# Why the set $\mathcal{S} = \left\{x\in\mathbb{Q} : x>0 , x^2<2 \right\}$ doesn't have the least-upper-bound in $\mathbb{Q}$?

Why this set $$\mathcal{S} = \left\{x\in\mathbb{Q} :x>0 , x^2<2 \right\}$$ does not have a least upper bound in $\mathbb{Q}$ ? I am not seeking for a proof, but rather want to get a clear sense that this set has no least upper bound. I know that $\sqrt2$ would the lub if the set we are working on was real numbers. But if we consider the set of rational numbers it has no lub because $\sqrt2$ is irrational? How can it be explained more clearly?

And also can you give me a nice source that explains this argument ?

If you are working in a context where the basic facts about the real numbers are already known, then you can simply apply those; given any rational upper bound $u$ to $S$, there exists another rational number $u'$ satisfying $\sqrt{2} < u' < u$, and thus $u'$ is a smaller upper bound.

Since $u$ was arbitrary, no rational upper bound can be a least upper bound.

When not working in such a context, coming up with a well-motivated proof isn't particularly enlightening (better to learn the ideas in the context of doing real analysis rather than when working under restricted conditions); the following trick derived from Newton's method to find a root of $x^2 - 2$ is instead used:

Define a function on nonzero rationals

$$f(x) = \frac{x + \frac{2}{x}}{2}$$

If $x^ 2> 2$, then it's pretty easy to see that $f(x) < x$. Furthermore, $f$ has the distinctive property that

$$f(x)^2 - 2 = \frac{(x^2-2)^2}{(2x)^2}$$

which, among other things, shows that you always have $f(x)^2 > 2$ when $x$ is a nonzero rational.

Note, given my comment above, this argument is meant to be one that is easily verified; it is not intended to be particularly enlightening, to be a demonstration of the tools you'll be learning, or to be something you'd be expected to be able to come up with on your own.

• Aside: I wish the order of operations were universally understood, so that I could use the more legible $x+2/x$ in that numerator. – user14972 Sep 24 '17 at 14:43
• I have just come across with a definition that says: An ordered set $S$ is said to have the least upper-bound-property if the following is true: If $E$ $\subset S$ , $E$ is not empty, and $E$ is bounded above, then sup$E$ exists in S. In my question, can we use this definition to prove that the set S has no lub because there exists no such a set $E$ ? – Leyla Alkan Sep 24 '17 at 16:42

Let $a$ be a rational upper bound of $S$. Prove that $$b=\frac12\left(a+\frac2a\right)$$ is also a rational upper bound of $S$ and that $a>b$.

Hint: Let $s$ be the supremum of $S$, assuming it exists.

If $s>\sqrt2$, then we can find another upper bound between $s$ and $\sqrt 2$ (why)?

If $s< \sqrt2$, then we can find some $z\in S$ such that $z>s$. (Again why?)

You got it right. Because $\sqrt{2}$ is the infimum of $\mathcal{S}$ but it doesn't belong to $\mathcal{S}$, so $\mathcal{S}$ has no least-upper-bound.

More precise, you can assume that $A$ is the least-upper-bound of $\mathcal{S}$. So $x < A$ for all $x\in \mathcal{S}$ and $A\in \mathcal{S}$, i.e A is rational . But you always can take $A'$ is rational and $A^2 < A'^2 < 2$.