Q: What is the Largest real number that can be stored in binary using 16 bits?

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674

The largest possible value using 8 bits in binary is actually quite simple. Binary is a numerical system that only uses 2 numbers (1 and 0) to determine value. Our system is decimal. (0-9) Now, a 'bit' is one number from the binary system. It can either be 1 or 0. So, 8 bits means using 8 digits in binary. 1 is greater than 0, so the largest value is 11111111. (8 'one's)

The largest unsigned integer is 26 - 1 = 63, giving the range 0 to 63; The largest signed integer is 25 - 1 = 31, giving the range -32 to 31.

You get the largest number if you sort the digits, from largest to smallest.

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

674

31 - it's binary equivalent is 11111

255 as a decimal number (also known as a base 10 number) = 11111111 in binary (also known as a base 2 number). In binary, each digit is known as a bit, and 8 bits are known as 1 byte. 255 is the largest (positive) number you can make in binary using only 8 bits (1 byte).

The largest possible value using 8 bits in binary is actually quite simple. Binary is a numerical system that only uses 2 numbers (1 and 0) to determine value. Our system is decimal. (0-9) Now, a 'bit' is one number from the binary system. It can either be 1 or 0. So, 8 bits means using 8 digits in binary. 1 is greater than 0, so the largest value is 11111111. (8 'one's)

An N-bit integer holds 2N different values.For an unsigned integer, the range of values is 0..2N-1 thus.For a signed integer using 2s complement, the range is -2N-1..+2N-1-1.Therefore, the largest positive number that can be stored using 8 bits is 255.

The largest unsigned integer is 26 - 1 = 63, giving the range 0 to 63; The largest signed integer is 25 - 1 = 31, giving the range -32 to 31.

write a c++ program to convert binary number to decimal number by using while statement

11b which is 1*2 + 1*1 = 3 would be for two bits. But a byte is 8 bits, so 2 bytes is 16 bits. The largest binary number is [2^16 - 1], which is 65535 (base ten)

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

Using the digits 12279, the largest number that can be represented is 97,221.

In binary the largest number (using IEEE binary16) representable would be: 0111 1111 1111 1111 (grouping the bits in nybbles* for easier reading). This is split as |0|111 11|11 1111 1111| which represents: 0 = sign 111 11 = exponent 11 1111 1111 = mantissa. Using IEEE style, the exponent is offset by 011 11, making the maximum exponent 100 00 This is scientific notation using binary instead of decimal. As such there must be a non-zero digit before the binary point, but in binary this can only ever be a 1, so to save storage it is not stored and the mantissa effectively has an extra bit, which for the 10 bits specified makes it 11 bits long. Thus the mantissa represents: 1.11 1111 1111 This gives the largest number as: 1.1111 1111 11 × 10^10000 (all digits are binary, not decimal.) This expands to 1 1111 1111 1100 0000 (binary) = 0x1ffc0 = 131,008 Note that this is NOT accurate in storage - there are 6 bits which are forced to be zero, making the number only accurate to ±32 (decimal): the second largest possible real would be 1 1111 1111 1000 000 = 0x1ff80 = 130,944 - the numbers are only accurate to about 4 decimal digits; the largest decimal real number would be 1.310 × 10^5, and the next 1.309 × 10^5 and so on. However, with proper IEEE, an exponent with all bits set is used to identify special numbers, which makes the largest possible 0111 1101 1111 1111 which is 1.1111 1111 11 × 10^1111 = 1111 1111 1110 0000 = 0xffe0 = 65504 accurate to ±16, ie the largest is about 6.55 × 10^4. * a nybble is half a byte which is directly representable as a single hexadecimal digit.

IF you are asking what that binary number is in decimal form... it would be 7. The question though seems to be asking waht that decimal number is in binary. You want to know what 111 is in binary? 1101111. Try using google. "111 in binary" as a search phrase gives you the answer.