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? exponent $=00000000$ and mantissa $=0000000000000000000000000$ exponent $=00000000$ and mantissa $=0000000000000000000000001$ exponent $=00000001$ and mantissa $=0000000000000000000000000$ exponent $=00000001$ and mantissa $=0000000000000000000000001$ Digital Logic gatecse-2021-set2 digital-logic number-representation ieee-representation 1-mark + – Arjun asked Feb 18, 2021 retagged Nov 30, 2022 by Lakshman Bhaiya Arjun 10.0k views answer comment Share Follow See all 4 Comments See all 4 4 Comments reply Umakant_Mukhiya commented Oct 28, 2021 i edited by Umakant_Mukhiya Oct 28, 2021 reply Follow Share Since the smallest Normalized value is asked 1 $\leq$ Exponent $\leq$ 254 and Mantissa (can be anything). So, Option A, B cannot be possible If you compare Option C and D, Both Have Exponent = 1 so we will compare Mantissa Part Looking at Mantissa Part it is no Brainer that Option C should be the answer. 8 votes 8 votes ShouvikSVK commented Jan 3, 2022 reply Follow Share Point 1: Smallest and Normalized Number Option A Special Representation of “0” Option B: Denormalized Number Option C: Is Normalized and Smallest Positive number (also Smallest number > 0 ) The number is : 1*(2^-126) assuming the scale factor is 2 Option D: Smallest number + 1 4 votes 4 votes Ray Tomlinson commented Jan 26 reply Follow Share Is circled area value is correct ? 0 votes 0 votes Abhilash Garapati commented Feb 3 reply Follow Share No because it is explicit normalization( denormalization) but in question he is asking about implicit normalization 1 votes 1 votes Please log in or register to add a comment.
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 gatecse answered Apr 3, 2021 selected Apr 3, 2021 by Arjun gatecse comment Share Follow See all 5 Comments See all 5 5 Comments reply Show 2 previous comments SABAREESH V commented Dec 26, 2022 reply Follow Share Answer will be 1x2^-126 Not 2^-126 @Arjun @Lakshman Patel RJIT 0 votes 0 votes Abhrajyoti00 commented Jan 5, 2023 reply Follow Share @SABAREESH V What’s the difference between the two? 1 votes 1 votes Ray Tomlinson commented Jan 26 reply Follow Share @jatinmittal199510 I think it is $2^{n-1}$ ? 0 votes 0 votes Please log in or register to add a comment.
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 $1.0 * 2^{1-127}$ = $1.0 * 2^{-126}$ jatinmittal199510 answered Feb 18, 2021 jatinmittal199510 comment Share Follow See all 0 reply Please log in or register to add a comment.
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 harish3598 answered Feb 26, 2021 harish3598 comment Share Follow See all 0 reply Please log in or register to add a comment.