Let $D$ be a positive integer and the let the square root of $D$ be a real number. Assuming that the square root of D is not an integer (i.e. $D$ is not a perfect square), use Well-Ordering to prove that there exists some integer $a$ such that $a < sqrt(D) < a+1.$
My proof attempt begins with assuming that the square root of D is not an integer and letting $a$ be the least integer that satisfies the inequality $a < $sqrt(D)$ < a + 1$, then I'm not sure what to do. My intuition tells me to square both sides, but then how would I proceed with the proof.
Let $A = \{ n \in \mathbb{N} ~| ~ \sqrt{D} < n \}$.
By the Archimedean property, the set $A$ is not the empty set, and $A$ is a subset of $\mathbb{N}$.
Thus by the well ordering principle, $A$ has the least element, say $\alpha$.
The assumption for $D$ implies that $D \in \mathbb{N}$ and $D \neq m^2$ for every $m\in \mathbb{N}$.
Thus $D \in \mathbb{N}$ and $D \neq 1^2$, i.e., $D \neq 1$. Hence $1<D$, so $1<\sqrt{D}$.
Thus $1 \notin A$.
(If $1 \in A$, then $\sqrt{D}<1$. Contradiction.)
Hence $\alpha$ cannot be $1$, so $\alpha \in \mathbb{N} -\{1\} = \{2,3,4,\ldots\}$.
Thus $\alpha -1 \in \mathbb{N}$.
Because $\alpha$ is the least element of $A$, we know that $\alpha -1 \notin A$, i.e. , $\alpha-1 \leq \sqrt{D}$.
If $\alpha -1 = \sqrt{D}$, then $D = (\alpha-1)^2$, and D becomes a perfect square. contradiction.
Hence $\alpha -1 < \sqrt{D}$. And with $\sqrt{D}< \alpha$, we can conclude that $\alpha -1 < \sqrt{D} < \alpha$.