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 $\mathbb{Q}$.

Question

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.

Answer 1

(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.

Answer 2

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.