 Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2015/digital-systems-linear-algebra-introduction-to-formal-proof/5/b/solution

Both signs are positive, so perform addition of the aligned significands. The true exponents are

- 1

and

3

, a difference of four. Shift the smaller significand right four places, retaining its low bits:

\frac{5}{8} = 0.00010 1_{2} 2^{3}, 15 \frac{1}{2} = 1.11110 0_{2} 2^{3} .

Add at the common exponent:

1.11110 0_{2} + 0.00010 1_{2} 10.00000 1_{2} so 10.00000 1_{2} 2^{3} = 1.000000 1_{2} 2^{4} .

4

and the biased exponent to

E = 19 = 1001 1_{2}

. There are only seven fraction places to retain, so no floating-point rounding changes the result:

\frac{5}{8} + 15 \frac{1}{2} = 16 \frac{1}{8}, stored fields 010011 0000001000.

The complete word is 0100110000001000.

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