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I am trying to prove in a different way then how it was already proved on this website (another question). So yes this is sort of a duplicate.

Claim: $\mathbb{Q}$ does not have the least upper bound property

Let $S = \{r \in \mathbb{Q} : r > 0~~ \text {and} ~~r^2 < 2 \}$. Clearly $S$ is bounded above by 2 and $S$ is nonempty since $1 \in S$.

To prove S has no least upper bound it suffices to prove that if $t$ is an upperbound for $S$ then there exists a upper bound $t' \in \mathbb Q$ such that $t' < t$.

It can then be shown that for $n \in \mathbb{N}$ is sufficiently large then $$\left(t-\frac{1}{n} \right)^2 > 2$$ given that $t^2 > 2$

But I get stuck here. How would I go about proving this is true? any ideas would help. I appreciate it.

Thankyou

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Hint: any strictly increasing continuous function has a continuous inverse function. –  Robert Israel Jan 13 '12 at 19:27
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Or you can use $\sqrt{t} - \sqrt{s} = \frac{t-s}{\sqrt{t}+\sqrt{s}}$ –  Robert Israel Jan 13 '12 at 19:28
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@Robert: Your hints seem to require the construction of the reals. Otherwise, for instance, $\sqrt t$ has no meaning. But the failure of the least-upper-bound property is an intrinsic property of the rationals, and can be proved without using real numbers at all. –  TonyK Jan 13 '12 at 19:34
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2 Answers

up vote 1 down vote accepted

If $t^2 > 2$, then $t^2-2>0$ so there must be a positive integer $n>1$ so that $t^2-2>\frac{1}{n}$.

But $$(t-\frac{1}{2nt})^2 = t^2-\frac{1}{n} + \frac 1{4t^2n^2} > t^2-\frac{1}{n} > 2$$

So $t-\frac{1}{2nt}$ is smaller than $t$ and it's square is still greater than 2.

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I'm learning how to use the least upper bound propierty. So for example here, if $t$ is the least upper bound then there are two cases, $t$ might be in $S$ or it might not. I don't understand why you don't consider the first case, meaning that might be $t^{2}<2$. Also the case $t^{2}=2$ I guess is because you're assuming as proved that $2$ is irrational. Thank you so much for your help. –  Daniela Diaz Jun 28 '13 at 7:06
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@DanielaDiaz Yes, we are assuming that $\sqrt{2}$ has been proven to be irrational, or, spoken another way, there is no rational number $t$ such that $t^2=2$. –  Thomas Andrews Jun 28 '13 at 11:16
    
Thanks, Thomas. What about the case $t^{2}<2$. –  Daniela Diaz Jun 28 '13 at 11:26
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If you substitute 4 for 2 in your proof, then the conclusion will be false: $\{r \in \mathbb{Q} : r > 0$ and $r^2 < 4 \}$ does have a least upper bound in $\mathbb Q$ (because 4 is a perfect square). So somewhere you have to use the fact that if $r \in \mathbb Q$, then $r^2 \ne 2$.

So your proof should have two parts:

  1. Show that if $r \in \mathbb Q$, then $r^2 \ne 2$.
  2. Show that if $r \in \mathbb Q$ and $r^2 > 2$, then $\exists t \in \mathbb Q$ such that $2 < t^2 < r^2$.

If you can prove these (can you?), then the result follows: any upper bound $r$ will satisfy $r^2 > 2$ (because of 1), so it can't be a least upper bound (because of 2).

Edited to reply to comment: OK, if $r^2 > 2$, then $r^2 = 2 + \epsilon$ for some rational $\epsilon > 0$. Now look at $t = r - \epsilon/2r$.

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Yes, my question got up to part 2 (may have missed some details). I am looking for an idea/headstart on how to go about proving number 2. In fact I am trying to prove this with the inequality (involving n) in my questino. –  Tyler Hilton Jan 13 '12 at 19:40
    
@Tyler: See my edit. –  TonyK Jan 13 '12 at 19:57
    
You really need to look at $r-\frac{\epsilon}{2r}$. –  Thomas Andrews Jan 13 '12 at 21:40
    
@Thomas: Thanks, I fixed it now. –  TonyK Jan 14 '12 at 8:44
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