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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 $=0000000000000000000000001$

edited

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.

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

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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)

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In general, exponent bias = $2^{n-1}-1$, where n is the number of exponent bits.
Thanks for the explanation, and the reference at the bottom saved a ton of time 🙏

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

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