QUESTION 2 2.1 Express the decimal number 114 in binary form. 2.2 Determine the yalue of: \( \begin{array}{ll}2.2 .1 & 11101_{2}+111_{2} \\ 2.2 .2 & i^{12} \text { if } i^{2}=-1 \text { (Show all calculations). }\end{array} \)
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To convert the decimal number 114 into binary, we repeatedly divide by 2 and record the remainders. Starting with 114: - 114 ÷ 2 = 57, remainder 0 - 57 ÷ 2 = 28, remainder 1 - 28 ÷ 2 = 14, remainder 0 - 14 ÷ 2 = 7, remainder 0 - 7 ÷ 2 = 3, remainder 1 - 3 ÷ 2 = 1, remainder 1 - 1 ÷ 2 = 0, remainder 1 Now, we take the remainders from bottom to top: thus, \( 114_{10} = 1110010_{2} \). For \( 2.2.1 \): Adding \( 11101_{2} \) (which is 29 in decimal) and \( 111_{2} \) (which is 7 in decimal) gives: \[ 11101_{2} + 111_{2} = 100100_{2} \] This binary number is \( 36_{10} \). For \( 2.2.2 \): To find \( i^{12} \), we note that \( i^2 = -1 \). So, - \( i^3 = i \times i^2 = i \times -1 = -i \) - \( i^4 = (i^2)^2 = (-1)^2 = 1 \) Therefore, \( i^{12} = (i^4)^3 = 1^3 = 1 \).