Error

measured value: $x$ , exact value: $x_{0}$

Error’s operation

e (x \pm y) = e (x) \pm e (y) x (x y) = ye (x) + x_{0} e (y) e (\frac{x}{y}) = \frac{x e ( y ) + ye ( x )}{y y _{0}} (1.1) (1.2) (1.3)

For Eq.1.1, when two approximate value subtracted, its relative error will be increased.

For Eq.1.3, when the value is divided by decimals, its absolute error will be increased.

53 \div 2 26 \div 2 13 \div 2 6 \div 2 3 \div 2 1 \div 2 = 26 R 1 = 13 R 0 = 6 R 1 = 3 R 0 = 1 R 1 = 0 R 1

therefore, $(53)_{10} = (110101.)_{2}$

.7 \times 2 = .4 + 1 .4 \times 2 = .8 + 0 .8 \times 2 = .6 + 1 .6 \times 2 = .2 + 1 .2 \times 2 = .4 + 0 .4 \times 2 = .8 + 0

Therefore, $(0.7)_{10} = (.1 \overline{0110})_{2}$

precision	sign	exponent	mantissa
single	1	8	23
double	1	11	52
long double	1	15	64

the form of a normalized IEEE floating point number :

\pm 1. b_{1} b_{2} \dots b_{m} \times 2^{p} (3.1)

double precision number of 1

Machine epsilon: $ϵ_{ma t c h}$ is the distance between 1 and the smallest floating point number greater than 1. $ϵ_{ma t h c} = 2^{- 52}$ .
Truncate number method：
- Chopping
  - if the 53nd bit is 0, than truncate the number at 53nd bit.
- rounding
  - add 1 to bit 52 if bit 53 is 1, and add nothing to bit 52 if bit 53 is 0.
- Rounding to Nearest Rule
  - the 53 bit is 1, and is followed by other nonzero bits. either round up or round down to make 52 bit equal to 0.
Bit 1 : sign number
- 0: +
- 1: -
Bit 2: exponent sign number
- 1: the exponent number is greater than 1
- 0: the exponent number is less than 1