What Is the Name of the Type That Denotes Floating-point Numbers That Can Have Fractional Parts?

Fractional Part

Advanced PIC18 Projects

Dogan Ibrahim , in Motion-picture show Microcontroller Projects in C (Second Edition), 2014

Displaying the Fractional Part

The program in Figure 7.51 does not display the fractional part of the temperature. The plan can exist modified to display the fractional part besides. In the new role, the LSB byte of the converted data is taken into consideration and the fractional part is displayed as ".00", ".25", ".fifty", or ".75". The two most significant bits of the LSB byte are shifted right by 6 $.25. The partial function then takes one of the post-obit values:

Two Shifted LSB Bits	Fractional Role
00	.00
01	.25
10	.50
xi	.75

Figure 7.52 shows the modified program (XC8-SPI2.C).

Effigy 7.52. Modified MPLAB XC8 Program to Display Fractional Part too.

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Number Theory, Algebraic and Analytic

H.Due east. Rose , in Encyclopedia of Physical Science and Technology (Third Edition), 2003

Iv.A.iii Sets of Real Numbers Modulo ane

Let ((α)) denote the partial office of α, that is ((α))  =   α   −   [α]. P. 50. Chebyshev (1821–1894) was the first to enquire: Given α how is the set {((northα)):due north  =   1, 2,…} distributed in the unit interval? Using the methods discussed above, he was able to show that this set is dumbo in the unit of measurement interval and that the distribution is uniform provided α is irrational. His methods employ to a number of like issues.

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Filters

In Ethereal Package Sniffing, 2004

Floating Point Numbers

Floating signal numbers are different from integer numbers in that they contain fractional parts. Even if the number to the right of the decimal point is 0 (or decimal comma, if your locale uses commas instead of periods), information technology's withal a fractional part of the number. Floating point numbers can exist positive or negative. Ethereal provides two types of floating point numbers: regular floating bespeak numbers, and double-precision floating point numbers. The departure between the 2 is that double-precision floating point numbers can more accurately represent numbers than regular floating betoken numbers because more digits can be stored. In practice, all of Ethereal's floating point numbers are of the double-precision blazon.

Floating indicate numbers, either regular or double-precision, are not frequently constitute in protocols, simply they do exist. For example, the who protocol, which is the format of the letters sent by the rwhod program on UNIX systems announcing load averages and current logins, has floating point numbers. Some instance display filters are:

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Non-integral mathematics

Larry D. Pyeatt , William Ughetta , in ARM 64-Fleck Associates Linguistic communication, 2020

8.1.ii Decimal to arbitrary base

When converting from base ten into another base, the integer and fractional parts are treated separately. The base conversion for the integer office is performed in exactly the same was equally in Section one.3.2, using repeated division by the base b. The fractional office is converted using repeated multiplication. For instance, to catechumen the decimal value 5.6875₁₀ to a binary representation:

1.

Convert the integer portion, v₁₀ into its binary equivalent, 101_two.

2.

Multiply the decimal fraction by two. The integer part of the effect is the first binary digit to the right of the radix point.

Because $\mathrm{ten}=0.6875\times \mathrm{ii}=1.375$ , the kickoff binary digit to the right of the binimal point is 1. So far, we have ${5.625}_{10}={101.1}_{\mathrm{two}}$ .

iii.

Multiply the fractional part of x by 2 once once again.

Because $x=0.375\times \mathrm{ii}=0.75$ , the second binary digit to the right of the binimal point is 0. So far, we have ${5.625}_{10}={\mathrm{101.ten}}_{\mathrm{ii}}$ .

4.

Multiply the fractional part of x by ii once again.

Considering $x=0.75\times 2=1.50$ , the 3rd binary digit to the right of the binimal point is 1. So now we have $5.625=101.101$ .

five.

Multiply the fractional function of x past 2 once more.

Considering $x=0.5\times 2=1.00$ , the fourth binary digit to the right of the binimal point is 1. Then now we take $\mathrm{v.625}=101.1011$ .

6.

