0.000 000 050 in scientific notation or standard form is: 5.0 x 10-8
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1 + 1,111,111,111,111,111,111,111,111,111,111,111,111,111,111,111,111,111,111,111,111,111,111,111,111,111,111,111,111,111 = 1,111,111,111,111,111,111,111,111,111,111,111,111,111,111,111,111,111,111,111,111,111,111,111,111,111,111,111,111,112 Unless it is binary, in which case: 1 + 111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 11111 1111 1111 1111 1111 = 1000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000
G'day,It depends upon your word size and notations use.Irrespective of that, a floating point notation has three parts:Sign Bit | Exponent | MantissaSign Bit is always just one bit.Size of Exponent depends upon a particular selection for a notation. Also, the encoding of Exponent could be in different notation, such as 2's comp, biased, excess, etc.Similarly size of Mantissa depends upon the notation.Here, for your question, I will take 32-bit word for FP and use the following convention.Sign Bit | Exponent | Mantissa1 | 8 bits | 23 bits37 in 2's comp is 100101This could be written as 100101.0000____In normalised form, we start populating the Mantissa using first 1 from the leftMantissa: 1001 0100 0000 0000 0000 000If the radix point were at the extreme left of Mantissa, if would have to be moved 6 places to the right to get the original number. So we encode 6 in excess notation using 8 bits. This forms our Exponent.Exponent: 1000 0110Finally, the number is positive, hence the sign bit will be 0, interpreted as per 2's compSign Bit: 0Put this all together.0 1000 0110 1001 0100 0000 0000 0000 00is 37 in floating point.This is correct to the best of my knowledge. If, however, someone has some correction, please do.Cheers,A
0000 0000 0000 0000
300 = 256 + 32 + 8 + 4 = Binary 0000 0001 0010 1100
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.