showing that a set S in not a rational interval

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



Therefore for any rational less than $\sqrt{2}$ I can generate a bigger one that's still less than $\sqrt{2}$.

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



Therefore for any rational less than $\sqrt{2}$ I can generate a bigger one that's still less than $\sqrt{2}$

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 $\{b \in \mathbb{Q} : b^2 < 2\}$, you can find $c \in \mathbb{Q} > b$ such that $c^2 < 2$

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

$c=b+\epsilon$ with $\epsilon \in \mathbb{Q}, 0 < \epsilon < 1$

so

$c^2 = (b+\epsilon)^2 = b^2+2b\epsilon + \epsilon^2 = b^2 +\epsilon(2b+\epsilon) < b^2 + \epsilon(2b+1)$

Now what condition must hold on $\epsilon$?

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 $\{b \in \mathbb{Q} : b^2 < 2\}$, you can find $c \in \mathbb{Q} > b$ such that $c^2 < 2$

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

$c=b+\epsilon$ with $\epsilon \in \mathbb{Q}, 0 < \epsilon < 1$

so

$c^2 = (b+\epsilon)^2 = b^2+2b\epsilon + \epsilon^2 = b^2 +\epsilon(2b+\epsilon) < b^2 + \epsilon(2b+1)$

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

$c^2 < b^2 + \epsilon (2b + 1) < 2 + \epsilon (2b + 1)$

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 $\sqrt{2} < b \Rightarrow 2 < b^2$ then we could find $u \in \mathbb{Q}$ with $2 < u < b^2$ (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 $x > b$, contradicting the suggested hypothesis.

I suggested that you consider $c = b+\epsilon$ where $0 < \epsilon < 1$ 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 $c^2 < 2$.

I showed that *$c^2 < b^2 + \epsilon(2b+1)$*. You want, however, $c^2 < 2$.

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:

$\epsilon < \text{something depending only on b}$

[I've edited this about 8 times - I hope it now makes sense]

..

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].

I think I'd approach it like this:

1. We have $S=\{ x \in \mathbb{Q} : x^2 < 2 \}$

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 $\{x \in \mathbb{Q} : 0 < x \le b \}$ 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:

$b < \sqrt{2} \Rightarrow b^2 < 2$
$b = \sqrt{2} \Rightarrow b^2 = 2$
$b > \sqrt{2} \Rightarrow b^2 > 2$

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 $b^2 < 2$, 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 $b^2 < 2$, we can find $c > b$ with $c \in \mathbb{Q}, c^2 < 2$ 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 $0 < x < b$, remember. So consider $0 < c = b+\epsilon$ with $0 < \epsilon < 1$. Then

$c^2 = (b+\epsilon)^2 = b^2 + 2b\epsilon + \epsilon^2 = b^2 +\epsilon(2b+\epsilon) < b^2 +\epsilon(2b+1)$

Now if $\epsilon < \frac{2-b^2}{2b+1}$ then $c^2 < b^2 + \epsilon(2b+1) < b^2 +\frac{2-b^2}{2b+1}(2b+1) = 2$

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

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 $b^2 < 2$ will allow us to define the interval.

5. We now must rule out the case with $b^2 > 2$. 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 $b^2 > 2$ then you can find $b^2 > c^2 > 2$ 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 $b^2 > 2$ 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.

I will then have found $b^2 < b^2 + \dfrac{1}{n} < 2$

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.

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 $b^2 > 2$. 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 $x \in \mathbb{Q} : x < b, x^2 > 2$.

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

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 $b < x$ with $x \in \mathbb{Q}$ and $x^2 < 2$.

This is not the same as finding $b^2 < u < 2$ 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 $b^2 < p^2/q^2 < 2$ with $p,q \in \mathbb{Q}$ since then $x = \sqrt{p^2/q^2} = p/q \in \mathbb{Q}$.

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?]

$\Rightarrow x > \dfrac{(b-2)(b+1)}{2b - 1}$

$\Rightarrow x^2 > \dfrac{(b-2)^2 (b+1)^2}{(2b - 1)^2}$

Now, if we assume $b \in S$ then $x \in S$ as $x < b$ and $x$ is rational.

However, $x^2 > 2$ therefore $x \not\in S$

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