# Dedekind cuts: Showing that the set B has no smallest element

Let $A$ and $B$ be subset of the rational numbers defined as follows:

$A = \{p \in \mathbb{Q} \mid p\leq0 \lor p^2<2\}$

$B = \{q \in \mathbb{Q} \mid q>0 \land q^2 >2\}$

1. Show that the set $B$ has no smallest element.
2. Show that the set $A$ fails to have a least upper bound property

My attempt at question 1 is that we try to find an element $x \in \mathbb{Q}$ such that $x$ has form $x= q - \delta$, where $q \in B, \delta > 0$ and $q - \delta < q$. I know that $x^2 = (q - \delta)^2 > 2$.

This is the point where I get stuck. I'm trying to define $\delta$ in terms of $q$ and then expanding the term $(q-\delta)^2$ to show that it is greater than 2. I am not too sure if this is the correct and most efficient way to do prove that B has no smallest element.

Help and advice is very much appreciated. Thank you

• It would certainly work. Show that $q^2-2\delta-\delta^2>2$ iff $\delta(2-\delta)<q^2-2$. Since $0<q^2-2$ and $\delta(2-\delta)\mathop{\to}_{\delta\to 0}0$, one such rational $\delta>0$ exists. – Jonathan Y. Sep 17 '13 at 0:08
• Thanks for the quick response. I'm having a hard time trying to find an explicit equation of delta in terms of q that will make this inequality true. Any tips to find such a delta? – tamefoxes Sep 17 '13 at 0:16
• Well, if we accept that $\delta<2$ than one has $\delta(2-\delta)<2\delta$, so it would suffice that $\delta<\frac{q^2-2}{2}$. One such $\delta$ is $\min\{1,\frac{q^2-2}{4}\}$. – Jonathan Y. Sep 17 '13 at 0:19
• Ahhhh, that makes perfect sense. Thank you very much for all the help you've provided Johnathan. – tamefoxes Sep 17 '13 at 0:24

Suppose we are given a positive rational $r$ such that $r^2\gt 2$. We want to produce a smaller positive rational $s$ such that $s^2\gt 2$.
We will produce $s$ by taking a little off $r$, say by using $s=r-\epsilon$, where $\epsilon$ is a small positive rational.
So we need to make sure that $(r-\epsilon)^2$ is still $\gt 2$.
Calculate. We have $$(r-\epsilon)^2-2=(r^2-2)-2r\epsilon+\epsilon^2\gt (r^2-2)-2r\epsilon.$$ If we can make sure that $(r^2-2)-2r\epsilon\ge 0$ we will have met our goal. That can be done by choosing $\epsilon=\frac{r^2-2}{2r}$. That leads to the choice $$s=r-\frac{r^2-2}{2r}=\frac{r^2+2}{2r}$$
• You are welcome. I hope I kind of managed to do what motivated me to answer. Finding an $s$ that works comes from a more or less natural exploration. – André Nicolas Sep 17 '13 at 0:43