For part (a), I begin by trying to prove $S$ is empty implies the square root of $D$ is irrational. If we take the contrapositive of this implication, this is equivalent to proving that if the square root of $D$ is rational then $S$ is not empty. Let $D$ be a positive integer and suppose $D$ is rational. Then it follows that $D^2$ is a perfect square. Moreover, S = n*D^1/2 and since n and D are positive integers, then by definition, S is non-empty. Conversely, we wish to prove that the square root of D is not rational implies S is empty. But this trivially follows from the product of a rational and irrational number. I'm not sure if my reasoning is sound, but constructive criticism would be appreciated.
For part (b), I'm not sure how exactly to use Well-Ordering to prove the inequality. I suppose my initial thought was to square both sides of the inequality and then rearrange some terms. Because the square root of D is not rational, it follows D^2 is not a perfect square. Then we can bound it appropriately?
For part (c), I think what we should begin with is to consider the number m*(D^1/2 - a). From there, I do not know how to proceed. Any help would greatly be appreciated.
Answer
In part (a) you are right about the contrapositive of the implication but you proved it wrong because you said that "and since $n$ and $D$ are positive integers, then by definition, S is non-empty" but I don't see that this is true because $\sqrt D \in \mathbb Q$ doesn't imply that $D\in \mathbb Z$
So instead you can say that if $\sqrt D \in \mathbb Q \implies \sqrt D= \frac{a}{b}$ where $a,b\in \mathbb Z$
(Note that $a$ and $b$ are positive here since a square root is always positive so $a,b\in \mathbb N$)
Which gives that $b\sqrt D=a \in \mathbb Z$ so S is not empty since $b\in S$.
Now for the sufficient condition, yes it is trivial since if $\sqrt D \not \in \mathbb Q \implies n\sqrt D \not \in \mathbb Q, \forall n\in \mathbb N \implies S=\phi$
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