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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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    @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.2012-01-13

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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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    Thanks, Thomas. What about the case t^{2}<2.2019-04-16
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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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    @Thomas: Thanks, I fixed it now.2012-01-14