What is the difference 2^-1024 and 1/(2^1024) in ieee- double precision format
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From: Vivian Richy on 26 Jan 2010 01:30 Hey , can anyone explain me the difference between 2^-1024 and 1/(2^1024) in ieee double precision format. Thanks in advance
From: James Tursa on 26 Jan 2010 02:26 "Vivian Richy" <vivianrichy(a)gmail.com> wrote in message <hjm25t$4ok$1(a)fred.mathworks.com>... > Hey , > can anyone explain me the difference between 2^-1024 and 1/(2^1024) in ieee double precision format. > Thanks in advance Sounds like homework again, but I will answer. IEEE double precision has a range for the normalized numbers. You can get it easily using MATLAB as follows: >> realmax ans = 1.797693134862316e+308 >> realmin ans = 2.225073858507201e-308 If you look at your first number, it is outside this range. In particular, it is one of the so-called denormalized numbers with exponent bits exactly 000. e.g., >> 2^-1024 ans = 5.562684646268004e-309 >> format hex >> realmin ans = 0010000000000000 % exponent bits 001 >> 2^-1024 ans = 0004000000000000 % exponent bits 000 So you can see it is a factor of 4 less than the smallest normalized number. Now look at realmax: >> format hex >> realmax ans = 7fefffffffffffff The exponent bits are 7fe hex, a decimal value of 2046. The IEEE double exponent has a bias of 1023, so this bit pattern actually represents a value of 2^(2046-1023) = 2^1023. This, along with the mantissa bits being all set, means realmax is very close to, but slightly less than, 2^1024 because the mantissa is a number slightly less than 2. Calculating 2^1024 exceeds realmax, so you get infinity. When you divide by it, you get zero. >> 2^1024 ans = Inf >> 1/2^1024 ans = 0 James Tursa
From: Vivian Richy on 26 Jan 2010 02:46
"James Tursa" <aclassyguy_with_a_k_not_a_c(a)hotmail.com> wrote in message <hjm5ed$269$1(a)fred.mathworks.com>... > "Vivian Richy" <vivianrichy(a)gmail.com> wrote in message <hjm25t$4ok$1(a)fred.mathworks.com>... > > Hey , > > can anyone explain me the difference between 2^-1024 and 1/(2^1024) in ieee double precision format. > > Thanks in advance > > Sounds like homework again, but I will answer. IEEE double precision has a range for the normalized numbers. You can get it easily using MATLAB as follows: > > >> realmax > ans = > 1.797693134862316e+308 > >> realmin > ans = > 2.225073858507201e-308 > > If you look at your first number, it is outside this range. In particular, it is one of the so-called denormalized numbers with exponent bits exactly 000. e.g., > > >> 2^-1024 > ans = > 5.562684646268004e-309 > >> format hex > >> realmin > ans = > 0010000000000000 % exponent bits 001 > >> 2^-1024 > ans = > 0004000000000000 % exponent bits 000 > > So you can see it is a factor of 4 less than the smallest normalized number. > > Now look at realmax: > > >> format hex > >> realmax > ans = > 7fefffffffffffff > > The exponent bits are 7fe hex, a decimal value of 2046. The IEEE double exponent has a bias of 1023, so this bit pattern actually represents a value of 2^(2046-1023) = 2^1023. This, along with the mantissa bits being all set, means realmax is very close to, but slightly less than, 2^1024 because the mantissa is a number slightly less than 2. Calculating 2^1024 exceeds realmax, so you get infinity. When you divide by it, you get zero. > > >> 2^1024 > ans = > Inf > >> 1/2^1024 > ans = > 0 > > James Tursa Hey James, itz actually statement in the homework problem. But i couldn't get it. I thought they are the same. Thank you very much for your help. Richy |