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+16 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
asked in Digital Logic by Boss (18k points)
retagged by | 2k views
But in case if u use explicit then option c

Please explain

2 Answers

+22 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


it 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:
answered by Boss (18k points)
selected 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
+8 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.

answered by Active (1.1k points)
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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