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Q: What is the effect of increasing the number of bits on 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.

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

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.

when the bit rate increases bandwidth increases.

The number is divided by 4.

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.

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).

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.

A 32 binary number is a number stored by a computer in 32 bits. it can represent: 1) An unsigned number in the range 0 to 4,294,967,295 2) A signed number in the range -2,147,483,648 to 2,147,483,647 3) A single precision IEEE floating point number with 1 sign bit, 8 exponent bits and 23 mantissa bits give an accuracy of about 7.2 decimal digits and a range of ± 10^-38 to 10^38

Decimal Cases * * * () * () In programming, a floating point number is expressed as . In general, a floating-point number can be written aswhere * M is the fraction mantissa or significand. * E is the exponent. * B is the base, in decimal case . Binary Cases As an example, a 32-bit word is used in MIPS computer to represent a floating-point number: 1 bit ..... 8 bits .............. 23 bits representing: * The implied base is 2 (not explicitly shown in the representation). * The exponent can be represented in signed 2's complement (but also see biased notation later). * The implied decimal point is between the exponent field E and the significand field M. * More bits in field E mean larger range of values representable. * More bits in field M mean higher precision. * Zero is represented by all bits equal to 0: Normalization To efficiently use the bits available for the significand, it is shifted to the left until all leading 0's disappear (as they make no contribution to the precision). The value can be kept unchanged by adjusting the exponent accordingly. Moreover, as the MSB of the significand is always 1, it does not need to be shown explicitly. The significand could be further shifted to the left by 1 bit to gain one more bit for precision. The first bit 1 before the decimal point is implicit. The actual value represented isHowever, to avoid possible confusion, in the following the default normalization does not assume this implicit 1 unless otherwise specified. Zero is represented by all 0's and is not (and cannot be) normalized. Example: A binary number can be represented in 14-bit floating-point form in the following ways (1 sign bit, a 4-bit exponent field and a 9-bit significand field): * * * * * with an implied 1.0: By normalization, highest precision can be achieved. The bias depends on number of bits in the exponent field. If there are e bits in this field, the bias is , which lifts the representation (not the actual exponent) by half of the range to get rid of the negative parts represented by 2's complement. The range of actual exponents represented is still the same. With the biased exponent, the value represented by the notation is:Note: * Zero exponent is represented by , the bias of the notation; * The range of exponents representable is from -126 to 127; * The exponent (with all zero significand) is reserved to represent infinities or not-a-number (NaN) which may occur when, e.g., a number is divided by zero; * The smallest exponent is reserved to represent denormalized numbers (smaller than which cannot be normalized) and zero, e.g., is represented by: Normalization: If the implied base is , the significand must be shifted multiple of q bits at a time so that the exponent can be correspondingly adjusted to keep the value unchanged. If at least one of the first q bits of the significand is 1, the representation is normalized. Obviously, the implied 1 can no longer be used. Examples: * Normalize . Note that the base is 4 (instead of 2)Note that the significand has to be shifted to the left twobits at a time during normalization, because the smallest reduction of the exponent necessary to keep the value represented unchanged is 1, corresponding to dividing the value by 4. Similarly, if the implied base is , the significand has to be shifted 3 bits at a time. In general, if , normalization means to left shift the significand q bits at a time until there is at least one 1 in the highest q bits of the significand. Obviously the implied 1 can not be used. * Represent in biased notation with bits for exponent field. The bias is and implied base is 2.The biased exponent is , and the notation is (without implied 1): or (with implied 1): * Find the value represented in this biased notation: The biased exponent is 17, the actual exponent is , the value is (without implied 1):or (with implied 1):Examples of IEEE 754: * -0.3125The biased exponent is , * 1.0The biased exponent is , * 37.5The based exponent: , . * -78.25The biased exponent: , * As the most negative exponent representable is -126, this value is a denorm which cannot be normalized: by GAURAV PANDEY & VIJAY MAHARA..........AMRAPALI INSTITUTE...................

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.

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

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