Since the fractional part of x is at present zero, we know that all remaining digits will exist zero.

The procedure for obtaining the fractional part can exist accomplished hands using a table, equally shown below:

Operation	Result
	Integer	Fraction
.6875 × ii = 1.375	i	.375
.375 × 2 = 0.75	0	.75
.75 × 2 = i.v	1	.5
.5 × 2 = 1.0	ane	.0

Putting it all together, ${5.6875}_{10}={101.1011}_2$ . Subsequently converting a fraction from base of operations 10 into some other base, the upshot should exist verified by converting back into base ten. The results from the previous example can exist expanded as follows:

$1\times {\mathrm{ii}}^{\mathrm{ii}}+0\times {\mathrm{ii}}^1+1\times 2^0+\mathrm{ane}\times 2^{-\mathrm{one}}+0\times 2^{-2}+\mathrm{one}\times 2^{-\mathrm{iii}}+1\times 2^{-\mathrm{iv}}=\mathrm{iv}+0+1+\frac{1}{2}+0+\frac{1}{8}+\frac{\mathrm{one}}{16}={5.6875}_{\mathrm{ten}}$

Converting decimal fractions to base 16 is achieved in a very similar manner. To convert 842.234375_x into base of operations xvi, nosotros first catechumen the integer portion past repeatedly dividing past 16 to yield 34A. We then repeatedly multiply the fractional role, extracting the integer portion or the event each time as shown in the table below:

Functioning	Consequence
	Integer	Fraction
.234375 × xvi = 3.75	3	.75
.75 × xvi = 12.0	12	.0

In the second line, the integer office is 12, which must exist replaced with a hexadecimal digit. The hexadecimal digit for 12₁₀ is C, so the fractional office is 3C. Therefore, ${842.234375}_{10}=34\mathrm{A}.3{\mathrm{C}}_{16}$ The issue is verified past converting back into base of operations 10 every bit follows:

$\mathrm{three}\times {16}^2+4\times {16}^1+10\times {16}^0+\mathrm{three}\times {16}^{-1}+12\times {16}^{-\mathrm{ii}}=768+64+\mathrm{x}+\frac{3}{16}+\frac{12}{256}={842.234375}_{\mathrm{ten}}$

This tabular method works for fractional note as well every bit radix-betoken notation. For instance, when converting $15{\frac{11}{24}}_{10}$ to base vi, nosotros convert the integer portion by repeatedly dividing past half dozen, which results on 23. Then nosotros convert the partial portion as follows:

Operation	Upshot
	Integer	Fraction
$\frac{11}{24}\times 6=2\frac{3}{\mathrm{four}}$	ii	$\frac{\mathrm{three}}{4}$
$\frac{\mathrm{iii}}{\mathrm{iv}}\times 6=4\frac{1}{\mathrm{two}}$	4	$\frac{1}{\mathrm{two}}$
$\frac{1}{\mathrm{two}}\times \mathrm{half\; dozen}=3$	3	0

Therefore, $15{\frac{11}{24}}_{\mathrm{ten}}={23.243}_6$ . To check the issue, nosotros can convert information technology back into base ten every bit follows:

$1\times 6^1+3\times {16}^0+2\times 6^{-1}+4\times {\mathrm{vi}}^{-\mathrm{ii}}+3\times 6^{-2}=15+\frac{2}{\mathrm{half\; dozen}}+\frac{\mathrm{iv}}{36}+\frac{3}{216}=\mathrm{xv}+\frac{72}{216}+\frac{24}{216}+\frac{3}{216}=\mathrm{xv}+\frac{99}{216}=\mathrm{fifteen}\frac{\mathrm{xi}}{24}$

8.1.2.one Bases that are powers-of-two

Converting fractional values betwixt binary, hexadecimal, and octal can exist accomplished in the same manner as with integer values. However, care must be taken to marshal the radix point properly. As with integers, converting from hexadecimal or octal to binary is accomplished past replacing each hex or octal digit with the corresponding binary digits from the advisable table shown in Fig. 1.3.

