The set $\{\frac{1}{x^2-3}: x\in\mathbb{Q}\}$ is bounded? Explain.

Question

The set $\{\frac{1}{x^2-3}: x\in\mathbb{Q}\}$ is bounded? Explain.

I got this question some weeks ago in an Introduction to Real Analysis exam. In the exam review, the professor mentioned it could be solved with supremum and infimum concepts. Indeed, the exam (technically) was related to sequences (so when I saw this question I thought it could be approached with sequences techniques).

However, I didn't progress so much. This is my approach:

By hyphotesis we know $x\in \mathbb{Q}\implies x=\frac{p}{q}$, where $ q\neq 0$. Considering some bounds, we have: $$\frac{1}{x^2-3}<\frac{1}{x^2}\leq \frac{1}{x} \implies \frac{1}{\left(\frac{p}{q}\right)^2-3}<\frac{1}{\left(\frac{p}{q}\right)^2}\leq \frac{1}{\left(\frac{p}{q}\right)}$$ So, we have $\sup \left(\frac{1}{x^2-3}\right)= \frac{1}{x}$. $\therefore$ The sequence is bounded $\forall x\in\mathbb{Q}$.

I know my approach is a little bit... bad. I got a 2/20 score in this question, because if $x\in (0,1)$ then the bounds are not true and is not true $\forall x \in \mathbb{Q}$.

What could be a good procedure to solve this question? Thanks in advance.

Answer 1

There are three problems.

1. $1/(x^2 - 3) < 1/x^2$ does not hold for $x > \sqrt{3}$. Moreover, the RHS is undefined when $x = 0$.

2. $1/x^2 < 1/x$ does not hold for $x \in (0,1)$, as you mentioned. Moreover, the RHS is undefined when $x = 0$.

3. Even if $1/(x^2 - 3) < 1/x$ was true, this would not show that $\{ 1/(x^2 - 3) : x \in \mathbb{Q}\}$ was bounded. This might be a problem with your understanding of the word "bounded"? A subset $S$ of $\mathbb{R}$ is said to be bounded if there exist $a, b \in \mathbb{R}$ such that $a \leq s \leq b$ for all $s \in S$. You simply did not produce such constants, so your proof stood no chance of being correct.

In fact, the set in question is not bounded. You should try to prove this by producing a sequence $x_1, x_2, \dots$ of rational numbers such that $\lim_{n \to \infty} \frac{1}{x^2 - 3} = \infty$.

Let me extract a couple morals from this story:

1. Make sure you justify every claim you make: you could have caught errors #1 and #2 if you had done so.

2. Before you start writing a proof, spell out to yourself exactly what you need to show. In this case, you should say to yourself "I need to produce real numbers $a$ and $b$ such that $a \leq \frac{1}{x^2 - 3} \leq b$ for all $x \in \mathbb{Q}$, or show that no such $a, b$ exist". This has two advantages: first, understanding what you need to show often helps you figure out how to approach a proof, and second, if you know exactly what you need to show, you'll never end a proof without deriving the correct conclusion (this was error #3).

Edit: Missed that the first inequality also fails, corrected.

Answer 2

Hint: Heron's method starting at $x_0=1$ produces a sequence of rationals converging to $\sqrt3$.