showing that a set S in not a rational interval

No. Why do you care whether b is in S?

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.

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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 $x < b \Rightarrow x \in P$

...

similar to the BMO maths problem in 2005 or 2006 i think or 2004.


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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 $f: x \mapsto x^2$?)

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?)

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?

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 $\epsilon < b - \sqrt{2}$?

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?

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.

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.

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 $(\pm \sqrt{2}, 2)$

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.

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?