Square Root of a Product, Quotient, and Power
Why This Matters
Square roots appear everywhere in mathematics — from the Pythagorean theorem to quadratic equations. To simplify complex expressions involving square roots, we need powerful properties that let us break apart (or combine) roots of products, quotients, and powers. In this lesson, we'll prove three fundamental theorems and learn to apply them fluently.
Theorem 1: Square Root of a Product
Proof:
We need to show that $ \sqrt{a} \cdot \sqrt{b} $ is the square root of $ ab $. By definition, $ \sqrt{ab} $ is the unique non-negative number whose square equals $ ab $.
Step 1: Check that $ \sqrt{a} \cdot \sqrt{b} \geq 0 $.
Since $ a \geq 0 $ and $ b \geq 0 $, we have $ \sqrt{a} \geq 0 $ and $ \sqrt{b} \geq 0 $, so their product $ \sqrt{a} \cdot \sqrt{b} \geq 0 $. ✓
Step 2: Check that $ \left(\sqrt{a} \cdot \sqrt{b}\right)^2 = ab $.
$$ \left(\sqrt{a} \cdot \sqrt{b}\right)^2 = \left(\sqrt{a}\right)^2 \cdot \left(\sqrt{b}\right)^2 = a \cdot b = ab \quad ✓ $$
Conclusion: Since $ \sqrt{a} \cdot \sqrt{b} $ is non-negative and its square equals $ ab $, by the definition of square root, $ \sqrt{a} \cdot \sqrt{b} = \sqrt{ab} $. Q.E.D.
Extension to Three or More Factors:
The property extends naturally. For $ a \geq 0, b \geq 0, c \geq 0 $:
$$ \sqrt{abc} = \sqrt{a} \cdot \sqrt{b} \cdot \sqrt{c} $$Why? Apply Theorem 1 twice:
$$ \sqrt{abc} = \sqrt{(ab) \cdot c} = \sqrt{ab} \cdot \sqrt{c} = \sqrt{a} \cdot \sqrt{b} \cdot \sqrt{c} $$