For example, to convert $\mathrm{5AC}.\mathrm{43}{\mathrm{B}}_{\mathrm{16}}$ to binary, we just replace "5" with "0101," replace "A" with "1010," replace "C" with "1100," replace "4" with "0100," replace "3" with "0011," replace "B" with "1011," So, using the tabular array, we can immediately see that ${\mathrm{5AC}.\mathrm{43B}}_{\mathrm{xvi}}={010110101100.010000111011}_2$ . This method works exactly the same for converting from octal to binary, except that it uses the tabular array on the right side of Fig. 1.3.

Converting fractional numbers from binary to hexadecimal or octal is also very easy using the tables. The process is to split the binary string into groups of $.25, working outwards from the radix point, then replace each group with its hexadecimal or octal equivalent. For instance, to convert 01110010.1010111₂ to hexadecimal, just carve up the number into groups of four $.25, starting at the radix bespeak and working outwards in both directions. It may be necessary to pad with zeroes to make a complete group on the left or right, or both. Our example is grouped every bit follows: $|0111|0010.1010|1110{|}_2$ . Now each group of four bits is converted to hexadecimal by looking upwards the corresponding hex digit in the table on the left side of Fig. one.3. This yields $\mathrm{72}.\mathrm{A}{\mathrm{Eastward}}_{\mathrm{16}}$ . For octal, the binary number would be grouped as follows: $|001|110|010.101|011|100{|}_{\mathrm{two}}$ . Now each grouping of three bits is converted to octal by looking upwards the respective digit in the table on the right side of Fig. 1.3. This yields ${\mathrm{162.534}}_{\mathrm{8}}$ . Note that the conversion to octal required the addition of leading and trailing zeroes.

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Number Representation and Computer Arithmetic

Behrooz Parhami , in Encyclopedia of Data Systems, 2003

V Fixed-Point Numbers

A fixed-betoken number consists of a whole or integral office and a fractional part, with the 2 parts separated by a radix point (decimal indicate in radix x, binary betoken in radix two, and then on). The position of the radix signal is almost always implied and thus the bespeak is non explicitly shown. If a stock-still-point number has grand whole digits and l fractional digits, its value is obtained from the formula:

(4) $x={\Sigma }_{i=-1}^{k-1}{x}_i{r}^i={(x_{k-1}{x}_{\mathrm{one\; thousand}-\mathrm{two}}\dots x_{\mathrm{i}}{x}_0\cdot {\mathrm{10}}_{-1}{x}_{-\mathrm{two}}\dots x_{-l})}_r$

In other words, the digits to the right of the radix betoken are given negative indices and their weights are negative powers of the radix. For example:

$\mathrm{ii.375}=(1\times 2^{\mathrm{i}})+(0\times 2^0)+(0\times {\mathrm{two}}^{-\mathrm{one}})+(\mathrm{i}\times 2^{-\mathrm{two}})+(1\times 2^{-\mathrm{three}})={(10.011)}_{\text{two}}$

In a (g + l)-digit radix-r stock-still-point number system with k whole digits, numbers from 0 to γ ^k −r ^–fifty , in increments of r ^–l can be represented. The step size or resolution r ^–l is oftentimes referred to every bit ulp, or unit in least position. For case, in a (2 + 3)-flake binary stock-still-point number arrangement, we accept ulp = two^−three and the values 0 = (00.000)_two through ii² −2⁻ⁱⁱⁱ = 3.875 = (11.111)₂ are representable. For the same total number k + 50 of digits in a stock-still-betoken number system, increasing g will lead to enlarged range of numbers, whereas increasing l leads to greater precision. Therefore, there is a trade-off betwixt range and precision.

Signed fixed-signal numbers tin be represented by the same methods discussed for signed integers: signed-magnitude, biased format, and complement method. In particular, for 2's-complement format, a negative value −x is represented equally the unsigned value 2 ^k –x. Figure 3 shows encodings of positive and negative integers in the (1 + 3)-scrap stock-still-point 2'southward-complement format. Notation that the positive values 0 to 7/eight (or 2 ^{chiliad–ane} −2^–l , in general) have the standard binary encoding, whereas negative values −ane/8 to −1 (or −2 ^–l to −2 ^g−1, in general) are transformed to unsigned values by adding 2 (or ii ^k , in full general) to them.

