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Let $E=\left\{ r\in\mathbb{Q}:r^{2}<3\right\}$. Prove that $\sup E=\sqrt{3}$.

Since $E$ is bounded from above by $\sqrt{3}$ and is nonempty, $\alpha:=\sup E$ must exist by the Least Upper Bound Principle. Now I am stuck. Suppose $\alpha<\sqrt{3}$. Then what?

Edit: It just hit me. If $\alpha < \sqrt{3}$ then since the rationals are dense in $\mathbb{R}$, there exists $q\in\mathbb{Q}$ such that $\alpha < q <\sqrt{3}$ and hence $\alpha^2 which contradicts the fact that $\alpha$ is the sup of $E$

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    Well, how do you know the rationals are dense in $\Bbb R$? That's a strong result.2012-06-28
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    @Peter: Perhaps they either proved that in class or were told to use it freely. I don't image that one would assign such an exercise otherwise (this exercise is trivial compared to proving density of rationals).2012-06-28
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    @William But it can be proven given $m/n<\sqrt 3$, there is another $p/q$ dependent on $m/n$ such that $m/n

    2012-06-28

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    @Peter: Yup, and using exactly that one proves density of rationals. Notice, I'm not saying that density of $\mathbb{Q}$ is nontrivial, but only that the exercise above follows trivially from density of $\mathbb{Q}$.2012-06-28
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    @William Sure. ${}{}$2012-06-28
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    What's your definition for real numbers? Different difinitions result in different proofs.2012-06-28

1 Answers 1

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Let $r\in \bf E$ , $r = \dfrac m n $.

Then we need to prove that there always exists a $\dfrac{p}{q}$ such that

$$ \frac{m}{n}<\frac p q < \sqrt 3$$

By exploting the special properties of $\sqrt 3 $,we can do it:

We start with

$$\eqalign{ & r < \sqrt 3 \cr & r + 1 < \sqrt 3 + 1 \cr & \frac{1}{{r + 1}} > \frac{1}{{\sqrt 3 + 1}} \cr & \frac{1}{{r + 1}} > \frac{{\sqrt 3 - 1}}{2} \cr & \frac{{r + 3}}{{r + 1}} > \sqrt 3 \cr} $$

But now we've gotten a number greater than $\sqrt 3 $. So what we'll do is apply the process again, and the relation will be reversed:

$$\frac{{Q + 3}}{{Q + 1}} < \sqrt 3 $$

Letting ${Q = \frac{{r + 3}}{{r + 1}}}$ will give

$$\frac{{2r + 3}}{{r + 2}} < \sqrt 3 $$

All we need to do now is prove that

$$r < \frac{{2r + 3}}{{r + 2}}$$

But since $r+2>0$ we get that

$$\eqalign{ & {r^2} + 2r < 2r + 3 \cr & {r^2} < 3 \cr} $$

which is true by hypothesis. Thus, given any rational $r$, there exists another rational $q$ such that

$$r

where $$q = \frac{{2r + 3}}{{r + 2}}$$

Just to make things clear, I'll add some extra information.

The recursion defined as $r_0=1$ $$r_{n+1}=\frac{2 r_n+3}{r_n +2} $$ converges monotonically (increasing) to $\sqrt 3 $. This in particular means that given any $\epsilon>0$ there is an $r \in \rm E$ such that $$\sqrt 3 -\epsilon < r$$ which means $\sqrt 3 $ is the supremum of the set.

This also plays the role of showing that $3$ is irrational.

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    Have you discriminate this case: there exists rational numbers $r_1 such that $r_n\le \alpha<\sqrt3$ holds for all $n>0$?2012-06-28
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    @FrankScience I think this shows that no rational number can be the supremum you are looking for - there is always another rational number greater than the one you chose and less that $\sqrt 3$. If $\alpha$ is not rational and less than $\sqrt 3$ then consider $\beta = \sqrt 3 - \alpha$. Choose an integer $N > \beta$ which you can do by "Archimedes Axiom" and consider rationals with denominator $N$. This gives you a rational between $\alpha \text { and } \sqrt 3$, so that $\alpha$ can't be the sup.2012-06-28
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    One more step is required namely let $r\notin E$ and $r>0$ be a rational number then $\exists\ q\in \mathbb Q$ such that $\sqrt 3< q< r$. Check that $q=\frac{2r+3}{r+2}$2012-06-28
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    @MarkBennet Yes, but I think it is not trivial in Peter's proof. I have mentioned that, the different definitions of real numbers result in different proof, and notice that, Archimedes Axiom implies the original problem directly.2012-06-28
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    @MarkBennet For example, $m$ is such a large positive integer that $m>1/\epsilon$, then consider the $n/m$ for all nonnegative integer $n$, and $n_0=\sup\left\{\;n\in \Bbb Z\;\big|\;n/m\leq\sqrt 3\;\right\}$, we have $n_0$ is a nonnegative integer and $|n_0/m-\sqrt 3|\leq1/m<\epsilon$.2012-06-28
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    @MarkBennet I think the Archimedes Axiom should be proved by least upbound principle. For example, consider $n_0=\sup\left\{\;n\;\big|\;n\le\alpha\;\right\}$, prove that $n_0\in\Bbb Z$ and show that $n_0+1>\alpha$.2012-06-28
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    @FrankScience - You need Archimedes Axiom, or something similar, to prove this. How else are you going to find a rational number between two real numbers, or show that the rationals are dense in the reals?2012-06-28
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    @MarkBennet As I've shown, Achimedes Axiom results from the completeness of real numbers. A definition of real number at [Wikipedia](http://en.wikipedia.org/wiki/Construction_of_the_real_numbers#Synthetic_approach)2012-06-28
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    @FrankScience You can show that the recursion obtained, namely $r_{n+1}=\frac{2r_n+3}{r_n+2}$ converges to $\sqrt 3$ (which means there is not such an $\alpha$). This in particular means that given any $\epsilon >0$ there is an $r \in \rm E$ such that $\sqrt 3 - \epsilon < r$, which proves that the supremum is $\sqrt 3$.2012-06-28
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    @Kuashik I don't follow. Why do you think that is required?2012-06-28
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    @PeterTamaroff I get it. I think $r_{n+1}=(r_n+3/r_n)/2$ might be better. Incidentally, what's the largest speed of convergence when we use $r_{n+1}=(\alpha r_n+\beta)/(\gamma r_n+\delta)$? Is it yours?2012-06-29
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    @Peter Tamaroff: I take supremum of a subset of $\mathbb R$ is a least upper bound of that subset.2012-06-29