retagged by
10,000 views
20 votes
20 votes

The format of the single-precision floating point representation of a real number as per the $\text{IEEE 754}$ standard is as follows:

$$\begin{array}{|c|c|c|} \hline \text{sign} & \text{exponent} & \text{mantissa} \\ \hline \end{array}$$

Which one of the following choices is correct with respect to the smallest normalized positive number represented using the standard?

  1. exponent $=00000000$ and mantissa $=0000000000000000000000000$
  2. exponent $=00000000$ and mantissa $=0000000000000000000000001$
  3. exponent $=00000001$ and mantissa $=0000000000000000000000000$
  4. exponent $=00000001$ and mantissa $=0000000000000000000000001$
retagged by

3 Answers

Best answer
17 votes
17 votes

In IEEE 754 representation all $1s$ in exponent field is reserved for special numbers

  • $+$ (when sign bit is positive) and $-$ (when sign bit is negative) infinities when all manitssa bits are zeroes.
  • SNAN (Signaling Not A Number): when leading mantissa bit is $0$ and at least one other mantissa bit is non-zero
  • NAN (Quiet NAN): when leading mantissa bit is $1$

More read on QNAN vs SNAN: https://stackoverflow.com/questions/18118408/what-is-the-difference-between-quiet-nan-and-signaling-nan

Also, all $0s$ for exponent field is reserved for denormalized numbers (small numbers between $0$ and $\pm1$ which cannot be represented using normalized numbers). That is, a normalized IEEE 754 represented number (both single and double precision) must have at least one bit set in the exponent field and for the smallest exponent this will be the right most bit. Now, to make it the smallest positive normalized number in single-precision format, we can have all mantissa bits $0$ which will give the numerical value as $1.\underbrace{000\dots0}_{23 \text{ zeroes}} \times 2^{1-127} = 2^{-126}.$ (Here, $1$ before "." is implied in IEEE 754 representation for every normalized numbers and $127$ is the exponent bias used to have negative exponents without an explicit sign bit)

Reference: https://steve.hollasch.net/cgindex/coding/ieeefloat.html 

selected by
15 votes
15 votes
The answer is c.

In normalized form, you cannot have exponent as all one or all zero. So, to make the smallest number, make exponent one and mantissa equal to 0. So the number would become $1.0 * 2^{1-127}$ = $1.0 * 2^{-126}$
4 votes
4 votes
As per the question, We should know that in normalised it is not possible to have exponents all 0’s So A and B options get eliminated from here. But Mantissa All 1’s is possible so. The answer should be C

 

Answer:- C
Answer:

Related questions

13 votes
13 votes
3 answers
1
Arjun asked Feb 18, 2021
4,753 views
If $x$ and $y$ are two decimal digits and $(0.1101)_2 = (0.8xy5)_{10}$, the decimal value of $x+y$ is ___________