a. \( 1=\frac{1}{2}+\square \) c. \( 1=\frac{1}{16}+\square \)

To solve for the square in both equations, we want to isolate it. In the first equation, \( 1 = \frac{1}{2} + \square \), subtract \( \frac{1}{2} \) from both sides to get \( \square = 1 - \frac{1}{2} \), which simplifies to \( \square = \frac{1}{2} \). For the second equation, \( 1 = \frac{1}{16} + \square \), subtract \( \frac{1}{16} \) from both sides: \( \square = 1 - \frac{1}{16} \). To solve this, convert 1 to a fraction with a denominator of 16: \( 1 = \frac{16}{16} \), so \( \square = \frac{16}{16} - \frac{1}{16} = \frac{15}{16} \). The solution gives us \( \square = \frac{15}{16} \).