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It is clear that an irrational number such as is also

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Unformatted text preview: t is clear that an irrational number such as is also represented most accurately by a normalized representation: signi cand bits should not be wasted by storing leading zeros. However, the number zero is special. It cannot be normalized, since all the bits in its representation are zero. The exponent E is irrelevant and can be set to zero. Thus, zero could be represented as 0 E = 0 0.0000000000000000000000 : The gap between the number 1 and the next largest oating point number is called the precision of the oating point system, 1 or, often, the machine precision, and we shall denote this by . In the system just described, the next oating point bigger than 1 is 1:0000000000000000000001 with the last bit b22 = 1. Therefore, the precision is = 2;22. Exercise 1 What is the smallest possible positive normalized oating point number using the system just described? 1 Exercise 2 Could nonzero numbers instead be normalized so that 2 m < 1? Would this be just as good? It is quite instructive to suppose that the computer word size is much smaller than 32 bits and work out in detail what all the possible oating numbers are in such a case...
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