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Lets denote the least upper bound by $\alpha \in \mathbb{Q}$ and $\delta > 0$ be a small number.

Now $\alpha^2 \neq 2$ because there is no such rational $\alpha$. If $\alpha^2 > 2$ then $(\alpha +\delta)^2 >2$ and so $\alpha$ is not a least upper bound.

I can't obtain a valid reason why $\alpha^2 < 2$ can not be the case.

  • If $\alpha^2<2$, you should be able to find a rational $\varepsilon>0$ so that $(\alpha+\varepsilon)^2$ is still smaller than $2$. – Quang Hoang Oct 02 '15 at 11:58
  • This must be a duplicate. – azimut Oct 02 '15 at 12:06
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    One might mention that if $x>\sqrt{2}$ is a rational number then $$y=\frac12x+\frac1x$$ is another rational number such that $$\sqrt{2}<y<x.$$ The "no least upper bound" property follows. – Did Oct 02 '15 at 12:59

1 Answers1

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At first i can't follow your argument for why if $\alpha^2>2$ than $\alpha$ can't be a least upper bound, because you only give another upper bound which is greater.

If $\alpha^2 <2$ then you have to find a rational number $q$ such that $\alpha^2< q^2< 2$.

First we show that if $\alpha^2>2$ then $\alpha$ can't be a least upper bound. Because of $\alpha^2 >2$ there is a $\varepsilon >0$ such that $\alpha^2 > 2+\varepsilon$. Due to archimedes wlog we may assume this $\varepsilon$ be be rational. We now want to find a $\delta >0$ such that $\alpha-\delta$ is another upper bound, which would contradict that $\alpha$ is the least upper bound. We now have \begin{align*} (\alpha -\delta)^2 &= \alpha^2 - \delta(2 \alpha -\delta)\\ & > 2 + \varepsilon - \delta (2 \alpha - \delta) \end{align*} which tells us we want to have that $$\varepsilon > \delta(2\alpha-\delta)$$ to show that for this choice of $\delta$ the number $\alpha-\delta$ will be an upper bound for our set. This calculation even shows, that not every $\varepsilon$ should be neglected. Even though you could solve this inequality exactly due to the easy terms, it is even more easy (and should be trained) to make some more estimates.

As $\delta >0$ (and $\alpha >0$ we choose $\delta$ such that $$\varepsilon > 2\delta \alpha$$ in other words $\delta < \frac{\varepsilon}{2 \alpha}$ we see that

$$\varepsilon > 2\delta \alpha > \delta( 2\alpha -\delta)$$ and hence $\alpha-\delta$ is a lower upper bound.

Dominic Michaelis
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  • Yes, I see. That part in nonsense. I got as far as saying that $(\alpha + \delta)^2 < 2 + 2\alpha\epsilon + \epsilon^2$ I know that $epsilon $ is small to can I "ignore" the terms with epsilon in? –  Oct 02 '15 at 12:05
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    How about $(\alpha -\delta)^2=\alpha^2- 2 \alpha \delta +\delta ^2 $ and now choose $\delta$ small enough that this is still strictrly greater than $2$, feel free to ask if you need additional help, but ignoring the $\varepsilon$-term is crucial in this phase of your mathematical education ;) – Dominic Michaelis Oct 02 '15 at 12:10
  • I am still struggling with the final case. Can I just leave it as that and then state than we CAN choose a epsilon/delta sufficiently small? –  Oct 02 '15 at 12:12
  • @HMPARTICLE even though I wouldn't be more explicit, I recommend you to be more explicit, as I don't think how exactly you can choose it. – Dominic Michaelis Oct 02 '15 at 12:22
  • I have said that I would like to find $\delta$ sufficiently small such that $\alpha^2 + 2\alpha\delta + \delta^2 < 2$ and rearranged to get $\delta(2(\alpha + 2\delta) < 2 - \alpha^2$ but this doesn't look like its getting me anywhere :( –  Oct 02 '15 at 12:38
  • @HMPARTICLE gonna make my answer a bit mor explicit – Dominic Michaelis Oct 02 '15 at 12:39
  • @HMPARTICLE check my answer again – Dominic Michaelis Oct 02 '15 at 12:53
  • I don't understand why $$\varepsilon > \delta( 2\alpha -\delta) > 2\delta \alpha$$ –  Oct 04 '15 at 20:15
  • I would have thought that we can ignore the delta square term as delta is assumed to be small? Is that correct? –  Oct 04 '15 at 20:53
  • @HMPARTICLE oh gee did make a mistake the terms should be changed, – Dominic Michaelis Oct 04 '15 at 21:27
  • @HMPARTICLE changed it now it works – Dominic Michaelis Oct 04 '15 at 21:31
  • Thanks, I have done the same for $\alpha ^ 2 < 2$ and I get the same result, If we take $\delta < \epsilon / 2 \alpha$ we can show that alpha is not a upper bound. Is that the result I should arrive at? you have helped a great deal, thank you. –  Oct 04 '15 at 21:33
  • @HMPARTICLE yeah that is the desired result – Dominic Michaelis Oct 05 '15 at 08:30