Here is the solution to your question: b) Using two's complement, work out $37_{10} - 17_{10}$, giving your answer in decimal notation. (4 marks) This is equivalent to $37_{10} + (-17_{10})$. We will use an 8-bit representation. Step 1: Convert $37_{10}$ to 8-bit binary. $$37_{10} = 00100101_2$$ Step 2: Convert $17_{10}$ to 8-bit binary. $$17_{10} = 00010001_2$$ Step 3: Find the two's complement of $17_{10}$ to represent $-17_{10}$. Original binary for $17_{10}$: $00010001_2$ One's complement (invert all bits): $11101110_2$ Two's complement (add 1 to one's complement): $$\begin{array}{c} \quad 11101110_2 \\ + \quad 00000001_2 \\ \hline \quad 11101111_2 \end{array}$$ So, $-17_{10}$ in 8-bit two's complement is $11101111_2$. Step 4: Add $37_{10}$ and the two's complement of $-17_{10}$. $$\begin{array}{c} \quad 00100101_2 \quad (37_{10}) \\ + \quad 11101111_2 \quad (-17_{10} \text{ in two's complement}) \\ \hline 1 \ 00100100_2 \end{array}$$ The leftmost '1' is a carry-out. In two's complement arithmetic, a carry-out from the most significant bit is discarded. Step 5: The result is $00100100_2$. Convert this binary number back to decimal. $$0 \times 2^7 + 0 \times 2^6 + 1 \times 2^5 + 0 \times 2^4 + 0 \times 2^3 + 1 \times 2^2 + 0 \times 2^1 + 0 \times 2^0$$ $$= 0 + 0 + 32 + 0 + 0 + 4 + 0 + 0$$ $$= 36_{10}$$ The final answer is $\boxed{36_{10}}$.