2. Show that a) \( \sqrt[n]{a} \cdot \sqrt[n]{b}=\sqrt[n]{a b} \)

To prove that \[ \sqrt[n]{a} \cdot \sqrt[n]{b} = \sqrt[n]{a b}, \] we can use the properties of exponents and radicals. Here's a step-by-step demonstration: ### Step 1: Express Radicals as Exponents Recall that the \(n\)-th root of a number can be expressed as that number raised to the power of \(1/n\). Therefore, we can rewrite the radicals as exponents: \[ \sqrt[n]{a} = a^{1/n} \quad \text{and} \quad \sqrt[n]{b} = b^{1/n}. \] ### Step 2: Multiply the Expressions Now, multiply the two expressions: \[ \sqrt[n]{a} \cdot \sqrt[n]{b} = a^{1/n} \cdot b^{1/n}. \] ### Step 3: Use the Property of Exponents When multiplying two expressions with the same exponent, you can multiply the bases and keep the exponent: \[ a^{1/n} \cdot b^{1/n} = (a \cdot b)^{1/n}. \] ### Step 4: Convert Back to Radical Notation Finally, convert the expression back to radical form: \[ (a \cdot b)^{1/n} = \sqrt[n]{a b}. \] ### Conclusion Putting it all together, we have shown that: \[ \sqrt[n]{a} \cdot \sqrt[n]{b} = \sqrt[n]{a b}. \] This property is a fundamental rule of radicals and exponents, allowing us to combine roots of the same degree through multiplication.