Proof that any upper bound of $(0, \sqrt 2]$ has to be greater than $\sqrt 2$ in the domain of rationals.

Question

I have to prove that rationals are not complete. And I do this by showing that there exists no lowest upper bound of $A = \{x \in \mathbb{Q} | x > 0 \wedge x^2 \leq 2\}$ in the domain of rationals.

To do that, I show that any upper bound of $A$ is greater $\sqrt 2$, then I show that given an upper bound, I can always find a lower one.

I have come up with a proof that any upper bound of $A$ has to be greater than $\sqrt 2$ and this method can be later extended to the second part of the proof. Can someone check whether my work is correct?

Proof that any upper bound of $A$ has to be greater than $\sqrt 2$:

Let $x \in \mathbb{Q}$ be an upper bound of $A$ such that $x < \sqrt 2$.

We show that we can find a rational $y$ such that $y > x$ and $y < \sqrt 2$. Hence, $y$ lies in $A$ but is greater than $x$, which means that what we've assumed is false.

Consider the following algorithm to find $y$:

1. Write down the decimal representation of $x$ (Prefix and suffix with zeros if necessary). Since $x < \sqrt 2$, there must exist some digit of $x$ which is lesser than corresponding digit of $\sqrt 2$

2. Let $y$ be the number achieved by setting that digit to be the corresponding digit of $\sqrt 2$ and terminate there.

3. $y < \sqrt 2$ because it terminates ($y$ is just a prefix of $\sqrt 2$).

4. $y > x$ because it is same as $x$ until that digit, where we explicitly increased the digit to be the larger, original digit of $\sqrt 2$.

5. $y$ is rational because it terminates.

So, $x$ can't be an upper bound of $A$.

Then, a similar procedure can be used to prove that no lowest upper bound exists (by finding a $y$ such that $\sqrt 2 < y < \text {upper-bound}$).

(Note that since $\sqrt 2$ is irrational, $x, y$ or any rational number can't equal $\sqrt 2$. So any rational number is either greater or smaller than $\sqrt 2$. Any rational number can't equal $\sqrt 2$.)

Is this correct?

Thanks

Answer 1

let $S=\{x:(0<x\leq\sqrt{2})\land x\in Q\}$

First of all any upper bound $a$ of $S$ has to satisfy $a\geq \sqrt{2}$, by definition.

So the least upper bound $q$ has to be the least element in the set of upper bounds.

We call it supremum or $\sup$.

Your direction of proof is a correct reasoning.

However i would like to put it more rigorously.

assume on the contrary that there exists: $q=\sup_{x\in Q}{(0<x\leq\sqrt{2})},\;q\in Q$

let wlog, $q=\frac{m}{n}$ and $gcd(m,n)=1$

then by definition of $\sup$ : $(\forall x\in S)(x\leq q)\land (\forall \epsilon>0)(\exists y)(y\in S \land (y+\epsilon>q))$

now $\sqrt{2}\notin Q$

the proof's basis is that: $(\exists x\in S)(x>q) \lor (\exists \epsilon>0)(\forall y \in S)(y+\epsilon\leq q)$

since $\sqrt{2}$ is irrational we can always pick an open interval of 2 finite rational expansions of $\sqrt{2}$ $(a,b)$ s.t. $\sqrt{2}\in(a,b)$ and $b-a\leq 10^{-p}$ for whatever $p\in N$ we choose.

so we can always choose a $p$ s.t. $\epsilon<\frac{10^{-p}}{2}$ and $y+\epsilon\leq\sqrt{2}, \forall y\in S$.