Proving $S:=\left\{1\Big/\left(x^2-3\right)\mid x\in\Bbb Q\right\}$ is unbounded

Question

Let $S:=\left\{\dfrac1{x^2 -3} :x\in\Bbb Q\right\}$. Prove carefully that $S$ is unbounded

My proof

By contradiction, if we can find some $M$ so that $|s| \leqslant M\ \ \forall s \in S$, that is, if

$$ \dfrac{1}{|x^2-3|} \leqslant M\ \forall x\in\Bbb Q\ . $$ Since $x$ is any rational number, we can choose $x$ so that $|x^2-3| = \dfrac{1}{M+1}$. Therefore, we obtain

$$ \dfrac{1}{\frac{1}{M}+1} \leqslant M \implies M+1 \leqslant M $$

A contradiction.

I think my proof is correct, but I have lost some points. What is my mistake?

Answer 1

One could show that $S$ is neither bounded from above nor bounded from below. However, we are only asked to sow that it is not bounded. This can be done ab ovo without any sophisticated converging sequences a la Heron formula.

So assume there exists $M\in \Bbb R$ with $|s|\le M$ for all $s\in S$.

Pick $n\in\Bbb N$ with $n>2M$. Then $A_n:=\{\,x\in\Bbb N\mid x^2>3n^2\,\}$ is a non-empty (e.g., $2n\in A_n$) proper (e.g., $1\notin A_n$) subset of $\Bbb N$ and we can let $m=\min A_n$. Then $$m^2>3n^2$$ and $(m-1)^2\le 3n^2$. We readily see (as in the proof of irrationality of $3$) that even stronger $$(m-1)^2<3n^2.$$ From $(2n)^2>3n^2$, it is also clear that $$ m\le 2n.$$

With $x_1=\frac{m-1}n$ and $x_2=\frac mn$, we have $$\begin{align}\frac1{x_2^2-3}-\frac1{x_1^2-3}&=\frac{n^2}{m^2-3n^2}-\frac{n^2}{\underbrace{(m-1)^2-3n^2}_{<0}}\\ &=\frac{n^2}{m^2-3n^2}+\frac{n^2}{3n^2-(m-1)^2} \\ &=\frac{n^2(3n^2-(m-1)^2+m^2-3n^2)}{(m^2-3n^3)(3n^2-(m-1)^2)}\\ &=\frac{(2m-1)n^2}{(m^2-3n^3)(3n^2-(m-1)^2)}.\end{align}$$ By the arithmetic-geometric inequality, the denominator is $$ \begin{align}(m^2-3n^3)(3n^2-(m-1)^2)&\le\left(\frac{m^2-3n^3+3n^2-(m-1)^2}{2}\right)^2\\&=\frac{(2m-1)^2}{4}\end{align}$$ so that $$2M\ge \frac1{x_2^2-3}-\frac1{x_1^2-3}\ge\frac{4n^2}{2m-1} \ge \frac{4n^2}{4n-1}>n.$$ Contradiction.

Answer 2

The emphasis is on the word 'carefully'. You did not say what exactly your $x$ is. You can take $x=\sqrt {3+\frac 1 {M+1}}$. Now you have to make sure that $x$ is rational. For this you have to choose $M$ in particular way. There is a sequence of rationals $r_n$ tending to $\sqrt 3$. Take $M=\frac 1 {r_n^{2}-3}-1$. If $n$ is large enough this would serve as a bound for $S$. (Why?). In this case we get $x=r_n$ which is rational.

Answer 3

The carefulness here lies rather in choosing your $x$ to be in $\mathbb{Q}$.

You can choose a rational $x_M$ such that

$$0 < x_M^2-3 < \frac 1{M+1}$$

Now, you get your contradiction

$$M+1 < \frac 1{x_M^2-3} \leq M \quad\bf{\text{Contradiction!}}$$