Figure three. Schematic representation of four-scrap 2's-complement encoding for (1 + 3)-fleck fixed-signal numbers in the range [−1, + 7/eight].

The two important properties of 2'southward-complement numbers, previously mentioned in connection with integers, are valid here too; namely, the leftmost bit of the number acts as the sign fleck, and the value represented past a item chip blueprint tin can be derived by considering the sign bit as having a negative weight. Hither are 2 examples:

${(01.011)}_{2\text{'}\text{southward}-\text{compl}}=(-0\times 2^1)+(1\times 2^0)+(0\times 2^{-\mathrm{one}})+(1\times 2^{-\mathrm{two}})+(1\times 2^{-3})=+1.375$

${(11.011)}_{\mathrm{two}\text{'}\text{due south}-\text{compl}}=(-\mathrm{one}\times 2^1)+(1\times 2^0)+(0\times 2^{-\mathrm{one}})+(\mathrm{ane}\times 2^{-2})+(\mathrm{one}\times 2^{-3})=-0.625$

Conversion of fixed-signal numbers from radix r to another radix R is done separately for the whole and fractional parts. Converting the whole part was discussed in Department I. To catechumen the fractional role, we can again employ arithmetic in the new radix R or in the quondam radix r, whichever is more convenient. With radix-R arithmetic, we simply evaluate a polynomial in r ⁻¹ whose coefficients are the digits x _i The simplest way to do this is to view the fractional part as an 50-digit integer, convert this integer to radix R, and divide the result by r ^l .

To perform radix conversion using arithmetic in the quondam radix r, we repeatedly multiply the fraction y by the new radix R, noting and removing the integer part in each step. These integer parts correspond to the radix-R digits Ten _–i, offset from Ten _−one. For case, we convert 0.175 to radix two as follows:

Reading the recorded integer parts from top to bottom, we find 0.175 = (0.00101100)_ii. This equality is estimate because the result did non converge to 0. In general a fraction in i radix may not have an exact representation in another radix. In any instance, we simply carry out the process above until the required number of digits in the new radix have been obtained.

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The M.729-Based Spoken language Information Hiding Approach

Wu Zhijun , in Information Hiding in Speech Signals for Secure Communication, 2015

6.1.2.5 Finding Fractional Lag

If information technology is the first subframe and the integer lag satisfies T > 84, the fractional part of the pitch period is set to zero. Otherwise, the fractions around T are evaluated by interpolating the correlation values of Eq. (6.12) in the following way [41,51] 41 51 :

(6.thirteen) $R_0(t,f)=\sum _{i=0}^{3}R[t-i]\mathrm{westward}13[3f+\mathrm{iii}i]+\sum _{i=0}^{3}R[t+i+\mathrm{i}]\mathrm{west}13[\mathrm{iii}-3f+3i]$

for $f=0,\frac{1}{3}$ , and $\frac{\mathrm{ii}}{3}$ . In bodily implementation, v fractional values are considered:

$f=-\frac{2}{3},-\frac{\mathrm{i}}{3},0,\frac{1}{\mathrm{three}},\text{and}\frac{2}{3}\text{.}$

The G.729 adopts such a strategy for the adaptive codebook search that is intended to be computationally efficient with some cede on overall accurateness [51], especially for brusque pitch periods (t + f < 40). It can be concluded that the approach is suboptimal in the sense that the best possible solution is not found, with the benefit of reducing computational brunt. In do, yet, the method in combination with other components of the coder produces synthetic speech of high quality.

