Question

Express (118)10 and (-49)10 in 8-bit binary one’s complement form and then add the numbers. What would be the representation (-0)10 in 16-bit binary one’s complement? (be sure to show your work).

Answer 1

Answer:

$118_{10}= 0110111$ 2) $-49_{10}=110001_{2}$ 3) $0_{10}=0:16 \Rightarrow 0_{10}=0_{16}$

Explanation:

1) Expressing the Division as the summation of the quotient and the remainder

for

118, knowing it is originally a decimal form:

118:2=59  +(0), 59/2 =29 + 1, 29/2=14+1, 14/2=7+0, 7/2=3+1, 3/2=1+1, 1/2=0+1

$118_{10}= 0110111$

2) $-49_{10}$

Similarly, we'll start the process with the absolute value of -49 since we want the positive value of it. Then let's start the successive divisions till zero.

|-49|=49

49:2=24+1, 24:2=12+0,12:2=6+0,6:2=3+0,3:2=1+1,1:2=0+1

100011

$-49_{10}=110001_{2}$

3) $(-0)_{10}$

The first step on that is dividing by 16, and then dividing their quotient again by 16, so on and adding their remainders. Simply put:

$(-0)_{10}=0:16=0 \Rightarrow (0)_{10}=0_{16} \:or\\(0)_{16}=0000000000000000$