GATE CSE 2021 | Set 2 | Question: 4

Question

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?

• A. exponent $=00000000$ and mantissa $=0000000000000000000000000$
• B. exponent $=00000000$ and mantissa $=0000000000000000000000001$
• C. exponent $=00000001$ and mantissa $=0000000000000000000000000$
• D. exponent $=00000001$ and mantissa $=0000000000000000000000001$

Comments

Case	Exponent (E)	Mantissa (F)	Represents	Formula
Normalized	Not all 0s or all 1s (1 to 254)	Any value	"Normal" numbers	$(-1{)}^S\times 1.F\times 2^{(E-127)}$
Denormalized	All 0s (0)	Non-zero	Very small numbers near zero	$(-1{)}^S\times 0.F\times 2^{-126}$
Zero	All 0s (0)	All 0s	Zero	$\pm 0$
Infinity	All 1s (255)	All 0s	Infinity	$\pm \infty$
NaN	All 1s (255)	Non-zero	Not a Number (e.g., 0/0)	NaN

 

Normalized: The Exponent field (E) cannot be all 0s (00000000) or all 1s (11111111). 

---

the circle give value for denormalized

---

Also Refer :
Digital Logic: GATE CSE 2008 | Question: 4
Digital Logic: GATE CSE 2024 | Set 2 | Question: 4

Answer 1

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 

Comments

Answer will be 1x2^-126 

Not 2^-126

 

@Arjun @Lakshman Patel RJIT

---

@SABAREESH V What’s the difference between the two?

---

@jatinmittal199510 I think it is $2^{n-1}$ ?

Answer 2

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 $1.0 * 2^{1-127}$ = $1.0 * 2^{-126}$

Answer 3

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 4

Correct answer is option C ✅ and Here is the Details and Simplest Explanation 

Answer 5

Ans =c