Now consider adding 3 to 3 223 we get

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: to 3 2;23. We get ( 1:1000000000000000000000 )2 21 + ( 0:0000000000000000000001j1 )2 21 = ( 1:1000000000000000000001j1 )2 21 : This time, the result is not an IEEE single precision oating point number, since its signi cand has 24 bits after the binary point: the 24th is shown beyond the vertical bar. Therefore, the result must be correctly rounded. Rounding down gives the result (1:10000000000000000000001)2 21, while rounding up gives (1:10000000000000000000010)2 21. In the case of rounding to nearest, there is a tie, so the latter result, with the even nal bit, is obtained. For another example, consider the numbers 1, 2;15 and 215, which are all oating point numbers. The result of adding the rst to the second is (1:000000000000001)2, which is a oating point number the result of adding the rst to the third is (1:000000000000001)2 215, which is also a oating point number. However, the result of adding the second number to the third is (1:000000000000000000000000000001)2 215 which is not a oating point number in the IEEE single precision format, since the fraction eld would need 30 bits to represent this number exactly. In this example, usin...
View Full Document

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.

Ask a homework question - tutors are online