FloatingPointRepresentation

Then if our general real number x is positive and

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Unformatted text preview: l real number x is positive, (and normalized or subnormal), with x = (b0:b1b2 : : :b23b24b25 : : :)2 2E we have x; = (b0:b1b2 : : :b23)2 2E : Thus, x; is obtained simply by truncating the binary expansion of m at the 23rd bit and discarding b24, b25, etc. This is clearly the closest oating point number which is less than x. Writing a formula for x+ is more complicated since, if b23 = 1, nding the closest oating point number bigger than x will involve some bit \carries" and possibly, in rare cases, a change in E . If x is negative, the situation is reversed: it is x+ which is obtained by dropping bits b24, b25, etc., since discarding bits of a negative number makes the number closer to zero, and therefore larger (further to the right on the real line). 12 The IEEE standard de nes the correctly rounded value of x, which we shall denote round(x), as follows. If x happens to be a oating point number, then round(x) = x. Otherwise, the correctly rounded value depends on which of the following four rounding modes is in e ect: Round down. round(x) = x; : R...
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This note was uploaded on 02/12/2014 for the course MATH 4800 taught by Professor Lie during the Spring '09 term at Rensselaer Polytechnic Institute.

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