convert (98.48) to binary
Answers
Answer:
division = quotient + remainder;
98 ÷ 2 = 49 + 0;
49 ÷ 2 = 24 + 1;
24 ÷ 2 = 12 + 0;
12 ÷ 2 = 6 + 0;
6 ÷ 2 = 3 + 0;
3 ÷ 2 = 1 + 1;
1 ÷ 2 = 0 + 1;
2. Construct the base 2 representation of the integer part of the number.
Take all the remainders starting from the bottom of the list constructed above.
98(10) =
110 0010(2)
3. Convert to the binary (base 2) the fractional part: 0.48.
Multiply it repeatedly by 2.
Keep track of each integer part of the results.
Stop when we get a fractional part that is equal to zero.
#) multiplying = integer + fractional part;
1) 0.48 × 2 = 0 + 0.96;
2) 0.96 × 2 = 1 + 0.92;
3) 0.92 × 2 = 1 + 0.84;
4) 0.84 × 2 = 1 + 0.68;
5) 0.68 × 2 = 1 + 0.36;
6) 0.36 × 2 = 0 + 0.72;
7) 0.72 × 2 = 1 + 0.44;
8) 0.44 × 2 = 0 + 0.88;
9) 0.88 × 2 = 1 + 0.76;
10) 0.76 × 2 = 1 + 0.52;
11) 0.52 × 2 = 1 + 0.04;
12) 0.04 × 2 = 0 + 0.08;
13) 0.08 × 2 = 0 + 0.16;
14) 0.16 × 2 = 0 + 0.32;
15) 0.32 × 2 = 0 + 0.64;
16) 0.64 × 2 = 1 + 0.28;
17) 0.28 × 2 = 0 + 0.56;
18) 0.56 × 2 = 1 + 0.12;
19) 0.12 × 2 = 0 + 0.24;
20) 0.24 × 2 = 0 + 0.48;
21) 0.48 × 2 = 0 + 0.96;
22) 0.96 × 2 = 1 + 0.92;
23) 0.92 × 2 = 1 + 0.84;
24) 0.84 × 2 = 1 + 0.68;
We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit) and at least one integer that was different from zero => FULL STOP (losing precision...)