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In the IEEE floating point representation the hexadecimal value $0\text{x}00000000$ corresponds to

  1. The normalized value $2^{-127}$
  2. The normalized value $2^{-126}$
  3. The normalized value $+0$
  4. The special value $+0$
asked in Digital Logic by Veteran (59.7k points) | 2.9k views
0
what is the difference between Option A and D

2 Answers

+32 votes
Best answer
answered by Boss (30.9k points)
edited by
+11 votes
answered by Veteran (369k points)
0
sir why not A??
this 32 bit ieee representation also correspond to normalize 2^(-127)
using bias=(2^(8-1))-1  value= (-1)^0 * 1.(23 0's) * 2^(0-bias)=1*2^(-127)
1 implicit here and not stored.  @arjun sir plz reply
+1
srry got the point, we can't take all zero in exponent part in case of ieee 754 single precision, min value that can be stored is 2^(-126), which in case of when exponent field contain 1 in LSB position and all zero in mantissa part, so given format reserved for special value/ thanks
0
Answer is not A because Exponent and Mantisa both are zero which is equal to special value +0.
0
Very Nice and useful link it clears all IEEE Floating Point Representation Doubts.

Thank you Sir!.
0
Answer is not A because range of single precision normalized form is $\pm 2^{-126}$ to $\left (2 -2^{(-23)} \right ) \times 2 ^{127}$
0
But it is hexadecimal number.How can we convert a hexadecimal number to IEEE format?(we convert from only binary number)
Answer:

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