Multiplying Positives Makes a Positive¶

Proposition

\[ \forall a > 0, \forall b > 0 \quad (a, b \in \mathbb{R}): \quad ab > 0. \]

Verifying for Rationals¶

Lemma:

\[\begin{split} \begin{aligned} & \forall a > 0, \forall b > 0 \quad (a \in \mathbb{R}, b \in \mathbb{N}): \quad ab > 0 \\\\ & \forall a < 0, \forall b > 0 \quad (a \in \mathbb{R}, b \in \mathbb{N}): \quad ab < 0 \end{aligned} \end{split}\]

Proof (by induction on $b$):

We prove the first part of the lemma. The second part follows similarly.

Base Case: Let $b = 1$. Then:

\[ ab = a \cdot 1 = a > 0 \quad (\text{Identity Element}) \]
Inductive Step: Assume for some $n \in \mathbb{N}$, we have:

\[ a n > 0 \]

We need to show that $a (n + 1) > 0$.

\[ a(n + 1) = a n + a \quad (\text{Distributive Property}) \]

since $an > 0$ and $a > 0$, we have:

\[ a n + a > 0 \]

Thus, $a(n + 1) > 0$.

By induction, the statement holds for all $b \in \mathbb{N}$.

Similar reasoning applies for $a < 0$ and $b \in \mathbb{N}$, yielding $ab < 0$.

\[\tag*{$\blacksquare$}\]

Theorem 1:

\[ \forall a > 0, \forall b > 0 \quad (a \in \mathbb{R}, b \in \mathbb{Q}): \quad ab > 0 \]

Proof (by contradiction):

Suppose the theorem is false, i.e.:

\[ \exists a_0 > 0, \exists b_0 > 0 \quad (a_0 \in \mathbb{R}, b_0 \in \mathbb{Q}): \quad S.T. \quad a_0 b_0 < 0 \]

Let $b_0 = \frac{m}{n}$, where $m, n \in \mathbb{N}$.

\[ \therefore a_0 b_0 = \frac{m a_0}{n} \]

\[ \therefore n a_0 b_0 = m a_0 \]

Since $a_0 > 0$ and $m \in \mathbb{N}$, Thus by the lemma:

\[ n a_0 b_0 = m a_0 > 0 \]

However, since $a_0 b_0 < 0$ (assumption) and $n \in \mathbb{N}$, we also have by the lemma:

\[ n a_0 b_0 < 0 \]

This leads to a contradiction. Therefore, our assumption is false, and the theorem holds.

\[\tag*{$\blacksquare$}\]

The Proof¶

Before extending to the real numbers, we need to establish one key points:

the limit of a bounded sequence of positive rational numbers is strictly positive.

Theorem 2:

If a real sequence $S_n \to S$ as $n \to \infty$, then:

\[ S_n > c > 0 \quad (\forall n \ge N) \implies S \ge c > 0 \]

Proof (by contradiction):

Assume the opposite, i.e.

\[ S < c \]

\[ \therefore c - S > 0 \]

Select $\epsilon = c - S$, we have:

\[ \exists M \in \mathbb{N} \quad S.T. \quad \forall n \ge M: \quad |S_n - S| < c - S \]

\[ \therefore S_n < c - S + S = c \]

According to the assumption, we also have:

\[ S_n > c \quad (\forall n \ge N) \]

Thus $\forall n \ge \max(N, M)$, $S_n$ should be both greater than $c$ and less than $c$, which is a contradiction.

Therefore, our assumption is false, and the theorem holds.

\[\tag*{$\blacksquare$}\]

Now we can prove the proposition.

Proof:

Construct a sequence of rational numbers $b_n \in \mathbb{Q}$ such that:

\[\begin{split} \begin{aligned} & \forall n \in \mathbb{N}: b_n > 0 \\\\ & \lim_{n \to \infty} b_n = b \end{aligned} \end{split}\]

This is possible by the density of $\mathbb{Q}$ in $\mathbb{R}$. For the same reason, we can select:

\[ q \in \mathbb{Q} \quad S.T. \quad 0 < q < b \]

\[ \because \lim_{n \to \infty} b_n = b \]

\[ \therefore \exists N \in \mathbb{N} \quad S.T. \quad \forall n \ge N: \quad b_n > q \]

\[ \therefore b_n - q > 0 \]

Since $a > 0$, $b_n - q > 0$, and $b_n - q \in \mathbb{Q}$, applying Theorem 1:

\[ a (b_n - q) > 0 \]

\[ \therefore a b_n > a q > 0 \quad \forall n \ge N \]

Define function:

\[ f(x) = a x \]

Since $f(x)$ is continuous for $x \in \mathbb{R}$, we have:

\[ \lim_{n \to \infty} f(b_n) = \lim_{n \to \infty} a b_n = ab \]

Let sequence $S_n = a b_n$. Then $S_n \to ab$ as $n \to \infty$. Also since $S_n > a q > 0$ for all $n \ge N$, applying Theorem 2:

\[ ab \ge a q > 0 \]

\[\tag*{$\blacksquare$}\]

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