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Q: What is the effect of increasing the of bits used to store the exponent?

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The mantissa holds the bits which represent the number, increasing the number of bytes for the mantissa increases the number of bits for the mantissa and so increases the size of the number which can be accurately held, ie it increases the accuracy of the stored number.

It is somewhat complicated (search for the IEEE floating-point representation for more details), but the basic idea is that you have a few bits for the base, and a few bits for the exponent. The numbers are stored in binary, not in decimal, so the base and the exponent are the numbers "a" and "b" in a x 2b.

28-bits

Floating point numbers are typically stored as numbers in scientific notation, but in base 2. A certain number of bits represent the mantissa, other bits represent the exponent. - This is a highly simplified explanation; there are several complications in the IEEE floating point format (or other similar formats).Floating point numbers are typically stored as numbers in scientific notation, but in base 2. A certain number of bits represent the mantissa, other bits represent the exponent. - This is a highly simplified explanation; there are several complications in the IEEE floating point format (or other similar formats).Floating point numbers are typically stored as numbers in scientific notation, but in base 2. A certain number of bits represent the mantissa, other bits represent the exponent. - This is a highly simplified explanation; there are several complications in the IEEE floating point format (or other similar formats).Floating point numbers are typically stored as numbers in scientific notation, but in base 2. A certain number of bits represent the mantissa, other bits represent the exponent. - This is a highly simplified explanation; there are several complications in the IEEE floating point format (or other similar formats).

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.

Related questions

when the bit rate increases bandwidth increases.

The mantissa holds the bits which represent the number, increasing the number of bytes for the mantissa increases the number of bits for the mantissa and so increases the size of the number which can be accurately held, ie it increases the accuracy of the stored number.

Binary bits are necessary to represent 748 different numbers in the sense that binary bits are represented in digital wave form. Binary bits also have an exponent of one.

It is somewhat complicated (search for the IEEE floating-point representation for more details), but the basic idea is that you have a few bits for the base, and a few bits for the exponent. The numbers are stored in binary, not in decimal, so the base and the exponent are the numbers "a" and "b" in a x 2b.

The exponent field for a float data type according to the IEEE-754 Standard is comprised of 8 bits, a whole number range of 0-255.

40 bits or 5 byrtes

It wears down the high bits (and the bits that come off fills up the low bits).

the largest binary number is 1.84467440737e19. to figure this out you put 2 to the exponent of the certain amount of bits. Eg: 2^64 equals the binary number

28-bits

Floating point numbers are typically stored as numbers in scientific notation, but in base 2. A certain number of bits represent the mantissa, other bits represent the exponent. - This is a highly simplified explanation; there are several complications in the IEEE floating point format (or other similar formats).Floating point numbers are typically stored as numbers in scientific notation, but in base 2. A certain number of bits represent the mantissa, other bits represent the exponent. - This is a highly simplified explanation; there are several complications in the IEEE floating point format (or other similar formats).Floating point numbers are typically stored as numbers in scientific notation, but in base 2. A certain number of bits represent the mantissa, other bits represent the exponent. - This is a highly simplified explanation; there are several complications in the IEEE floating point format (or other similar formats).Floating point numbers are typically stored as numbers in scientific notation, but in base 2. A certain number of bits represent the mantissa, other bits represent the exponent. - This is a highly simplified explanation; there are several complications in the IEEE floating point format (or other similar formats).

It depends on the particular implementation's representation of integer and floating point number. The IEEE 754-2008 standard provides four basic resolutions, 16 bits (not common), 32 bit, 64 bits, and 128 bits (also not common). At the same time, integers can be 8 bits, 16 bits, 32 bits, 64 bits (in 64 bit platforms and some libraries on 32 bit platforms) and 128 bits (not common). In general, if you want to keep resolution down to the units digit, you can store a larger number in an integer than you can in a floating point, due to overhead in the exponent, but, at the same time, due to the scalability of floating point numbers, you can store larger numbers in floating point numbers if you are willing to lose resolution on the low-order end.

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