1. ## Real Analysis Help!

Need help on this problem..how can I use the definition of inf to prove this without using the square root of 3. Thanks!

2. ## Re: Real Analysis Help!

Use a proof by contradiction. Assume that $s^2 > 3$ and show that there exists another rational $q < s$ with $q^2 > 3$. This contradicts the assertion that $s$ is the infimum of the set.

Spoiler:
You may like to use $q= {3(s+1) \over s+3}$.

Formulae such as this can be built for the square root of any rational $a$ by writing $x^2 = a$ and then adding $bx$ to both sides for some number $b$. You then factor the left and divide by the factor that is not identically $x$. You need the resulting right hand side to be an increasing function of $x$ for all $x > \sqrt{a}$. I believe this means that $b$ must be greater than $\sqrt{a}$.

Once you have constructed such a $q$, you will probably be able to use the same technique to prove that there is no rational that is the supremum of the complement of the given set, thus proving that $\sqrt{3}$ is irrational.

3. ## Re: Real Analysis Help!

The problem's claim is false, because $\inf\{ x \in \mathbb{Q} \ | \ x^2 > 3 \} = -\infty$.

For example, $-20 \in \mathbb{Q}$, and $(-20)^2 = 400 > 3$, so $-20 \in \{ x \in \mathbb{Q} \ | \ x^2 > 3 \}$,

and thus $s = \inf \{ x \in \mathbb{Q} \ | \ x^2 > 3 \} \le -20$, and so $s^2 \ge 400$. So it's just false that $s^2 \le 3$.

Of course, since negative infinity isn't a number, even what I wrote isn't correct - but you see the problem. (If you define the arithmetic of the extended reals (extended to plus and minus infinity) in the natural way, then what I wrote is ok.)

I suspect what was intended was one of these:

Let $s = \inf\{ x \in \mathbb{Q} \ | \ x \ge 0 \text{ and } x^2 > 3 \}$. Prove $s^2 \le 3$. (It will be that $s = \sqrt{3}$.)

or (basically equivalently)

Let $s = \inf\{ |x| \ | \ x \in \mathbb{Q} \text{ and } x^2 > 3 \}$. Prove $s^2 \le 3$. (It will be that $s = \sqrt{3}$.)

or (this is the easiest to prove)

Let $s = \inf\{ |x| \ | \ x \in \mathbb{Q} \text{ and } x^2 > 3 \}$. Prove $s^2 \ge 3$. (It will be that $s = \sqrt{3}$.)

or (less likely)

Let $s = \inf\{x \in \mathbb{Q} \ | \ x^2 < 3 \}$. Prove $s^2 \le 3$. (It will be that $s = -\sqrt{3}$.)

or (much less likely, since here s won't be real number, and this isn't finding the square root of 3 via real number completeness)

Let $s = \inf\{x \in \mathbb{Q} \ | \ x^2 > 3 \}$. Prove $s^2 \ge 3$. (It will be that $s = -\infty$.)