Question

that’s the same as $(1 \times 8) + (0 \times 4) + (1 \times 2) + (0 \times 1)$, or $8 + 0 + 2 + 0$. indeed, binary 1010 equals the decimal 10. now you try it: how would you represent the decimal number 6 in binary? choose 1 answer: \\(\boldsymbol{\text{a}}\\) 0101 \\(\boldsymbol{\text{b}}\\) 0111 \\(\boldsymbol{\text{c}}\\) 0110 \\(\boldsymbol{\text{d}}\\) 1000 check if you managed to figure that out, congratulations! if not, that’s totally expected: there are techniques that will help you convert between the number systems, and it’s much easier when you learn those techniques. converting decimal to binary here’s my favorite way to convert decimal numbers to binary: 1. grab a piece of paper or a whiteboard. 2. draw dashes for each of the bits. if the number is less than 16, draw 4 dashes. otherwise, for numbers up to 255, draw 8 dashes. bigger numbers than that require more bits and take a while to do by hand so let’s focus on the smaller numbers. 3. write the powers of 2 under each dash. start under the right most dash writing 1, then keep multiplying by 2. 4. now start at the left most dash and ask yourself: is the number greater...

Explanation:

Step1: Recall binary to decimal conversion

To convert a binary number \( b_{n}b_{n - 1}\dots b_{1}b_{0} \) to decimal, we use the formula \( \sum_{i = 0}^{n}b_{i}\times2^{i} \), where \( b_{i} \) is the \( i \)-th bit (0 or 1). We can reverse this to find the binary of a decimal number by finding the largest power of 2 less than or equal to the decimal number, subtracting it, and repeating.

For decimal 6, the powers of 2 are \( 2^{0}=1 \), \( 2^{1}=2 \), \( 2^{2}=4 \), \( 2^{3}=8 \) (but 8 > 6, so we stop at \( 2^{2}=4 \)).

Step2: Decompose 6 using powers of 2

We know that \( 6 = 4 + 2+ 0\times1 \). In terms of powers of 2: \( 6=1\times2^{2}+ 1\times2^{1}+0\times2^{0} \). Wait, also, we can consider 4 bits (since 6 < 16, as per the method in the text). The powers of 2 for 4 bits (from right, starting at \( 2^{0} \)) are \( 8(2^{3}) \), \( 4(2^{2}) \), \( 2(2^{1}) \), \( 1(2^{0}) \). But 8 > 6, so the left - most bit (for \( 2^{3} \)) is 0. Then for \( 2^{2}=4 \), 6 >= 4, so we put 1. Subtract 4 from 6, we get 2. For \( 2^{1}=2 \), 2 >= 2, so we put 1. Subtract 2 from 2, we get 0. For \( 2^{0}=1 \), 0 < 1, so we put 0. So the binary number (from \( 2^{3} \) to \( 2^{0} \)) is 0 (for \( 2^{3} \)), 1 (for \( 2^{2} \)), 1 (for \( 2^{1} \)), 0 (for \( 2^{0} \)), which is 0110.

We can also check each option:

• Option A: 0101. Convert to decimal: \( 0\times8 + 1\times4+0\times2 + 1\times1=4 + 1=5

eq6 \)

• Option B: 0111. Convert to decimal: \( 0\times8+1\times4 + 1\times2+1\times1=4 + 2+1 = 7

eq6 \)

• Option C: 0110. Convert to decimal: \( 0\times8+1\times4+1\times2 + 0\times1=4 + 2=6 \)
• Option D: 1000. Convert to decimal: \( 1\times8+0\times4+0\times2 + 0\times1 = 8

eq6 \)

Answer:

C. 0110