1. **State the problem:** Show that for any real numbers $a, b \neq 0$, the equality $$|ab| = |a||b|$$ holds. 2. **Recall the definition of absolute value:** For any real number $x$, $$|x| = \begin{cases} x & \text{if } x \geq 0 \\ -x & \text{if } x < 0 \end{cases}$$ 3. **Consider the four cases based on the signs of $a$ and $b$: ** - Case 1: $a > 0$, $b > 0$ - Case 2: $a > 0$, $b < 0$ - Case 3: $a < 0$, $b > 0$ - Case 4: $a < 0$, $b < 0$ 4. **Evaluate each case:** - Case 1: $a > 0$, $b > 0$ $$|a| = a, \quad |b| = b, \quad ab > 0 \implies |ab| = ab$$ So, $$|ab| = ab = a \cdot b = |a||b|$$ - Case 2: $a > 0$, $b < 0$ $$|a| = a, \quad |b| = -b, \quad ab < 0 \implies |ab| = -(ab) = -a b$$ So, $$|ab| = -ab = a(-b) = |a||b|$$ - Case 3: $a < 0$, $b > 0$ $$|a| = -a, \quad |b| = b, \quad ab < 0 \implies |ab| = -(ab) = -a b$$ So, $$|ab| = -ab = (-a) b = |a||b|$$ - Case 4: $a < 0$, $b < 0$ $$|a| = -a, \quad |b| = -b, \quad ab > 0 \implies |ab| = ab$$ So, $$|ab| = ab = (-a)(-b) = |a||b|$$ 5. **Conclusion:** In all cases, $$|ab| = |a||b|$$ holds true for all real $a, b \neq 0$. This completes the proof.