QUESTION 2 2.1 Express the decimal number 114 in binary form. 2.2 Determine the yalue of: $$ \begin{array}{ll}2.2 .1 & 11101_{2}+111_{2} \ 2.2 .2 & i^{12} ext { if } i^{2}=-1 ext { (Show all calculations). }\end{array} $$

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QUESTION 2 2.1 Express the decimal number 114 in binary form. 2.2 Determine the yalue of: $$ \begin{array}{ll}2.2 .1 & 11101_{2}+111_{2} \ 2.2 .2 & i^{12} 	ext { if } i^{2}=-1 	ext { (Show all calculations). }\end{array} $$

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### 2.1 Express the decimal number 114 in binary form.

To convert the decimal number 114 to binary, we can use the method of successive division by 2, keeping track of the remainders.

1. Divide 114 by 2: 
   - Quotient: 57, Remainder: 0
2. Divide 57 by 2: 
   - Quotient: 28, Remainder: 1
3. Divide 28 by 2: 
   - Quotient: 14, Remainder: 0
4. Divide 14 by 2: 
   - Quotient: 7, Remainder: 0
5. Divide 7 by 2: 
   - Quotient: 3, Remainder: 1
6. Divide 3 by 2: 
   - Quotient: 1, Remainder: 1
7. Divide 1 by 2: 
   - Quotient: 0, Remainder: 1

Now, we collect the remainders from bottom to top:

So, $$ 114_{10} = 1110010_{2} $$.

### 2.2 Determine the value of:

#### 2.2.1 $$ 11101_{2} + 111_{2} $$

First, we need to convert both binary numbers to decimal to perform the addition.

1. Convert $$ 11101_{2} $$ to decimal:
   $$
   1 \cdot 2^4 + 1 \cdot 2^3 + 1 \cdot 2^2 + 0 \cdot 2^1 + 1 \cdot 2^0 = 16 + 8 + 4 + 0 + 1 = 29_{10}
   $$

2. Convert $$ 111_{2} $$ to decimal:
   $$
   1 \cdot 2^2 + 1 \cdot 2^1 + 1 \cdot 2^0 = 4 + 2 + 1 = 7_{10}
   $$

Now, we can add the two decimal values:
$$
29 + 7 = 36_{10}
$$

Next, we convert $$ 36_{10} $$ back to binary:
1. Divide 36 by 2: 
   - Quotient: 18, Remainder: 0
2. Divide 18 by 2: 
   - Quotient: 9, Remainder: 0
3. Divide 9 by 2: 
   - Quotient: 4, Remainder: 1
4. Divide 4 by 2: 
   - Quotient: 2, Remainder: 0
5. Divide 2 by 2: 
   - Quotient: 1, Remainder: 0
6. Divide 1 by 2: 
   - Quotient: 0, Remainder: 1

Collecting the remainders from bottom to top gives us:
$$
36_{10} = 100100_{2}
$$

Thus, $$ 11101_{2} + 111_{2} = 100100_{2} $$.

#### 2.2.2 $$ i^{12} $$ if $$ i^{2} = -1 $$

To find $$ i^{12} $$, we can use the property of powers of $$ i $$:
- $$ i^1 = i $$
- $$ i^2 = -1 $$
- $$ i^3 = i^2 \cdot i = -i $$
- $$ i^4 = (i^2)^2 = 1 $$

The powers of $$ i $$ repeat every 4 terms. Therefore, we can find $$ i^{12} $$ by reducing the exponent modulo 4:
$$
12 \mod 4 = 0
$$

Thus, $$ i^{12} = i^0 = 1 $$.

### Final Answers:
- **2.1**: $$ 114_{10} = 1110010_{2} $$
- **2.2.1**: $$ 11101_{2} + 111_{2} = 100100_{2} $$
- **2.2.2**: $$ i^{12} = 1 $$
- **2.1**: 114 in binary is 1110010. - **2.2.1**: $$ 11101_{2} + 111_{2} = 100100_{2} $$ - **2.2.2**: $$ i^{12} = 1 $$

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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