Binary floating point subtraction

Question

(In binary environment)
0.100011 * 2^6 - 0.111001 * 2^3

= 0.100011 * 2^6 - 0.000111001 * 2^6

= 0.100011000 * 2^6 + 1.111000111 * 2^6 (convert left part into 2's complement)

= 10.011011111 * 2^6

However, answer was 0.011011111 * 2^6

Where did the first '1' go? Or am I right?

Sorry for lack of specific explanation about my question. But I got serious headache from solving so many question like this one.

2's complement works only with limited lengths where you ignore any carry-over if it would go outside of the length. Try to do the computation on integers to see how it works. It's more or less a computation mod $2^n$. — Gyro Gearloose, Apr 10 '16 at 14:43
So , you mean that my calculation step was correct? And first '1' is carry bit which may be ignored by limiting length of expressing binary? — 구마왕, Apr 10 '16 at 15:09
Yes, technically, you just skip the '1' that is outside the scope. — Gyro Gearloose, Apr 10 '16 at 15:10

score 1 · Accepted Answer · answered Apr 10 '16 at 15:13

Two's complement works for integers when you have a fixed storage space for the value. In this problem you are working with non-integers and don't appear to have a fixed length per value so I'd just use regular subtraction.

$$0.100011 \times 2^6 - 0.111001 \times 2^3$$

$$=100011 - 111.001$$

If it makes it easier you can use a pseudo two-complement approach to do the subtraction and write $111.001$ as $1000.000-000.111$

$$=100011-1000+0.111$$

$$=11011.111$$

Binary floating point subtraction

1 Answers1