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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

MangoPizza
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    Often these sorts of properties of rational numbers are studied as a preliminary to defining the real numbers and their properties. If that's the context here, we wouldn't assume that $\sqrt{2}$ exists and has a decimal representation. – aschepler Oct 01 '22 at 18:06
  • Hi! Your use of the phrase "the upper bound of $A$" [emphasis added] is a bit odd. Of course $A$ has many upper bounds. Are you saying you're trying to show that $A$ does not contain any number that upper bounds it? If so, then shouldn't we define $A \stackrel{\text{def}}{=} {x \in \mathbb{Q} \mid x > 0 \wedge x^2 \leq 2}$? Without the $\leq$, we can't really expect $A$ to contain its own upper bound; $B \stackrel{\text{def}}{=} {x \in \mathbb{Q} \mid x > 0 \wedge x^2 < 4}$ also doesn't contain its upper bound, and $\sqrt4$ is certainly rational. – Brian Tung Oct 01 '22 at 19:05
  • Of course, that creates a problem with your proof, because it begins with a supposition that $x < \sqrt2$. In the first place, as @aschepler points out, we can't necessarily assume $\sqrt2$ exists. In the second place, even if we posited such a thing, that supposition was previously true by hypothesis, but now you would have to show that $\sqrt2$ isn't rational. – Brian Tung Oct 01 '22 at 19:10
  • @BrianTung Hello! I actually meant to prove that any upper bound of $A$ has to be greater than $\sqrt 2$. A number $x$ is said to be an upper bound of a set $A$ iff $(\forall a \in A)[a \le x]$ – MangoPizza Oct 01 '22 at 19:54
  • @BrianTung My main goal is to show that $A$ has no smallest upper bound in the rational set of numbers. To do that, I intend to show that 1. Any upper bound of $A$ is greater than $\sqrt 2$. 2. Then I assume some rational $x$ is an upper-bound of $A$, and show that I can get some $\sqrt 2 < y < x$. Since $\sqrt 2 < y$, $y$ is an upper-bound of the set $A$. And by showing this, we have shown that there exists no smallest upper-bound of $A$. – MangoPizza Oct 01 '22 at 19:57
  • @BrianTung Indeed $A = {x \in \mathbb{Q} | x > 0 \wedge x^2 \leq 2}$. That is a mistake, sorry. – MangoPizza Oct 01 '22 at 19:59
  • I have fixed both of the issues. Sorry again for the mistakes – MangoPizza Oct 01 '22 at 20:00
  • I think that your approach may be over-thinking the problem. It is straightforward to prove that $\sqrt{2}$ exists and is not rational. You also (presumably) have been given the result that both the rational numbers and irrational numbers are dense in $\Bbb{R}.$ Assume that the original hypothesis is false. Then, there exists a rational upper bound $M$ that is $\leq \sqrt{2}.$ This implies, since $\sqrt{2}$ irrational, that $M < \sqrt{2}.$ However, the density of the rationals means that there exists an $M_1$ rational such that $M < M_1 < \sqrt{2}.$ : contradiction. – user2661923 Oct 01 '22 at 20:42
  • Your approach and thinking behind it is correct. For example any upper bound must be greater than $\sqrt{2}$. But the proper way to handle this problem is to use rationals only (see remark by @aschepler). Assume that $x>0$ is rational. Then we have either $x^2<2$ (ie $x\in A$) or $x^2>2$. For first case show that no member of $A$ can be an upper bound ie any upper bound must satisfy $x^2>2$. This is same as showing that $A$ has no maximum. Next show that the set of upper bounds given by $x^2>2$ has no minimum. – Paramanand Singh Oct 02 '22 at 02:42
  • The proofs indicated in my last comment are available at many places on this website. See https://math.stackexchange.com/a/2080551/72031 – Paramanand Singh Oct 02 '22 at 02:43

1 Answers1

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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 $y+\epsilon\leq\sqrt{2}, \forall y\in S$.

ryaron
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