subset of rationals doesn't have supremum proof understanding I'm trying to understand a proof I'm reading regarding rational numbers and supremums.
(*) Every nonempty bounded from above set of real numbers M has a least upper bound which is denoted supM.
We'll show that (*) is not true for the set of rational numbers.
Let $M = \{x \in \mathbb Q : x^2 \le 2\}$ . The number 2 is an upper bound of M which implies that the set M is bounded from above. Let l be the least upper bound of M and lets assume that $l \in \mathbb Q$. It is clear that $l > 1$. We'll show that $l^2 < 2$ is impossible. And really, then $m=\frac{2+2l}{2+l} \in \mathbb Q$. Lets look at 
$$
2 - (\frac{2+2l}{2+l})^2 = \frac{2(2+l)^2-(2+2l)^2}{(2+l)^2}=\frac{2-l^2}{(2+l)^2} > 0
$$
Which implies that $m^2 < 2$ and $m \in \mathbb Q$. But $ m-l>0 \implies m > l$, which shows that l cannot be an upper bound of M.
From there, there's an assumption that $l > 2$ and the proof continues. I think that if I get the first part I'll understand the rest of the proof.
My questions are:


*

*I feel like the statement is confusing. Isn't the set of rational a subset of the set of real numbers? Is the statement poorly written or is there something I'm missing? What is the main idea of the statement and the proof?

*What's the idea of using the set M where every element is rational with square less than or equal to 2? 

*Where did m come from in this proof?It feels like the author knew something which for me needs some derivation.


I'm sorry if my questions are somehow vague or unclear, but I'm trying to wrap my head around this statement and proof. Maybe if I get even the main idea the rest of the text will come clear to me.
Thanks in advance!
 A: The confusion is because the statement is not very clear. It may well be misinterpreted in a way that it holds for $\mathbb Q$ as well.
What it is missing is that it should require the supremum to be in the set of real numbers, the same set as $M$ is subset of. This in turn makes the statement to be disproved to be reformulated as we would require replacement of both sets.
If we reformulate it:

(*) Every nonempty bounded from above set $M\subset \mathbb R$ has a least upper bound in $\mathbb R$ which is denoted $\sup M$.
We'll show that (*) is not true if we replace both occurences of $\mathbb R$ with $\mathbb Q$.

Now it's clearer why $\mathbb Q\subset\mathbb R$ won't help as we added the requirement that $\sup M\in\mathbb Q$ by replacing $\mathbb R$. The set certainly has a least upper bound, but it's not obvious that it would be a member of $\mathbb Q$.
The reason why they consider the set $M$ of numbers with squares less than $2$ is that the supposed supremum of such set would be $\sqrt2$ which is not a rational number.
As for the magic value of $m$ we start with the structure of the proof. What we're looking for is to exclude $l^2\ne 2$, we do this by showing that if $l^2<2$ we find a larger $m\in M$ which makes $l$ not an upper bound and if $l^2>2$ we find a smaller $m$ which still is an upper bound of $M$ that is $l$ is not the least upper bound. Then when we excluded $l^2\ne 2$ we are left with $l^2=2$ which can be excluded due to Aristotle-Euclid. 
Basically what we need to do in the first step construct a recursion formula for evere larger members of $M$. We know that this can be done if we have a recursion formula which limit is $\sqrt 2$. To find this we turn to numerical methods of solving equations, the secant method for solving $f(x) = x^2-2=0$ turns out to be useful here (since $f$ is convex and if we start with $f(a_0)<0$ and $f(b_0)>0$ we will get $b_j=b_0$ and $f(a_j)<0$). Taking $b_0=2$ and $a_0=l$ we get the step in the secant method:
$$m = c_1 = a_0 - {f(a_0)(b_0-a_0)\over f(b_0)-f(a_0)} = l - {f(l)(2-l)\over
(f(2)-f(l))} = l - {(l^2-2)(2-l)\over 2-(l^2 - 2)} = {2l+2\over 2+l}$$
of course knowing the properties of the secant method we know that this means that $m>l$ and $m^2<2$, but the above calculation shows this directly instead.
I guess the next step is the opposite to assume that if $l^2>2$ we find a contradiction too. Here we would turn to Newton-Raphsons method instead as it will yield points to the right of the solutions (if we start from the right).
We have $f'(x) = 2x$ so the step in NR method would be
$$m=x_1 = x_0 - {f(x_0)\over f'(x_0)} = l - {l^2-2\over 2l} = {l^2 + 2\over 2l}$$ 
Now doing the same trick as before we get:
$$m^2 - 2 = \left({l^2+2\over 2l}\right)^2-2 = \left({x^2-2\over 2l}\right)^2>0$$
also we can verify that $m<l$ by calculating $m-l$:
$$m-l = \cdots = {l^2-2\over 2l} > 0$$
Now we have excluded the possibilities $l^2>2$ and $l^2<2$ so what remains is only the case $l^2=2$ (which I assume you can prove has no rational solutions).
However often one doesn't do a proof this verbose since as we have seen we can given the expression of $m$ directly verify that it fulfills our requirements. That is the validity of the proof does not rely on us motivating how we found the expression for $m$ and since that is not needed in the proof that part is often left out of the proof.
A: The statement says that not every subset of Q has a sup in Q
indeed the sup of A is $\sqrt{2}$ which is not rational.
Imagine you have found a sup $l$ that is the "last" rational number such that $l^2<2$. The proof shows that it can be found a rational number $m$ greater than $l$, such that $m^2<2$, which is greater than $l$, so $l$ fails to be the sup.
$m$  is greater than $l$ because $l<\frac{2+2l}{2+l}$ indeed $2l+l^2<2l+2$ because $l^2<2$
and $m^2<2$ because
$$\frac{(2 l+2)^2}{(l+2)^2}<2\rightarrow 4l^2+8l+4<2l^2+8l+8$$
Hope this helps
