A002B07D16 -> binary A 1010 0 0000 0 0000 2 0010 B 1011 0 0000 7 0111 D 1101 A002B07D16 = 1010 0000 0000 0010 1011 0000 0111 11012 10 100 000 000 000 101 011 000 001 111 1012 -> octal 010 2 100 4 000 0 000 0 000 0 101 5 011 3 000 0 001 1 111 7 101 5 10 100 000 000 000 101 011 000 001 111 1012 = 240005301758 A002B07D16 = 24 000 530 1758
54 010 000 000 or 5.401*1010 in standard form
Floating point numbers are stored in scientific notation using base 2 not base 10.There are a limited number of bits so they are stored to a certain number of significant binary figures.There are various number of bytes (bits) used to store the numbers - the bits being split between the mantissa (the number) and the exponent (the power of 10 (being in the base of the storage - in binary, 10 equals 2 in decimal) by which the mantissa is multiplied to get the binary/decimal point back to where it should be), examples:Single precision (IEEE) uses 4 bytes: 8 bits for the exponent (encoding ±), 1 bit for the sign of the number and 23 bits for the number itself;Double precision (IEEE) uses 8 bytes: 11 bits for the exponent, 1 bit for the sign, 52 bits for the number;The Commodore PET used 5 bytes: 8 bits for the exponent, 1 bit for the sign and 31 bits for the number;The Sinclair QL used 6 bytes: 12 bits for the exponent (stored in 2 bytes, 16 bits, 4 bits of which were unused), 1 bit for the sign and 31 bits for the number.The numbers are stored normalised:In decimal numbers the digit before the decimal point is non-zero, ie one of {1, 2, ..., 9}.In binary numbers, the only non-zero digit is 1, so *every* floating point number in binary (except 0) has a 1 before the binary point; thus the initial 1 (before the binary point) is not stored (it is implicit).The exponent is stored by adding an offset of 2^(bits of exponent - 1), eg with 8 bit exponents it is stored by adding 2^7 = 1000 0000Zero is stored by having an exponent of zero (and mantissa of zero).Example 10 (decimal):10 (decimal) = 1010 in binary → 1.010 × 10^11 (all digits binary) which is stored in single precision as:sign = 0exponent = 1000 0000 + 0000 0011 = 1000 00011mantissa = 010 0000 0000 0000 0000 0000 (the 1 before the binary point is explicit).Example -0.75 (decimal):-0.75 decimal = -0.11 in binary (0.75 = ½ + ¼) → 1.1 × 10^-1 (all digits binary) → single precision:sign = 1exponent = 1000 0000 + (-0000 0001) = 0111 1111mantissa = 100 0000 0000 0000 0000 0000Note 0.1 in decimal is a recurring binary fraction 0.1 (decimal) = 0.0001100110011... in binary which is one reason floating point numbers have rounding issues when dealing with decimal fractions.
Since 010 = 10, 010 of 1% = 10 of 1% = 10% or 0.1
To convert a binary number to an octal number, you need to know how an octal number is represented in binary. It is like this: 0 = 000 4 = 100 1 = 001 5 = 101 2 = 010 6 = 110 3 = 011 7 = 111 As you can see, an octal number consists of 3 'bits' (either a 0 of a 1). Now, to convert a binary number to an octal number, you first have to group the binary digits into groups of 3 bits (starting from the right). Then, you convert every group of bits into octal numbers. This way you get your binary number into an octal one. For example: (1010100111010010)2 We group them into groups of 3 bits, starting from the right. 1 010 100 111 010 010 As you see, we have a single digit left. We must add 0's to make it a group of 3 bits. 001 010 100 111 010 010 Then we convert every group into an octal number, according to the table above. 001 = 1 010 = 2 100 = 4 111 = 7 010 = 2 010 = 2 And in this way, you converted a binary number into an octal one. (1010100111010010)2 = (124722)8
A002B07D16 -> binary A 1010 0 0000 0 0000 2 0010 B 1011 0 0000 7 0111 D 1101 A002B07D16 = 1010 0000 0000 0010 1011 0000 0111 11012 10 100 000 000 000 101 011 000 001 111 1012 -> octal 010 2 100 4 000 0 000 0 000 0 101 5 011 3 000 0 001 1 111 7 101 5 10 100 000 000 000 101 011 000 001 111 1012 = 240005301758 A002B07D16 = 24 000 530 1758
10 billion 10 million.
1 000 000 010 000 000 days = 2.73790929 × 1012 year
50, 040, 546, 092, 010, 000, 000, 000, 000
54 010 000 000 or 5.401*1010 in standard form
13,010 = 1.301 × 104
100 000 010
Bit is short for BInary digiT. It is binary because it can hold two values a 1 or a 0. If you had one bit colour you could get white (1) or Black (0).With 2 bits you can have 4:00 01 10 11with 3 bits you can have 8:000 001 010 011 100 101 110 111with 4 bits you can have 16:0000 ...... 1111with 16 bits you can have 65536:0000 0000 0000 0000 ........ 1111 1111 1111 1111With all the bits set to one the value is 65535 (in decimal) plus 1 for the all set to 0 (Zero) gives a total of 65536.
9, 010, 000 comes one after 9, 009, 999.
input: 76543210(8) output: 111 110 101 100 011 010 001 000(2)
Floating point numbers are stored in scientific notation using base 2 not base 10.There are a limited number of bits so they are stored to a certain number of significant binary figures.There are various number of bytes (bits) used to store the numbers - the bits being split between the mantissa (the number) and the exponent (the power of 10 (being in the base of the storage - in binary, 10 equals 2 in decimal) by which the mantissa is multiplied to get the binary/decimal point back to where it should be), examples:Single precision (IEEE) uses 4 bytes: 8 bits for the exponent (encoding ±), 1 bit for the sign of the number and 23 bits for the number itself;Double precision (IEEE) uses 8 bytes: 11 bits for the exponent, 1 bit for the sign, 52 bits for the number;The Commodore PET used 5 bytes: 8 bits for the exponent, 1 bit for the sign and 31 bits for the number;The Sinclair QL used 6 bytes: 12 bits for the exponent (stored in 2 bytes, 16 bits, 4 bits of which were unused), 1 bit for the sign and 31 bits for the number.The numbers are stored normalised:In decimal numbers the digit before the decimal point is non-zero, ie one of {1, 2, ..., 9}.In binary numbers, the only non-zero digit is 1, so *every* floating point number in binary (except 0) has a 1 before the binary point; thus the initial 1 (before the binary point) is not stored (it is implicit).The exponent is stored by adding an offset of 2^(bits of exponent - 1), eg with 8 bit exponents it is stored by adding 2^7 = 1000 0000Zero is stored by having an exponent of zero (and mantissa of zero).Example 10 (decimal):10 (decimal) = 1010 in binary → 1.010 × 10^11 (all digits binary) which is stored in single precision as:sign = 0exponent = 1000 0000 + 0000 0011 = 1000 00011mantissa = 010 0000 0000 0000 0000 0000 (the 1 before the binary point is explicit).Example -0.75 (decimal):-0.75 decimal = -0.11 in binary (0.75 = ½ + ¼) → 1.1 × 10^-1 (all digits binary) → single precision:sign = 1exponent = 1000 0000 + (-0000 0001) = 0111 1111mantissa = 100 0000 0000 0000 0000 0000Note 0.1 in decimal is a recurring binary fraction 0.1 (decimal) = 0.0001100110011... in binary which is one reason floating point numbers have rounding issues when dealing with decimal fractions.
.010 is thicker than .005