1. **State the problem:** We need to show that dividing $\frac{1}{2}$ by a fraction can sometimes result in a value less than $\frac{1}{2}$ and sometimes not. 2. **Recall the division of fractions rule:** Dividing by a fraction is the same as multiplying by its reciprocal. 3. **Part A: Find a fraction $f$ such that** $$\frac{1}{2} \div f < \frac{1}{2}$$ Using the rule: $$\frac{1}{2} \div f = \frac{1}{2} \times \frac{1}{f} = \frac{1}{2f}$$ We want: $$\frac{1}{2f} < \frac{1}{2}$$ Multiply both sides by $2f$ (assuming $f>0$): $$1 < f$$ So, any fraction $f$ greater than 1 will make the expression less than $\frac{1}{2}$. **Example:** Let $f = \frac{3}{2}$ (which is greater than 1). $$\frac{1}{2} \div \frac{3}{2} = \frac{1}{2} \times \frac{2}{3} = \frac{1}{3} < \frac{1}{2}$$ 4. **Part B: Find a fraction $f$ such that** $$\frac{1}{2} \div f \geq \frac{1}{2}$$ Using the same expression: $$\frac{1}{2f} \geq \frac{1}{2}$$ Multiply both sides by $2f$: $$1 \geq f$$ So, any fraction $f$ less than or equal to 1 will make the expression not less than $\frac{1}{2}$. **Example:** Let $f = \frac{1}{3}$ (which is less than 1). $$\frac{1}{2} \div \frac{1}{3} = \frac{1}{2} \times 3 = \frac{3}{2} > \frac{1}{2}$$ **Summary:** - If $f > 1$, then $\frac{1}{2} \div f < \frac{1}{2}$. - If $f \leq 1$, then $\frac{1}{2} \div f \geq \frac{1}{2}$. This shows Jerald's claim is only sometimes true.