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*Dictionary Data Structures

Stefan Edelkamp , Stefan Schrödl , in Heuristic Search, 2012

Multiplicative Hashing

In this approach the production of the key and an irrational number ϕ is computed and the partial part is preserved, resulting in a mapping into $\left[0,1\right)\subset \mathbb{R}$ . This can be used for a hash function that maps the key x to $\mathfrak{\{}0,\dots ,m-\mathrm{one}\mathfrak{\}}$ as follows:

$h\left(\mathrm{10}\right)=\mathfrak{\lfloor }m\left(x\varphi -\mathfrak{\lfloor }\mathrm{10}\varphi \mathfrak{\rfloor }\right)\mathfrak{\rfloor }.$

One of the all-time choices for ϕ for multiplicative hashing is $\left(\sqrt{5}-1\right)∕\mathrm{ii}\approx 0.61\mathrm{viii}0\mathrm{iii}39\mathrm{eight}87$ , the gilded ratio. As an example accept $k=1\mathrm{ii}3,456$ and $\mathrm{thousand}=\mathrm{ane}0,000$ ; and so $h\left(k\right)=\mathfrak{\lfloor }\mathrm{ane}0,000\cdot \left(\mathrm{i}\mathrm{ii}3\mathrm{four}56\cdot \varphi \right)\mathfrak{\rfloor }=41$ .

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Stock-still-Indicate vs. Floating-Betoken

Nasser Kehtarnavaz , in Real-Time Digital Signal Processing, 2005

six.5.1 Division

The floating-signal C67x DSP provides a reciprocal instruction RCPSP, which gives a expert gauge consisting of the correct exponent role and an authentic fractional part up to the eighth binary position. It is possible to extend the accuracy past using this instruction equally the seed signal v[0] for the iterative Newton-Raphson algorithm expressed past the equation

(half-dozen.ix) $v[\mathrm{north}+1]=5[n]*(\mathrm{ii.0}-x*\mathrm{five}[n]),$

where ten is the value whose reciprocal is to exist institute. Accurateness is increased by each iteration of this equation. Hence, on the floating-point C67x DSP, segmentation can be achieved past taking reciprocal of the denominator so by multiplying the reciprocal with the numerator.

On the fixed-point C6x DSP, however, no reciprocal pedagogy is bachelor. I way to compute reciprocal on the fixed-point C6x is to use the iterative Newton-Raphson equation. However, representing "ii" as part of the equation poses a difficulty when values are in Q15 format. This difficulty can be overcome by representing "ii" in Q13 instead of Q15 format, noting that the overall accuracy of the reciprocal is reduced to Q13. Information technology is important to choose an initial seed that volition allow convergence in as few steps equally possible. The code to implement reciprocal with an initial seed of "one" for iii iterations is as follows:

Another manner to implement sectionalisation is by using the conditional subtraction instruction SUBC. Table half dozen-2 shows four possible situations, given a 16-bit positive dividend ten and a 16-flake positive divisor y. If a dividend or divisor is negative, the absolute instruction ABS should exist used to convert them into positive values. The sign of the quotient, of course, will be the same as the sign of the product x * y, which is resolved by using the compare instruction CMPLT. The sign of the balance volition be the same every bit the sign of the dividend. As shown in Tabular array half-dozen-2, all the situations lead to ii cases: integer division and partial division. If x > y, the caliber volition be in integer format, and if 10 > y, it will be in Q15 format.

Table six-2. Partitioning types for different datatypes.

	Datatypes	Division type
x > y	integers or fractions	integer
ten < y	integers or fractions	fractional

To do integer partitioning, the SUBC teaching is repeated 16 times. Equally illustrated in Effigy half dozen-nine, the dividend is shifted until subtracting the divisor no longer gives a negative result. Then, for each subtraction that generates a positive result, the result is shifted and a 1 is placed in LSB. After xvi such subtractions, the caliber appears in the low, and the remainder in the high, portion of the dividend register. Such a binary division is obtained in the same manner as long-division.

