Formally show that the set $\mathcal{S} = \left\{x\in\mathbb{Q} : x>0 , x^2<2 \right\}$ does not have a least upper bound in $\mathbb{Q}$. So, while teaching in our Analysis class, our professor said that an informal proof of the above can be found in Rudin's Analysis book. And here is what was mentioned:

I mean, I get what is written here. But our professor said that the claim can be proved formally using $\leq$ ordering on $\mathbb{Q}$ and properties of field. I am a little bit unsure as to how to "formalize" this thing. Could someone clarify/explain.
 A: I am not sure if this is the answer that you are searching for, but in any case what I am about to say seems pretty useful.
Firstly, lets prove a lemma:
Lemma: For any 2 real numbers $a$ and $b$, $0<a<b$ we can find a rational number $r$ such that $r\in(a,b]$
Proof:
Let $C\in \mathbb{N}$ big enough, such that $C(b-a)>1 \Longleftrightarrow Cb-Ca>1$ so $\lfloor Cb\rfloor-\lfloor Ca\rfloor\ge1$
Let $r=\frac{\lfloor Cb\rfloor}{C}$, which is obviously rational.
We have:
$a<\frac{\lfloor Ca\rfloor+1}{C}\leq\frac{\lfloor Cb\rfloor}{C}\leq b$
So $r\in (a,b]$
Now, back to your problem, lets assume that $A$ contains an rational number $q$ such that $q$ is the largest number from $A$
We know that $q<\sqrt2$ so from the lemma we find $r\in (q,\sqrt2$] which is rational but $\sqrt2$ isn't rational so $r\in(q,\sqrt2)$, thus we have found a number from $A$ which is bigger than $q$ so we have contradiction.
For $B$, do the exact same, but use the ceiling function for the lemma.
A: (i). $S$ has no largest member. So no member of $S$ is $\lub(S).$
(ii).  $T=\{1/x:x\in S\}$ has no smallest member, and every $y\in T$ is an upper bound for $S$ because $T=\{y\in\Bbb Q^+: y^2>2\}$. So no $y\in T$ is   $\lub (S).$
(iii). $S\cup T=\Bbb Q^+$ because there is no $\sqrt 2$ in $\Bbb Q.$
Hero (Heron) of Alexandria (circa 10-70 A.D.) wrote of a method for approximating square roots: If $A>0$ and $y_0>0,$ let $y_{n+1}=\frac 1 2 (y_n+\frac {A}{y_n}).$  If $0<y_0\ne \sqrt A$ then $\sqrt A<y_{n+2}<y_{n+1}.$
In particular, if $A=2$ and if $y_0$ is any member of $T$ then $2<y_1^2<y_0^2$ and $y_1\in T.$  Proving this is one way to show that $\min (T)$ does not exist (and hence $\max (S)$ does not exist).
Isaac Newton generalized Heron's method (see Newton's Method) to approximate solutions to a wide variety of equations.
