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Given the following binary number in 32-bit (single precision) IEEE-754 format:
                          10111110110000000000000000000000
The decimal value closest to this floating-point number is

 

i'm getting : -0.75

answer given is : -0.375

please verify
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 1  01111101 10000000000000000000

$-1.1000 \times 2^{125-127}$

$-1.100 \times 2^{-2}$

$-0.011$

$-(0.25+0.125)$

$\color{darkblue}{-0.375}$

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In $\text{IEEE-754}$ single precision formate ($32$ bit)  binary number is represented as:

$\text{S(1 bit) E(8 bit) M(23 bits)}$,with implicit normalization and exponent is represented with $\text{Excess-127}$ code.

So here,

  • Sign bit= $1$ $\Rightarrow$ number is negative.
  • Exponent bit= $01111101=125$
  • Mantissa bit= $10000000000000000000000 = 1.1 \rightarrow(\text{implicit normalized form})$

$\therefore V= (-1)^S*1.M*\text{Base}^{E-127}$

$\Rightarrow V=(-1)^1*1.1*\text{2}^{125-127}$

$\Rightarrow V= -1*(1.1)_2*\text{2}^{-2}$

$\because (1.1)_2=(1.5)_{10}$

$\therefore V= -1* 1.5*2^{-2}= -0.375$

So correct decimal value is $-0.375.$

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