Proof that there always exists an integer $x_{0}$ such that $(x_{0} − 1)^2 < n < x_{0}^2$ I have to prove that for every natural number $n > 1$ that is not a perfect square, $x_{0}$ is the unique integer satisfying $(x_{0} − 1)^2 < n < x_{0}^2$.
I know I have to prove existence and uniqueness of $x_{0}$. If I can show that $(x_{0} − 1) < \sqrt{n} < x_{0}$ is true for all $n > 1$, I should be able to do the rest.
I was told that the least natural number principle may come in handy, but I fail to see its relevance to this proof because what if $n$ itself is the least element in a set and therefore $x_{0} - 1$ is not in the set?
 A: As the comments indicated, we need to replace at least one $<$ with $\le$. Then, however, this is true for 
$$[0, \infty) = \bigcup_{x \in \mathbb N_0} [x^2, (x+1)^2) = \bigcup_{x \in \mathbb N_0} (x^2, (x+1)^2] \cup \{0\}.$$
(This equality is immediate: If $x$ is maximal such that $x^2 \le n$ then $n < (x+1)^2$ and hence $n \in [x^2, (x+1)^2) \subseteq \bigcup_{x \in \mathbb N_0} [x^2, (x+1)^2)$.)
Also note that if we replace only the right $<$ with $\le$ we need to add $\{0\}$ as above, i.e. it then holds only for natural numbers $>0$. (Let's not get into the discussion whether $0$ should be regarded as a natural number here...)
A: The inequality should be $(x_0-1)^2 < n\ \le x_0^2$ since $n$ itself can be a perfect square.
Take $x_0=\left\lceil\sqrt n\right\rceil$. Since $n>1$, we have $x_0-1\ge0$, in which case $x_0-1 < \sqrt n \le x_0$ $\implies$ $(x_0-1)^2 < n\ \le x_0^2$.
Uniqueness follows from the fact that $x_0$ is positive, since if $x_0<0$ we would have $(x_0-1)^2>x_0^2$.
A: Given $n>1$, let $A=\{\,x\in\Bbb N\mid x^2\ge n\,\}$. Then $A$ is not empty because for example $n^2\ge n$ implies $n\in A$. As a subset of $\Bbb N$, $A$ has a minimal element $x_0$. Note that $x_0-1\in\Bbb N$ because certainly $x_0>1$. Now $x_0-1\in\Bbb N\setminus A$, $x_o\in A$ translates to $(x_0-1)^2<n\le x_0^2$, thus showing existence.
As for uniqueness, assume that also $(x_1-1)^2<n\le x_1^2$. Then $x_1\in A$ and so $x_1\ge x_0$. We may thus assume $x_1>x_0$, i.e., $x_1-1\ge x_0$. But then $(x_1-1)^2\ge x_0^2\ge n$, contradiction.
