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+23 votes

The decimal value $0.5$ in IEEE single precision floating point representation has

  1. fraction bits of $000\dots 000$ and exponent value of $0$
  2. fraction bits of $000\dots 000$ and exponent value of $−1$
  3. fraction bits of $100\dots 000$ and exponent value of $0$
  4. no exact representation
in Digital Logic by
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But in case if u use explicit then option c

Please explain
Why denormalized representation not used here ??

With E = 0

@jatin khachane 1 denormalized form is used when exponent is 0. Here exponent is -1.


@jatin khachane 1-Untill told, don't consider denormal format.


Single precision FP representation.

SIgn = 0.

Mantissa = 00000...

Exponent = $-1+127=126$ (The value of it is $-1$, anyway. Because $126$ is the biased version of it)

2 Answers

+31 votes
Best answer

(B) is the answer. IEEE 754 representation uses normalized representation when the exponent bits are all non zeroes and hence an implicit '1' is used before the decimal point.So, if mantissa is:


Ut would be treated as:


and hence, the exponent need to be $-1$ for us to get $0.1$ which is the binary representation of $0.5.$

More into IEEE floating point representation:

edited by
why didn't we used biasing here?

I mean to say saved exponent will be $127-1 = 126$
@thor yes, exponent bits stored will be that of "126", but question asks for the value and it is of "-1".
Nice explaination .... Thank you
I think Biasing has been done here because we have to add the bias value to 1 that will make it 127 +1 (128) and In IEEE floating point representation Exponent is stored in 2's complement representation (continuous run of 8 One's) and its value is -1
in ieee representation if exponent is zero it represents fraction then why option c is not right ?

@subhash If the exponent is all 0s, then the value is a denormalized number, which now has an assumed leading 0 before the binary point. This would have represented the number as $(−1)^s × 0.f × 2^{-126}$​​​​​​ but this is not what we need.

+18 votes

In IEEE biasing of exponent is must.

Step 1: decimal 0.5 --> binary 0.1

Step 2: normalize binary 0.1 --> 1.0 * 2-1

Step 3: exponent -1 + 127 = 126 = binary 01111110

Step 4: remove hidden digit from 1.0 --> 0 (1 is implicit in IEEE representation)

Step 5: 0.5 is positive - the sign bit is zero: 0

The next eight bits are the exponent: 01111110

The next 23 bits are the mantissa: 000000000000000000000

Binary result (32 bits): 10111111000000000000000000000000

I think there is something wrong with the question. Arjun Sir, please explain.

edited by
Question did not ask for the "exponent bits" but its value. So, "-1" is enough.
Thank You so much sir.

@ Rajendra Dangwal

sign bit is 0 so firt bit(MSB) in ur ans must be 0 i.e 00111111000000000000000000000000 


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