1. **Problem:** Prove the basic rules for exponents: a) $a^m \times a^n = a^{m+n}$ b) $\frac{a^m}{a^n} = a^{m-n}$ c) $(a^m)^n = a^{mn}$ d) $a^{-n} = \frac{1}{a^n}$ e) $a^{\frac{1}{n}} = \sqrt[n]{a}$ f) $a^{\frac{m}{n}} = \sqrt[n]{a^m}$ --- 2. **Formula and rules:** - Exponentiation means repeated multiplication. - When multiplying powers with the same base, add exponents. - When dividing powers with the same base, subtract exponents. - Raising a power to another power multiplies exponents. - Negative exponents represent reciprocals. - Fractional exponents represent roots. --- 3. **Proofs:** **a)** $a^m \times a^n = a^{m+n}$ By definition, $a^m = \underbrace{a \times a \times \cdots \times a}_{m\text{ times}}$ and $a^n = \underbrace{a \times a \times \cdots \times a}_{n\text{ times}}$. Multiplying these gives $\underbrace{a \times a \times \cdots \times a}_{m+n\text{ times}} = a^{m+n}$. **b)** $\frac{a^m}{a^n} = a^{m-n}$ Dividing cancels $n$ factors of $a$ from numerator and denominator: $$\frac{\underbrace{a \times \cdots \times a}_{m}}{\underbrace{a \times \cdots \times a}_{n}} = \underbrace{a \times \cdots \times a}_{m-n} = a^{m-n}$$ assuming $m \ge n$. **c)** $(a^m)^n = a^{mn}$ Raising $a^m$ to the $n$th power means multiplying $a^m$ by itself $n$ times: $$(a^m)^n = \underbrace{a^m \times a^m \times \cdots \times a^m}_{n\text{ times}} = a^{m + m + \cdots + m} = a^{mn}$$ **d)** $a^{-n} = \frac{1}{a^n}$ By the division rule, $a^0 = 1$ and $a^{m-n} = \frac{a^m}{a^n}$. Setting $m=0$ gives: $$a^{0-n} = a^{-n} = \frac{a^0}{a^n} = \frac{1}{a^n}$$ **e)** $a^{\frac{1}{n}} = \sqrt[n]{a}$ By definition, the $n$th root of $a$ is the number which when raised to $n$ gives $a$: $$\left(a^{\frac{1}{n}}\right)^n = a^{\frac{1}{n} \times n} = a^1 = a$$ So $a^{\frac{1}{n}}$ is the $n$th root of $a$. **f)** $a^{\frac{m}{n}} = \sqrt[n]{a^m}$ Using the previous rule: $$a^{\frac{m}{n}} = \left(a^m\right)^{\frac{1}{n}} = \sqrt[n]{a^m}$$ --- **Final answer:** All exponent rules are proven as stated.