assume b<−21:
⇒c>b+2b+12−b2
As we know that −2<b<−21 we know that 2−b2 is always positive. However, we'll be diving it by 2b+1 which is negative and so 2b+12−b2 is negative, meaning b+2b+12−b2 is always less than b and so c> "some number less than b". This doesn't seem helpful.
If we assume −21<b then c<b+2b+12−b2. We know that −21<b<2 and so 2b+12−b2 is always positive and so c< "some number more than b". This also doesn't seem helpful.
I must be missing something but I can't spot what.
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∈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" 2, but since it's rational, we're mentally thinking the only thing that can work is if b is very close to 2.
So the key argument is what happens when b is close to 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].
Since f(x)=x2 is an even function, we can say by symmetry that if S were a rational interval then it must look like S={x∈Q:−b≤x≤b} i.e. whereas the question suggested an interval of the form a≤x≤b, we don't, in fact, need to think about this any more.
2. Now we can concentrate on {x∈Q:0<x≤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 2. We must have one of:
b<2⇒b2<2 b=2⇒b2=2 b>2⇒b2>2
But b=2∈/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 b2<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 b2<2, we can find c>b with c∈Q,c2<2 i.e. now matter how close b2 is to 2, then we can always fit another square rational in between it and 2.
We are thinking only about the range 0<x<b, remember. So consider 0<c=b+ϵ with 0<ϵ<1. Then
c2=(b+ϵ)2=b2+2bϵ+ϵ2=b2+ϵ(2b+ϵ)<b2+ϵ(2b+1)
Now if ϵ<2b+12−b2 then c2<b2+ϵ(2b+1)<b2+2b+12−b2(2b+1)=2
So with that value for ϵ we can ensure that b<c and c2<2 and c∈Q (since 0<c=b+ϵ).
Note that it looked like I pulled the value for ϵ out of a hat, but I showed the justification in an earlier post.
So no rational b with b2<2 will allow us to define the interval.
5. We now must rule out the case with b2>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 b2>2 then you can find b2>c2>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.
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∈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" 2, but since it's rational, we're mentally thinking the only thing that can work is if b is very close to 2.
So the key argument is what happens when b is close to 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
Since f(x)=x2 is an even function, we can say by symmetry that if S were a rational interval then it must look like S={x∈Q:−b≤x≤b} i.e. whereas the question suggested an interval of the form a≤x≤b, we don't, in fact, need to think about this any more.
2. Now we can concentrate on {x∈Q:0<x≤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 2. We must have one of:
b<2⇒b2<2 b=2⇒b2=2 b>2⇒b2>2
But b=2∈/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 b2<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 b2<2, we can find c>b with c∈Q,c2<2 i.e. now matter how close b2 is to 2, then we can always fit another square rational in between it and 2.
We are thinking only about the range 0<x<b, remember. So consider 0<c=b+ϵ with 0<ϵ<1. Then
c2=(b+ϵ)2=b2+2bϵ+ϵ2=b2+ϵ(2b+ϵ)<b2+ϵ(2b+1)
Now if ϵ<2b+12−b2 then c2<b2+ϵ(2b+1)<b2+2b+12−b2(2b+1)=2
So with that value for ϵ we can ensure that b<c and c2<2 and c∈Q (since 0<c=b+ϵ).
Note that it looked like I pulled the value for ϵ out of a hat, but I showed the justification in an earlier post.
So no rational b with b2<2 will allow us to define the interval.
Ahh that's clever, thank you. I think I understand up to this point.
5. We now must rule out the case with b2>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 b2>2 then you can find b2>c2>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 b2>2 when isn't it defined in the question that b2 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.
Would it be possible for me to approach the problem like this?
If b2<2 let μ=2−b2
Now can't I find some number less than μ by expressing it as n1 which I can make as small as I like by increasing n sufficiently?
I will then have found b2<b2+n1<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 ϵ as n1?
Why do I need to worry about b2>2 when isn't it defined in the question that b2 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.
Would it be possible for me to approach the problem like this?
If b2<2 let μ=2−b2
Now can't I find some number less than μ by expressing it as n1 which I can make as small as I like by increasing n sufficiently?
I will then have found b2<b2+n1<2You'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.
What I don't understand is this point. Why do I need to worry about b2>2 when isn't it defined in the question that b2 must be less than 2 to be in the set S? What am I missing?
We want to show that no b∈Q exists such that the set S can be written in the form S={x∈Q:−b≤x≤b}
We have already dismissed the possibilities b2≤2. The only remaining possibility is that b2>2. We want to make the argument that no matter how close b2 may be to 2, it still isn't close enough to stop us finding a rational x∈Q:x<b,x2>2.
If we can do that, then by the first inequality, we have x<b⇒x∈S, but by the second x2>2⇒x∈/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 b2>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 b2<2 then we can find x>b,x2<2.
Note that you may claim that since both b2∈Q,2∈Q, then you can certainly find a rational u:2<u<b2. But in fact that's not good enough: we need to find a rational x2:2<x2<b2 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.
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
We want to show that no b∈Q exists such that the set S can be written in the form S={x∈Q:−b≤x≤b}
We have already dismissed the possibilities b2≤2. The only remaining possibility is that b2>2. We want to make the argument that no matter how close b2 may be to 2, it still isn't close enough to stop us finding a rational x∈Q:x<b,x2>2.
If we can do that, then by the first inequality, we have x<b⇒x∈S, but by the second x2>2⇒x∈/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<x2<b2 or 2<x<b
let x=b−ϵ where ϵ∈Q and 0<ϵ<1
⇒x2=(b−ϵ)2
⇒2<b2−2bϵ+ϵ2
⇒2<b2+ϵ(ϵ−2b) but 0<ϵ<1
⇒2<b2+ϵ(1−2b)
Unparseable latex formula:
\Right \epsilon > \dfrac{2 - b^2}{1 - 2b}
⇒x>b−1−2b2−b2
⇒x>2b−1(b−2)(b+1)
⇒x2>(2b−1)2(b−2)2(b+1)2
Now, if we assume b∈S then x∈S as x<b and x is rational.
However, x2>2 therefore x∈S
We therefore have a contradiction, meaning our original assumption was wrong, therefore b∈S if b2>2
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∈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?]
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∈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?]