Figure 6-9. SUBC segmentation example 33 by v.^†

To do fractional division, the same long-segmentation procedure is used. All the same, this fourth dimension the SUBC instruction is repeated only 15 times, due to Q15 format representations of the dividend and divisor. The code below shows the division for the fractional case, x and y are assumed to exist in Q15 formats and 10 < y:

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Techniques in Detached-Time Stochastic Control Systems

Gianni Ferretti , ... Riccardo Scattolini , in Command and Dynamic Systems, 1995

Two MODEL Conception

Consider the sampled information system of Fig. i, where

Fig. 1. Sampled data organization

$H\left(s\right)=H_r\left(\mathrm{south}\right){\mathrm{eastward}}^{-\tau \mathrm{south}}=\frac{N\left(s\right)}{D\left(\mathrm{south}\right)}{\mathrm{east}}^{-\tau \mathrm{south}}\text{,}$

H_r (south) is a proper rational transfer function and h is the sampling menstruation, and split the continuous fourth dimension filibuster τ into an integer and a fractional office: τ  = d′h  + εh, with 0   <   ε   <   ane. If (F,g,h) is a realization of H_r (s), the pulse transfer function betwixt u(k) and y(1000) is given by (z ^{−   1} is the astern shift operator):

$\begin{array}{l}\mathrm{Grand}\left(z^{-\mathrm{one}}\right)=\mathbf{h}{\left(z\mathbf{I}-\mathbf{\Phi }\right)}^{-\mathrm{one}}\left({\mathbf{\gamma }}_0+z^{-1}{\mathbf{\gamma }}_{\mathrm{ane}}\right)z^{-d\text{'}},\\ \mathbf{\Phi }=e^{\mathbf{F}h},{\mathbf{\gamma }}_0={\displaystyle \underset{0}{\overset{h-\epsilon h}{\int }}{e}^{\mathbf{F}x}dx\mathbf{chiliad},{\mathbf{\gamma }}_{\mathrm{i}}=e^{\mathbf{F}\left(h-\epsilon h\right)}}{\displaystyle \underset{0}{\overset{\epsilon h}{\int }}{e}^{\mathbf{F}x}}dx\mathbf{g}\text{.}\end{array}$

The sampled information system can be likewise described past a linear difference equation of order n  =   deg{D(s)}:

(i.1) $A\left(z^{-\mathrm{i}}\right)y\left(k\right)=B\left(z^{-\mathrm{ane}}\right)u\left(k-d\right)\text{,}$

(1.2) $A\left(z^{-1}\right)=1+a_{\mathrm{i}}{z}^{-\mathrm{one}}+\dots +a_{\mathrm{northward}}{z}^{-\mathrm{due\; north}}$

(1.three) $B\left(z^{-1}\right)=b_0+b_1{z}^{-1}+\dots +b_{\mathrm{due\; north}}{z}^{-n}$

beingness d  = d′   +   1. Note that the fractional delay ε h gives ascent to a unitary increase of the detached-fourth dimension delay, to an additional zip in the numerator of the pulse transfer function and to an additional parameter (b ₀) in the B-polynomial (1.3). Without loss of generality, from now on it volition be assumed b ₀  ≠   0.

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Number systems and codes

B. HOLDSWORTH BSc (Eng), MSc, FIEE , R.C. WOODS MA, DPhil , in Digital Logic Design (Quaternary Edition), 2002

1.3 Conversion between number systems

A number in whatsoever base can be divided into two parts, (a) the integral office to the left of the radix point , and (b) the fractional part to the correct of the radix point. The process of conversion to another base is different for the two parts of the number.

The decimal value of the integral part (Northward_I )₁₀ of a base b number is given past:

${(N_I)}_{\mathrm{ten}}=a_{n-\mathrm{ane}}{b}^{n-1}+a_{n-\mathrm{ii}}{b}^{n-2}+\dots a_1{b}^{\mathrm{i}}+a_0{b}^0$

Dividing both sides of the equation by the base of operations b gives:

$\frac{{[N_i]}_{\mathrm{ten}}}{b}=a_{n-\mathrm{one}}{b}^{n-\mathrm{two}}+a_{n-2}{b}^{\mathrm{due\; north}-3}+a_{n-3}{b}^{\mathrm{north}-3}+\dots a_1{b}^0+\frac{a_0}{b}$

