Dedekind cuts: Showing that the set B has no smallest element Let $A$ and $B$ be subset of the rational numbers defined as follows:
$A = \{p \in \mathbb{Q} \mid p\leq0 \lor p^2<2\}$
$B = \{q \in \mathbb{Q} \mid q>0 \land q^2 >2\}$


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*Show that the set $B$ has no smallest element.

*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

We try to discover something that might work.   
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
$$s=r-\frac{r^2-2}{2r}=\frac{r^2+2}{2r}$$
