Let A = {$r \in \mathbb{Q} \space | \space r > 0 \land r^2 > 3$}. Show that A has a lower bound in $\mathbb{Q}$ but no greatest lower bound in $\mathbb{Q}$. Give all details of the proof.

To show that A has a lower bound in $\mathbb{Q}$ is trivial. Clearly, zero is a lower bound of A in $\mathbb{Q}$, as 0 < r for all r in A, and 0 is rational.

Let's assume A has a greatest lower bound x in $\mathbb{Q}$. There are then only three possibilities:

$$\begin{aligned} Case \space 1: \space x^2 & = 3 \\ Case \space 2: \space x^2 & > 3 \\ Case \space 3: \space x^2 & < 3 \end{aligned}$$

Case 1: $x^2 = 3$

The first case is obviously not possible, since $\sqrt{3}$ is irrational. See proof here.

Case 2: $x^2 > 3$

For the second case, suppose

$$\begin{aligned} x & = \frac{p}{q} \space where \space p,q \in \mathbb{N} \space and \space x^2 > 3 \end{aligned}$$

So

$$\begin{aligned} p^2 & > 3q^2 \end{aligned}$$

Let's consider the rational number

$$\begin{aligned} \frac{n^2}{2n+1} \space where \space n \in \mathbb{N} \end{aligned}$$

which increases without bound as n increases. In particular, we can always pick a large enough n so that

$$\begin{aligned} \frac{n^2}{2n+1} & > \frac{3q^2}{p^2 - 3q^2} \\ n^2p^2 - 3n^2q^2 & > 6q^2n + 3q^2 \\ n^2p^2 & > 3q^2(n + 2n + 1) \\ n^2p^2 & > 3q^2(n + 1)^2 \\ \frac{n^2}{(n + 1)^2} \cdot \frac{p^2}{q^2} & > 3 \end{aligned}$$

Let

$$\begin{aligned} y = \frac{n}{(n + 1)} \cdot \frac{p}{q} \end{aligned}$$

Since $n < n + 1$,

$$\begin{aligned} \frac{p^2}{q^2} & > \frac{n^2}{(n + 1)^2} \cdot \frac{p^2}{q^2} > 3 \\ x^2 & > y^2 > 3 \end{aligned}$$

This means y is an element of A but less than x, which contradicts the assumption that x is a lower bound of A in $\mathbb{Q}$ ! So the supposition that $x^2 > 3$ must be false.

Case 3: $x^2 < 3$

For the third case, suppose

$$\begin{aligned} x & = \frac{p}{q} \space where \space p,q \in \mathbb{N} \space and \space x^2 < 3 \end{aligned}$$

So

$$\begin{aligned} p^2 & < 3q^2 \end{aligned}$$

Let's consider the rational number

$$\begin{aligned} \frac{n^2}{2n+1} \space where \space n \in \mathbb{N} \end{aligned}$$

which increases without bound as n increases. In particular, we can always pick a large enough n so that

$$\begin{aligned} \frac{n^2}{2n+1} & > \frac{p^2}{3q^2 - p^2} \\ 3n^2q^2 - n^2p^2 & > 2p^2n + p^2 \\ 3n^2q^2 & > p^2(n + 2n + 1) \\ 3n^2q^2 & > p^2(n + 1)^2 \\ 3 & > \frac{p^2}{q^2} \cdot \frac{(n + 1)^2}{n^2} \end{aligned}$$

Let

$$\begin{aligned} y = \frac{p}{q} \cdot \frac{(n + 1)}{n} \end{aligned}$$

Since $n < n + 1$,

$$\begin{aligned} 3 & > \frac{p^2}{q^2} \cdot \frac{(n + 1)^2}{n^2} > \frac{p^2}{q^2} \\ 3 & > y^2 > x^2 \end{aligned}$$

This means y is a lower bound of A in $\mathbb{Q}$ but greater than x, which contradicts the assumption that x is the greatest lower bound of A in $\mathbb{Q}$ ! So the supposition that $x^2 < 3$ must be false.

As shown in all cases above, it is not possible for A to have a greatest lower bound in $\mathbb{Q}$. This completes the proof.

# posted by rot13(Unafba Pune) @ 10:08 PM