The result of dividing by the base of operations is to leave the least significant digit of the number a ₀ as the rest afterwards the first division. Subsequent repeated divisions will produce remainders of a ₁, a _ii…a _north−1. Equally an example of the process of repeated segmentation by the required base the decimal number (100)₁₀ is converted below to its binary, octal and hexadecimal equivalents:

The decimal value of the partial office (Northward_F )_ten of a base b number is given past:

${({\mathrm{North}}_F)}_{\mathrm{x}}=a_{-1}{b}^{-1}+a_{-2}{b}^{-2}+\dots a_{-\mathrm{grand}}{b}^{-m}$

and if both sides are multiplied by the base of operations, then

$b{({\mathrm{Due\; north}}_F)}_{\mathrm{ten}}=a_{-1}+a_{-\mathrm{ii}}{b}^{-1}+\dots a_{-m}{b}^{-(m-1)}$

and, the first multiplication reveals the coefficient a ₋₁. Subsequent multiplications volition reveal the coefficients a _−ii, a ₋₃, … a _−m . As an instance of this process (0.265)₁₀ is converted to its corresponding binary, octal and hexadecimal forms below:

and the number (0.265)₁₀ is expressed to five binary, octal and hexadecimal places respectively.

Besides these conversions from decimal to binary, octal and hexadecimal, information technology is also possible to convert from both octal and hexadecimal to binary and vice versa.

The octal digits from 0 to 7 inclusive can each exist represented by three binary digits as shown in Figure 1.1(a). To discover the octal representation of a cord of binary digits it is divided into groups of three, starting time from the to the lowest degree significant digit. The octal equivalent for each grouping of three digits is so written down with the aid of the conversion tabular array as shown below:

Figure ane.1. (a) Octal/binary conversion tabular array (b) Hexadecimal/binary conversion table

$\begin{array}{cccc}(110& 001& 011& 100{)}_{\mathrm{ii}}\\ \downarrow & \downarrow & \downarrow & \downarrow \\ =(6& 1& 3& 4{)}_8\end{array}$

If the binary number has a partial part and so, to notice the octal fractional part, dissever the binary fractional number into groups of three beginning at the binary point and moving to the right. The corresponding octal equivalents for each grouping of 3 are then plant in the conversion table. For example:

$\begin{array}{ccccc}(100& 001& 010& 100& .010)\\ \downarrow & \downarrow & \downarrow & \downarrow & \downarrow \\ =(4& 1& 2& 4& .2)\end{array}$

Octal numbers can also be converted to binary past replacing each octal digit with the respective three binary digits from the conversion table. For example:

$\begin{array}{cccc}(4& \mathrm{three}& 2& .7{)}_8\\ \downarrow & \downarrow & \downarrow & \downarrow \\ =(110& 011& 010& .111{)}_{\mathrm{two}}\end{array}$

Similarly, each of the sixteen hexadecimal digits can be represented by four binary digits equally shown in Effigy i.ane(b). To convert a binary number into hexadecimal, divide the integral digits into groups of iv, beginning at the binary point and moving left; and divide the fractional digits into groups of four commencement at the binary point and moving right. Each grouping of four binary digits is then replaced by its hexadecimal equivalent from the conversion tabular array every bit illustrated in the following example:

$\begin{array}{llll}(1011\hfill & 1010\hfill & 0011\hfill & .0010{)}_{\mathrm{ii}}\hfill \\ =(\text{B}\hfill & \text{A}\hfill & 3\hfill & .2{)}_{16}\hfill \end{array}$

For the reverse conversion, each hexadecimal digit can be replaced by the appropriate iv binary digits from the conversion table. For case:

$\begin{array}{llll}\hfill (4\hfill & \hfill \text{F}\hfill & \hfill \text{C}\hfill & \hfill \mathrm{ii}{)}_{\mathrm{sixteen}}\hfill \\ \hfill =(0100\hfill & \hfill 1111\hfill & \hfill 1100\hfill & \hfill 0010{)}_2\hfill \end{array}$

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