Let $A=\left| \frac{x^2-2}{x^2+2} \right|$. Prove that $A$ is bounded, and that: $\inf A=0$, $\sup A=1$.

Question

$A = \left\{ \left|\frac{x^2-2}{x^2+2}\right| : x\in\mathbb Q \right\}$

1.) Prove that $A$ is bounded, and that: $\inf A=0$, $\sup A=1$.

2.) Does $\max A$ exist? Does $\min A$ exist?

I'm hoping someone could check my work/train of thought.

Well to show that $A$ is lower bounded is quite trivial, since $|x|\ge 0$, therefore:

$$\left| \frac{x^2-2}{x^2+2} \right|\ge 0$$

We find that $A$ is lower-bounded by and 0 and negative numbers, with $\inf A=0$.

To show it is upper bounded, we can do the following. We can test out all the bigger numbers and we get the following:

$$\lim_{x \to \infty}\left|\frac{x^2-2}{x^2+2}\right|=|1|=1.$$

Therefore we get that $$0\le\left|\frac{x^2-2}{x^2+2}\right|\le1$$

2.) In order to show that $\max A$ or $\min A$ exists, we can use the fact that if $\inf A=0$, and $0\in A$, then $\min A = 0$. Same thing with $\max A$.

In our lectures we proved that there is no solution to $x^2=2$ that is rational, therefore we can say that $0$ does not belong to $A$, and therefore $A$ has no minimum.

Showing that $1\in A$ is quite trivial since $\frac11$ is rational, therefore $\max A=1$.

Were my steps enough to fully solve this question? Have I done something wrong or left out steps? I'd appreciate any input.

Answer 1

For your first part:

Well to show that A is lower bounded is quite trivial, since $|x|\ge 0$, therefore: $\left| \frac{x^2-2}{x^2+2} \right|\ge 0$

There is no any connection between the causation and the conclusion. The reason why the conclusion holds because:

Analytically, one can let $x^2-2=0$ and solve to show that lower bound exists.

Since for any real number $K$, $|K| \geq 0$ then $0$ is the lower bound

We can test out all the bigger numbers and we get the following:

This statement is not so much a good condition for bounded function:

Consider, for instance, $f(x) = \frac{1}{x}$, when $x$ goes up to infinity and then $f(x)$ approaches $0$. But $f$ is not bounded.

For your second part:

In our lectures we proved that there is no solution to x^2=2 that is rational, therefore we can say that 0 does not belong to A, and therefore A has no minimum.

This is a good statement but if the domain is $\mathbb R$ then $A$ has minimum indeed.

Showing that 1∈A is quite trivial since 1/1 is rational, therefore maxA=1.

This is a wrong statement because the reason why 1 belongs to A is because $A(0) = \left| \frac{x^2-2}{x^2+2} \right| = 1$

Answer 2

To begin with, you're missing some symbols in your definition of $A$ -- what you probably mean is $$ A = \left\{ \left|\frac{x^2-2}{x^2+2}\right| : x\in\mathbb R \right\} $$ or (equivalently) to let $A$ be the range of $f$ where $$ f(x) = \frac{x^2-2}{x^2+2} $$

You have a correct argument that $0$ is a lower bound for $A$, but in order to prove that $\inf A=0$, you need to prove that $0$ is the greatest lower bound.

Fortunately this is easy in this case, because it turns out that $0\in A$, because $f(\sqrt 2)=0$. The fact that $f(x)=0$ has no rational solution matters only if you have a secret condition that $x$ in your original expression is supposed to be rational.

Edit: Since it turns out that you do have such a condition, $f(\sqrt 2)$ won't work, so you'll need something like an $\epsilon$-$\delta$ proof to show that no positive $\epsilon$ can be a lower bound. The slickest would be to say that you know $f$ is continuous on the reals, so for every $\epsilon>0$ there is a $\delta$ such that $f(x)<\epsilon$ for every $x\in[\sqrt2-\delta,\sqrt2+\delta]$, and this nonempty interval has to contain at least one rational.

---

Your upper bound argument doesn't really work. It happens to be true that if $f$ is continuous on all of $\mathbb R$ and $f(x)$ has a limit for $x\to\pm\infty$, then $f$ is bounded -- but your reasoning is not enough to show that, and the bound doesn't necessarily equal the limit. For example, for $$ g(x) = \frac{2x}{x^2+1} $$ the limit in both directions is $0$ but the range is $[-1,1]$.

So you need a better argument. It is clear that $f(0)=1$, so the supremum is at least $1$, but you still need to show that $f(x)$ cannot be greater than $1$.

Answer 3

The first part is roughly correct. In the second part, however, you seem to be assuming that $A$ is taking rationals values and only those.

That's actually not the case, since numerator and denominator are real numbers. Even if they were integers, the fraction wouldn't necessarily assume all the rational values.

So, to answer the second part, we have to verify if $0$ and $1$ are in the image of $A$. To do that we have to simply solve the equations $A = 0$ and $A = 1$.