How to prove $r^2=2$ ? (Dedekind's cut) Let a (Dedekind) cut $r=\{p \in \mathbb{Q} :p^2<2 \text{ or } p<0\}$ and a cut $2^*=\{t\in \mathbb{Q} : t<2\}$.
I want to prove $r^2=2^*$. I could show that $r^2 \subset 2^*$ easily, but I couldn't show that $2^* \subset r^2$. How to show that there is $p$, $p' \in r$ such that $t \leq  pp' <2$ or $t \leq p^2<2$?
 A: As pointed out by N.S. you need to establish that given any rational $t < 2$ there is a positive rational $a$ with $t < a^{2} < 2$.
We can do this by analyzing the Dedekind cut $$r = \{p \mid p \in \mathbb{Q}^{+}, p^{2} < 2\} \cup \{p \mid p \in \mathbb{Q}, p \leq 0\}$$ in more detail. Let $r' = \mathbb{Q} - r$ so that $$r' = \{p\mid p\in\mathbb{Q}^{+}, p^{2} > 2\}$$ (note that there is no rational $p$ with $p^{2} = 2$)
We show that given any rational $\epsilon > 0$ we can find a member $a \in r$ and $b \in r'$ such that $0 < b - a < \epsilon$. Clearly it is obvious that any $b \in r'$ is greater than any $a \in r$. So we only need to find specific $a \in r, b \in r'$ such that $b - a < \epsilon$. Now we see that $1 \in r, 2 \in r'$ and choose a positive integer $n > 1/\epsilon$. Consider the numbers $$1, 1 + \frac{1}{n}, 1 + \frac{2}{n}, \cdots, 1 + \frac{n - 1}{n}, 2$$ In this arithmetic progression there is a last number (say $a = 1 + \dfrac{k}{n}$)which belongs to $r$ and the next one (say $b = 1 + \dfrac{k + 1}{n}$) belongs to $r'$. Then we can see that $b - a = 1/n < \epsilon$.
Now let us put $\epsilon = 2 - t \in \mathbb{Q}^{+}$. By the reasoning given in previous paragraph we can find $a \in r, b \in r'$ such that $b - a < \epsilon / 4$ and we can also take both $a, b$ positive and less than $2$ so that $b + a < 4$. Thus $b^{2} - a^{2} = (b + a)(b - a) < 4\cdot\epsilon/4 = \epsilon$. This means that $(b^{2} - 2) + (2 - a^{2}) < \epsilon$ and since both $(b^{2} - 2)$ and $(2 - a^{2})$ are positive by definition of $r, r'$ it follows that $2 - a^{2} < \epsilon = 2 - t$. Thus we have $a^{2} > t$ and by definition of $r$ we have $a^{2} < 2$. So we have found a positive rational $a$ such that $t < a^{2} < 2$.
Note: The above derivation is taken from G H Hardy's Pure Mathematics and I find this argument the best in explaining the Dedekind cut defining $\sqrt{2}$.
A: Allow me to give a cop-out answer, but tell me if you are dissatisfied and I'll write something else.
Consider the following sequence: $1, 14/10, 141/100, 1414/1000, 14142/10000, \ldots$ which is created by lopping off part of the decimal representation of $\sqrt{2}$ (never mind that you probably can't talk explicitly about $\sqrt{2}$ yet, since Dedekind cuts likely mean you are constructing the real numbers; in any event, the rational numbers listed above certainly exist). Now consider the sequence formed by squaring each of these. This sequence of rational squares gets arbitrarily close to $2$ from below, so for any $2 > t \in \mathbb{Q}$ you can find an element of the sequence that exceeds $t$ but not $2$.
A: You basically want to prove the following:
Given any $t<2$ you can find some $p \in \mathbb Q$ such that 
$$t<p^2 <2 \,.$$
Pick $n$ so that $n^2 > (*)$, where I could enter the right formula here, but we will discover it later, at that point it will be clear where it comes from.
Let $m$ be so that 
$$\frac{(m-1)^2}{n^2} \leq t <  \frac{m^2}{n^2} $$
I claim that $\frac{m^2}{n^2}< 2$
Indeed, assume by contradiction that $\frac{m^2}{n^2}>2$. Then
$$\frac{m^2}{n^2}>2> t> \frac{(m-1)^2}{n^2} \Rightarrow \frac{2m-1}{n^2} > 2-t$$
Now, since $(m-1)^2 \leq tn^2$ we have
$$(2m-1)^2 \leq(3m-3)^2\leq 9tn^2$$
(the case $m=1$ can be easily be dealt with, or can be avoided by assuming that $t>1$.)
And thus
$$(2-t)^2< \frac{(2m-1)^2}{n^4} \leq \frac{9tn^2}{n^4}$$
and hence 
$$n^2 < \frac{9t}{(2-t)^2}$$
which contradicts 
$$n^2> (*)$$
P.S. This is a modification of the proof that between any two numbers you can find a rational, is just that the inequalities we needed to deal with are a little more complicated.  
