Altavoces nativos del esperanto

Conversions

Enumeration of the first numbers

The first numbers are written: decimal binary 0 0 1 1 2 10 3 11 4 100 5 101 One passes from a binary number to the following by adding 1, as into decimal, without forgetting reserves and by using the tables of following additions: 0+0=0 0+1=1 1+0=1 1+1=10 as follows: 11 + 1 ==== 100 Detail: 1 + 1 = 10 => one poses 0, and retains 1 1 + 1 (reserve) = 10 => one poses 0, and retains 1 0 + 1 (reserve) = 1 => 1

---- The binary Arithmétique (more simply binary calculation) is used by the electronic machines most current (computers, computers, etc) because the presence or the absence of current can be used to represent two figures 0 and 1.

0 represent the closed state

1 represents the open state

Any number can be written into binary (breaks up powers of 2 all in all), for example 35 is broken down into:

32 16 8 4 2 1 1 0 0 0 1 1

One finds there 32,2 and 1 and thus 32+2+1= 35.

Expression of a number

A decimal number with several figures such as 123 is expressed as follows: 1 * 100 (1 * 10²) + 2 * 10 (2 * 10¹) + 3 * 1 (3 * 10⁰) Its representation into binary is 1111011 and is expressed in the same way: 1 * 64 (1 * 2⁶) + 1 * 32 (1 * 2⁵) + 1 * 16 (1 * 2⁴) + 1 * 8 (1 * 2³) + 0 * 4 (0 * 2²) + 1 * 2 (1 * 2¹) + 1 * 1 (1 * 2⁰) continuation of 1010-10100

Decimal system towards the binary system

To develop the example above, the number: 45853 writing in decimal base comes from the sum of numbers written hereafter in decimal base. To tell the truth, to propose a method simpler to include/understand, it is necessary to find the power of 2 largest possible lower or equal to the starting number. One withdraws from the number of origin (RO) this power, by noting one 1, then one again seeks a multiple (RM) for the remainder (Rr).

1. RO= RM1+ Rr1
2. Rr1=RM2+Rr2
3. Rr2=RM3+Rr3

…

32.768 1 times 32.768 one even makes 2 multiplied 14 times by him is 2¹⁵ + 0 0 times 16.384 one even make 2 multiplied 13 times by him is 2¹⁴ + 8.192 1 times 8.192 idem 12 idem 2¹³ + 4.096 1 times 4.096 idem 11 idem 2¹² + 0 0 times 2.048 idem 10 idem 2¹¹ + 0 0 times 1.024 idem 9 idem 2¹⁰ + 512 1 times 512 idem 8 idem 2⁹ + 256 1 times 256 idem 7 idem 2⁸ + 0 0 times 128 idem 6 idem 2⁷ + 0 0 times 64 idem 5 idem 2⁶ + 0 0 times 32 idem 4 idem 2⁵ + 16 1 times 16 idem 3 idem 2⁴ + 8 1 times 8 idem 2 idem 2³ + 4 1 times 4 idem 1 idem 2² + 0 0 times 2 idem 0 idem 2¹ = 2 + 1 1 time 1 2⁰ = 1 =45 853

That is to say written in positional system and decimal notation (by writing the powers of 2): 45.853 = 1×2¹⁵ + 0×2¹⁴ + 1×2¹³ + 1×2¹² + 0×2¹¹ + 0×2¹⁰ + 1×2⁹ + 1×2⁸ + 0×2⁷ + 0×2⁶ + 0×2⁵ + 1×2⁴ + 1×2³ + 1×2² + 0×2¹ + 1×2⁰

Maybe in positional system and binary notation since the powers of 2 are not deferred 45.853 decimal is written 1011 0011 0001 1101 binary (separated by groups of 4 bits to air the reading). This number requires 16 bits for its writing (it lies between 2¹⁵ and 2¹⁶).

The other method to convert a decimal number bases 2 of them is to use successions of divisions by number 2. Thus, one a:

45853/2 = 22926 remainder 1 22926/2 = 11463 remainder 0 11463/2 = 5731 remainder 1 5731/2 = 2865 remainder 1 2865/2 = 1432 remain 1 1432/2 = 716 remainder 0 716/2 = 358 remainder 0 358/2 = 179 remainder 0 179/2 = 89 remainder 1 89/2 = 44 remainder 1 44/2 = 22 remainder 0 22/2 = 11 remainder 0 11/2 = 5 remainder 1 5/2 = 2 remainder 1 2/2 = 1 remainder 0 1/2 = 0 remainder 1 That is to say (by reading the remainders obtained in opposite direction): 1011001100011101

Between bases 2,8 and 16

Binary worms octal or hexadecimal

Bases 8 (octal) and 16 (hexadecimal) are multiple bases of base 2. These two bases were usually employed in data processing and for practical reasons; these bases being strongly related to base 2 and the numbers written in these bases being more " manipulables" (because of shorter writing) by human intellect. The writing of numbers in these bases is easily obtained by regrouping of figures of the writing of the number bases 2 of them.

Octal: base 8: $8 = 2^3$ , it is enough to gather starting from the line and per packages of 3 the binary digits (see Bāguà). Each package of 3 (the last having to be sometimes supplemented by 0 on the left), being the binary writing of a figure bases 8 of them (0₈=000, 1₈=001, 2₈=010, 3₈=011, 4₈=100, 5₈=101, 6₈=110, 7₈=111).

*10101101110₂ will be written 10.101.101 110 and by converting the value of each block into an octal digit, one obtains the octal number 2556₈.

Hexadecimal: base 16: $16 = 2^4$ , therefore one gathers starting from the line and per packages of 4 the binary digits. Each package of 4 bits being the binary representation of a figure bases 16 of them. One thus needs 16 digits, it was decided to use the 10 decimal digits plus the first 6 characters of the alphabet with following convention: A₁₆=10₁₀=1010₂, B₁₆=11₁₀=1011₂, C₁₆=12₁₀=1100₂, D₁₆=13₁₀=1101₂, E₁₆=14₁₀=1110₂ and F₁₆=15₁₀=1111₂.

*10101101110₂ will be written 101 0110 1110 and by converting the value of each block into decimal one obtains: 5,6,14 i.e. 56E₁₆.

One could easily extend this principle to all the bases which are powers of 2.

Towards the binary one

It is enough to convert the value of each figure in their binary format.

*1A2F₁₆ will be written 1,10=8+2, 2,15=8+4+2+1 is 1 1010 0010 1111₂

*156₈ will be written 1,5=4+1, 6=4+2 is 1 101 110₂

Count of the values of the groupings of binary digits

See too

binary Format of the data
Arithmetic
binary Prefix
bibi-binary Floating decimal point
System of Boby Lapointe
Yin and yang

External bonds

Simple: Numeral Binary system

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