TITLE: Show that the set $A=\{x\in \mathbb{Q}^+| x^2>2\}$ has no least element in $\mathbb{Q}$
QUESTION [5 upvotes]: Show that the set $A=\{x\in \mathbb{Q}^+| x^2>2\}$ has no least element in $\mathbb{Q}.$

I assume that $y=\inf A=$ least element. Then $y\in A$ or $y$ does not belongs to A. 
Case I: Let $y\in A$; 
Considering $k=\frac{y^2-2}{2y}$, I have shown $y-k \in \mathbb{Q}^+$ (as $y-k>0$ and rational). Also $(y-k)^2>2$ implying $y-k\in A$. But $y-k<y$ and $y=\inf A$, a contradiction. So $A$ has no least element in this case.
Case II: Let $y$ that does not belongs to $A$. Then either $y^2=2$ or $y^2<2$.
My questions:

Please help me to complete case II. 
Is there any other way to solve it?

REPLY [3 votes]: I think the real source of confusion here is the poorly worded question:

(1) Ambiguous: Show that the set $A=\{x\in \mathbb{Q}^+| x^2>2\}$ has no least element in $\mathbb{Q}.$

A least element is a property of an ordered set $X$. No other data is involved. There's no such relation as "least element of $X$ in $Y$" for some other set $Y$. So a much more sensible question would be simply:

(2) Good: Show that the set $A=\{x\in \mathbb{Q}^+| x^2>2\}$ has no least element.

That would make it clear that you only have to consider Case I. In my opinion, it is reasonable to interpret question (1) as meaning (2). This is assumed by the other answers on this page.
Or we could ask a different question:

(3) Also good, and harder: Show that the set $A=\{x\in \mathbb{Q}^+| x^2>2\}$ has no infimum in $\mathbb{Q}.$

It seems like the wording of question (1) made you assume that the author meant to ask (3). Perhaps they did! To answer (3), you do need to consider both Case I and Case II. Some of the other answers on this page address how you would do that.
As a final note, if we want to be charitable to the author, we could interpret (1) more straightforwardly to mean this:

(4) Silly: Show that the set $A=\{x\in \mathbb{Q}^+| x^2>2\}$ has no least element $y$ such that $y\in\mathbb{Q}.$

This is silly because $A$ is defined as a subset of $\mathbb{Q}$, so the question immediately reduces to (2). No one would actually write (4). But then, they shouldn't have written (1) in the first place.