 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/c/solution

E = 1

and a zero fraction, while the largest uses

E = 30

and all ten fraction bits equal to one. Hence

x_{m i n, normal} = 1. 0_{2} 2^{1 - 15} = 2^{- 14}, x_{m a x, normal} = (2 - 2^{- 10}) 2^{15} = 65504.

The corresponding words are 0 00001 0000000000 and 0 11110 1111111111. The stored exponent

31

is reserved and cannot extend the finite normal range.

E = 0

and the implicit leading bit becomes zero, while the scale remains

2^{1 - 15} = 2^{- 14}

. The least positive nonzero fraction is

0.000000000 1_{2} = 2^{- 10}

x_{m i n, subnormal} = 2^{- 10} 2^{- 14} = 2^{- 24}, word 000000 0000000001.

Using scale

2^{- 15}

here would be incorrect: the subnormal number convention uses the minimum normal scale, not the ordinary biased-exponent formula at

E = 0

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