Problem 2: True or Explain (Why Not) (15 pts)

1. The decimal number $-48$ is equal to the 8-bit signed magnitude binary number $1011\;0000$.

   Answer: True The magnitude of the binary number is equivalent to (16 + 32 = 48), and the sign bit is '1', making the 8-bit number negative.

2. There are $2^{k} -1$ possible k-bit unsigned binary numbers.

   Answer: False. In unsigned binary, the largest possible value is $U_{\rm max} = 2^{k} -1$, since it consists of $k$ ones. In fact there are $2^{k}$ possible $k$-bit strings, from all zeros to all ones; therefore, there are $2^{k}$ possible unsigned binary numbers.

3. The number $17_8$ is equal to $0A_{16}$.

   Answer: False. The easiest way to do this is to go from octal to binary to hex, padding the most significant bit with 0's as needed. So, $17_8$ can be written as $0000\; 1111_2$ or $0F_{16}$. Another way is to recognize that $17_8 = (8 + 7)_{10} = 15_{10} = F_{16}$.

4. The bit pattern $FA_{16}$ is equivalent to the negative base 10 integer $-26$ when interpreted as a 2's complement number.

   Answer: False If we convert $FA_{16}$ to base 2, and treat it as a 2's complement number, then we have $1111\;1010$, which is negative since the sign bit is 1. However, this mere mortal can't identify the magnitude directly; so, we can take the two's complement of the number in this form, to get
   

   $$0000\;0101\;\; +\;\; 0000\;0001\;=\; 0000\; 0110\; =\; 6_{10}.$$
   
   

   Thus, the original bit pattern is the 2's complement representation of $-6_{10}$, which is not $-26$.

5. Light travels approximately 1 meter in 1 nanosecond. (Note: The speed of light is $3 \times 10^{8}$ meters per second.)

   Answer: Do the math to get $0.3$ meters.