Questions: b. Use part (a) to explain why multiplication in base two by 100 results in appending 00 to the number. c. Use the distributive property of multiplication over addition and parts (a) and (b) to compute 110 two × 11 two. a. In base two, the number is 10 two. Consider a three-digit number in base two: abc two · 10 two = (a · 10 two + b · 10 two + c · 10 two) = a · 10 two + b · 10 two + c · 10 two + 00 · 00 two = abc00 two

b. Use part (a) to explain why multiplication in base two by 100 results in appending 00 to the number.
c. Use the distributive property of multiplication over addition and parts (a) and (b) to compute 110 two × 11 two.
a. In base two, the number is 10 two. Consider a three-digit number in base two:

abc two · 10 two = (a · 10 two + b · 10 two + c · 10 two)
= a · 10 two + b · 10 two + c · 10 two + 00 · 00 two
= abc00 two

Transcript text: b. Use part (a) to explain why multiplication in base two by $100_{\text {two }}$ results in appending 00 to the number. c. Use the distributive property of multiplication over addition and parts (a) and (b) to compute $110_{\text {two }} \times 11_{\text {two }}$. a. In base two, the number $\square$ is $10_{\text {two }}$ Consider a three-digit number in base two: \[ \begin{aligned} \mathrm{abc}_{\mathrm{two}} \cdot 10_{\mathrm{two}} & =\left(\mathrm{a} \cdot 10_{\mathrm{two}}+\mathrm{b} \cdot 10_{\mathrm{two}}+\mathrm{c} \cdot 10_{\text {two }}\right) \\ & =\mathrm{a} \cdot 10_{\mathrm{two}}+\mathrm{b} \cdot 10_{\mathrm{two}}+\mathrm{c} \cdot 10_{\mathrm{two}}+00 \cdot 00 \text { two } \\ & =\mathrm{abc00}_{\mathrm{two}} \end{aligned} \]

Solution

Solution Steps

Solution Approach

Part (a): Understand the multiplication of a three-digit binary number by $10_{\text{two}}$. This is equivalent to shifting the binary number left by one position, which appends a zero to the end.
Part (b): Explain that multiplying by $100_{\text{two}}$ (which is 4 in decimal) results in shifting the binary number left by two positions, appending two zeros to the end.
Part (c): Use the distributive property of multiplication over addition to compute $110_{\text{two}} \times 11_{\text{two}}$. Convert the binary numbers to decimal, perform the multiplication, and convert the result back to binary.

Step 1: Convert Binary Numbers to Decimal

Convert the binary numbers $110_{\text{two}}$ and $11_{\text{two}}$ to their decimal equivalents: \[ 110_{\text{two}} = 6_{\text{ten}} \] \[ 11_{\text{two}} = 3_{\text{ten}} \]

Step 2: Perform Multiplication in Decimal

Multiply the decimal equivalents: \[ 6_{\text{ten}} \times 3_{\text{ten}} = 18_{\text{ten}} \]

Step 3: Convert the Result Back to Binary

Convert the decimal result back to binary: \[ 18_{\text{ten}} = 10010_{\text{two}} \]