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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.

  • $\begingroup$ Can you please type your question out instead of pasting an image? Also, can you use MathJax to typeset your formulae? $\endgroup$ – Morgan Rodgers Jan 29 '19 at 7:08
  • $\begingroup$ Sorry I just need help $\endgroup$ – Sanjoy Kundu Jan 29 '19 at 19:06

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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