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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
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5 Comments

But in case if u use explicit then option c

Please explain
1
Why denormalized representation not used here ??

With E = 0
0

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

2

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

1
$(0.5)_{10}=(0.1)_{2}=(1.0*2^{-1})_{2}$

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)
5

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2 Answers

37 votes
 
Best answer

(B) is the answer. $\text{IEEE 754}$ 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:

$0000 \dots 0$

Ut would be treated as:

$1.000 \ldots 0$

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:
http://steve.hollasch.net/cgindex/coding/ieeefloat.html

edited by

7 Comments

why didn't we used biasing here?

I mean to say saved exponent will be $127-1 = 126$
22
@thor yes, exponent bits stored will be that of "126", but question asks for the value and it is of "-1".
36
Nice explaination .... Thank you
0
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
0
in ieee representation if exponent is zero it represents fraction then why option c is not right ?
0

@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.

0
In HEX it is 3F000000
0
19 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

4 Comments

Question did not ask for the "exponent bits" but its value. So, "-1" is enough.
7
Thank You so much sir.
0

@ Rajendra Dangwal

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

3
Exponent = −1+127=126 (The value of it is −1, anyway. Because 126 is the biased version of it)
0
Answer:

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