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The number $0.085$ represented in $IEEE-754$ single-precision format will be:

  1.   $0 01111011 01111100001010001111011$
  2.   $0 01111011 01011100001010001111011$
  3.   $0 01111011 01010100001010000111011$
  4.   $0 01111011 01111100001010001111011$
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2 Answers

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Best answer

$\begin{align}0.085 \times 2 &= 0.17  \to 0 \\ 0.17 \times 2 &= 0.34 \to 0\\
0.34 \times 2 &= 0.68 \to 0 \\
0.68 \times 2 &= 1.36 \to 1\\
0.36 \times 2 &= 0.72 \to 0\\
0.72 \times 2 &= 1.44 \to 1\\
0.44 \times 2 &= 0.88 \to 0\\
0.88 \times 2 &= 1.76 \to 1\\
0.76 \times 2 &= 1.52 \to 1\end{align}$

So, $0.085_{10} = (0.000101011\dots)_2$
$=1.01011 \times 2^{-4}$ (normalized form)

In IEEE 754 representation, we skip the leading 1 and also the exponent has a bias of +127 for single precision. So, here the mantissa will be $0.01011\dots$ and exponent is $-4+127 = 123_{10} = (1111011)_2 = (01111011)_2$ (exponent is of 8 bits in IEEE 754 representation).

So, the only option matching is B.

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

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option B

2 Comments

Are all the options correctly represented?

Mantissa of given solution and option B) is not matching.  Also, why have you shown only 19 mantissa bits?

It should be 23 bits. right?
1
1

0.08510=(0.000101011…)= 1.01011×2−4=1.01011×2−4 

we are shifting decimal in 4bit right position so expression become 2^-4

1
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