No largest element of Dedekind cuts I'm practicing proving that various sets are (or aren't) Dedekind cuts.  I am struggling with showing the property that there is no largest element.  For example, for a simple Dedekind cut:
$A=\{{a\in\mathbb{Q}:a^2<2}$ or ${a<0}\}$
what is a good general method for showing that:
if ${a\in A}$, there is a ${b}$ in ${A}$ with ${a<b}$ 
 A: Here is how one would approach this problem if one has no prior experience with such problems.
You are given $a\in A$ and you wish to find $b\in A$ with $a<b$. There are two cases to consider: first $a\leq 0$ and then we can simply take $b=1$.
The real challenge is to deal with case $a>0$. The problem can now be stated as

Let $a$ be a positive rational number with $a^2<2$. Prove that there is another positive rational number $b>a$ such that $b^2<2$.

Thus we want to have $b$ such that $a^2<b^2<2$. Now let us put $b=a+h$ with $h>0$ and we need to find $h$ which ensures $b^2=(a+h)^2<2$. Thus we need $$a^2+2ah+h^2<2$$ or $$h(2a+h)<2-a^2$$ Let's now observe that the left hand side decreases if $h$ decreases and hence we need some sort of small value of $h$ here. Let's arbitrarily restrict $h<1$ (you can choose $h<2$ or whatever you wish). Then we have $$h(2a+h)<h(2a+1)$$ and if we further ensure that $h(2a+1)<2-a^2$ then our job is done. Thus we need two conditions on $h$ namely $h<1$ and $$h<\frac{2-a^2}{2a+1}$$ and combining this we get $$0<h<\min\left(1,\frac{2-a^2}{2a+1}\right) $$ and then $b=a+h$ is our required element of $A$.
Let's check how this works out in practice. Let $a=1.4$ so that $a^2=1.96<2$. Now we have $$\frac{2-a^2}{2a+1}=\frac{0.04}{3.8}<1$$ and we need to find a positive $h$ less than above value. Clearly one such value is $$h=\frac{0.04}{4}=0.01$$ and hence $b=a+h=1.41$ works fine. Check that $1.41^2<2$. You can try another example with $a=1.41$. The key is that the entire argument in last paragraph is symbolic and deals with any positive value of $a$ with $a^2<2$. And thus given any positive $a\in A$ we are able to find another $b\in A$ with  $b>a$.
You should try the slightly harder problem using the same approach :

Let $a$ be a positive rational number such that $a^3<3a+8$. Prove that there is another positive rational number $b$ with $b>a$ and $b^3<3b+8$ so that the set $$B=\{x\mid x\in\mathbb {Q}, x>0,x^3<3x+8\}$$ has no greatest member. 

