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Consider the unsigned 8-bit fixed point binary number representation, below,

$$b_7 \: \: b_6 \: \: b_5 \: \: b_4 \: \: b_3 \: \: \cdot b_2 \: \: b_1 \: \: b_0$$

where the position of the primary point is between $b_3$ and $b_2$ . Assume $b_7$ is the most significant bit. Some of the decimal numbers listed below cannot be represented exactly in the above representation:

  1. $31.500$
  2. $0.875$
  3. $12.100$
  4. $3.001$

Which one of the following statements is true?

  1. None of $i, ii, iii, iv$ can be exactly represented
  2. Only $ii$ cannot be exactly represented
  3. Only $iii$ and $iv$ cannot be exactly represented
  4. Only $i$ and $ii$ cannot be exactly represented
in Digital Logic by Boss (17.5k points)
edited by | 3.1k views

4 Answers

+15 votes
Best answer
$31.500$ can be represented as $11111.100$ in the above mentioned fixed form representation

$0.875$ can be represented as $00000.111$.

We can't represent $3$ and $4$.

Hence, C is the correct answer.
by Active (1.3k points)
edited by
+25 votes
$b_{7}  b_{6} b_{5} b_{4} b_{3} . b_{2}b_{1}b_{0}$

Before decimal point means $b_{7}  b_{6} b_{5} b_{4} b_{3}$

$b_{7}  b_{6} b_{5} b_{4} b_{3} \rightarrow {11111}_2 \rightarrow {31}_{10}$

$b_{7}  b_{6} b_{5} b_{4} b_{3} \rightarrow {00000}_2 \rightarrow {0}_{10}$

It means we can represent all the values from $0_{10}$ to $31_{10}$

After decimal point means  $b_{2}b_{1}b_{0}$

 $3$ bits, means total $2^3$ combinations are possible

${111}_2  \mapsto {0.875}_{10}$

${110}_2 \mapsto {0.750}_{10}$

${101}_2 \mapsto {0.625}_{10}$

${100}_2 \mapsto {0.500}_{10}$

${011}_2 \mapsto {0.375}_{10}$

${010}_2 \mapsto {0.250}_{10}$

${001}_2 \mapsto {0.125}_{10}$

${000}_2 \mapsto {0.000}_{10}$

only the  values written above can be represented in $3$ bit binary form,

Now check the values after the decimal point in the given option

${0.500}_{10}$ can be represented

${0.875}_{10}$ can be represented

${0.100}_{10}$ cannot be represented

${0.001}_{10}$ cannot be represented

So, $(iii)$ and $(iv)$ cannot be exactly represented
by Boss (10.8k points)
edited by
0

Thanku pankaj_vir   A Great Explanation 

0
You are welcome
0
nice one
+17 votes

I. $(31.500)_{10}$

$31$ can be represented in binary as $(11111)_2$

$(.500)_{10}$ can be represented in binary as $(.1)_2$

∴ $\color{green}{(31.500)_{10} \rightarrow (11111.1)_2}$

II. $(0.875)_{10}$

$0$ can be represented in binary as $0$

$(.875)_{10}$ can be represented in binary as $(.111)_2$

∴ $\color{green}{(0.875)_{10} \rightarrow (.111)_2}$

III. $(12.100)_{10}$

$(12)_{10}$ can be represented in binary as $(1100)_2$

$(.100)_{10}$ can be represented in binary as $(.0001100....)_2$

∴ $\color{green}{(12.100)_{10} \rightarrow (1100.0001100)_2}$

IV.$(3.001)_{10}$

$(3)_{10}$ can be represented in binary as $(11)_2$

$(.001)_{10}$ can be represented in binary as $(.0000....)_2$

∴ $\color{green}{(3.001)_{10} \rightarrow (11.0000....)_2}$

We can easily be seen that $III. \& IV.$ can't be represented in the given format. 

by Boss (17.8k points)
edited by
+1
I think we only need to find binary value of value after point because as we can see there are min 5 bits given so max 63 value will be there and in option every value less than that
0

@Rishi yadav

Your logic is correct but with 5 bits we can represent decimal value 31 at maximum not 63..

+7 votes
There is no problem before decimal point.

But after decimal point using 3 digits we can represent minimum of 0.125 we can't represent exact value further.

Hence 3 and 4 can't be represented

Hence correct answer is option C
by Boss (18.8k points)
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