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In the standard $\text{IEEE}$ $754$ single precision floating point representation, there is $1$ bit for sign, $23$ bits for fraction and $8$ bits for exponent. What is the precision in terms of the number of decimal digits?

  1. $5$            
  2. $6$                    
  3. $7$                   
  4. $8$
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23 votes

Option C

Explanation

IEEE-754 Format   $(-1)^{S}  1.M^{E-127}$
$S \to \text{sign}$
$M \to \text{Mantissa}$
$E \to \text{Exponent}$

Precision is represented  by   ' 1.M '  where M is 23 bits and in total 24 bits are used for representing a precised number .

With 24 bits we can represent  $2^{24}$ numbers $(0 - 2^{24} -1).$  What is that value in decimal ?

Largest number that can be represented in any base $x$ is $x^{\text{no. of digits}} -1$ (starting from 0) . So,

$\text{base X}^{\text{no. of digits}} = \text{baseY}^{\text{no. of digits}}$

binary is base 2 and decimal is base 10

So, $2^{24} = 10^x$ . We need to find $x$

taking log  on both sides,

$\log_{2}2^{^{24}}$ = $\log_{2}10^{^{x}}$

$24=x \log_{2}10$

$x=7.22$

so we need minimum 7 digits in decimal

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Precision is the number of digits we can represent accurately.

$1.2 \times 10^8$ - precision is just 2 and rest we have 8 zeroes which might not be accurate.

So, in IEEE 754 floating point representation we have 23 precision bits and we also have one implied bit before '.' making it 24 precision bits. With 24 bits we can represent $\lfloor 24 \log 2 \rfloor = 7$ decimal digits.
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In IEEE 754 single precision format, the precision is 24 bits or 7 decimal digits. ($24 *log_{10}(2) ≈ 7.22$)

In IEEE 754 double precision format, the precision is 53 bits or 15 decimal digits. ($53 *log_{10}(2) ≈ 15.95$)

Notice that in bits, the precision is $|Mantissa| + 1$

Answer:

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