Chapter 1, Problem 27CRP

a.

Explanation of Solution

Convert base ten representations $7$ to its equivalent binary representation:

• To find the conversion of base ten representation $7$ to its equivalent binary form, divide the value continuously by $2$ until the quotient is equal to zero. Start with the bottom and write the sequence of remainders from bottom to the top:

$\begin{array}{ccc}2& 7& \\ 2& 3& 1\end{array}$

b.

Explanation of Solution

Convert base ten representations $11$ to its equivalent binary representation:

• To find the conversion of base ten representation $11$ to its equivalent binary form, divide the value continuously by $2$ until the quotient is equal to zero. Start with the bottom and write the sequence of remainders from bottom to the top:

$\begin{array}{ccc}2& 11& \\ 2& 5& 1\\ 2\end{array}$

c.

Explanation of Solution

Convert base ten representations $16$ to its equivalent binary representation:

• To find the conversion of base ten representation $16$ to its equivalent binary form, divide the value continuously by $2$ until the quotient is equal to zero. Start with the bottom and write the sequence of remainders from bottom to the top:

$\begin{array}{ccc}2& 16& \\ 2& 8& 0\\ 2& 4& 0\end{array}$

d.

Explanation of Solution

Convert base ten representations $17$ to its equivalent binary representation:

• To find the conversion of base ten representation $17$ to its equivalent binary form, divide the value continuously by $2$ until the quotient is equal to zero. Start with the bottom and write the sequence of remainders from bottom to the top:

$\begin{array}{ccc}2& 17& \\ 2& 8& 1\\ 2& 4& 0\end{array}$

e.

Explanation of Solution

Convert base ten representations $31$ to its equivalent binary representation:

• To find the conversion of base ten representation $31$ to its equivalent binary form, divide the value continuously by $2$ until the quotient is equal to zero. Start with the bottom and write the sequence of remainders from bottom to the top:

$\begin{array}{ccc}2& 31& \\ 2& 15& 1\\ 2& 7& \mathrm{1</}\end{array}$