Binary Scientific Notation Solutions 2022

Solutions
Sol-a
Part(i) In binary scientific notation, a decimal number is expressed as
( -1)S × (1 + M) × 2E, where S = sign bit, M = Mantissa, E =
(actual Exponent+Bias)
Here, 3.1*10-9 = (0.0000000031)10 = (0.11)2* 2-28 = (1.1)2 * 2-29 = (1+0.1)2 * 2-29
Comparing with the standard form, in 4-bit sign magnitude representation, the
mantissa =1000
But, the exponent = -29+127(in case of single precision) = 98 which needs more
than 4 bits for its binary notation. Hence, it cann’t be expressed in binary
scientific notation with only 4-bit sign magnitude representation. If expressed, it
would cause overflow error due to smaller allocation of register to store the bits.
Part(ii) In binary scientific notation, a decimal number is expressed as
( -1)S × (1 + M) × 2E, where S = sign bit, M = Mantissa, E =
(actual Exponent+Bias)
3.1*10-9 = (-1)0 * (1+0.1)2 * 2-29
In single precision i.e. 32-bit word register, 1 bit is allocated for sign bit, 8
bits for exponent and 23 bits for Mantissa. Sign bit = 0. Mantissa =
00000000000000000000001. The exponent = actual exponent +bias = -
29+127 = 98 = (01100010)2
Hence, the binary word is,
S E M
0 01100010 00000000000000000000001
Part(iii) (1011)2 in 2’s complement , is the representation of the decimal
number ‘x’ that can be found out by subtracting (0001)2 from the given number
and then, taking the 1’s complement of the resulting number .
Hence, x = 1’s complement of {(1011)2-(0001)2 } = 1’s complement of (1010)2
=negative of (0101)2 = −¿5.
Part(iv) In 4-bit 2’s complement representation, (-1)10 can be represented as
follows:
We know that (1)10 = (0001)2.

A negative number is represented as 2’s complement of its magnitude
representation.
The 2’s complement is represented as 1+one’s complement representation of
the number.
So, (-1)10 = (0001)2+(1110)2 = (1111)2
Sol-b
Part(i) Multiplying 0.4062510 by 2, we get 0.812502.
The integer part of 0.812502 = 0
Again, multiplying the fractional part of last product, 0.812502 by 2 gives
1.625004
The integer part of 1.625004 = 1
Multiplying the fractional part of last product, 0.625004 by 2 gives 1.250008
The integer part of 1.250008 = 1
Multiplying the fractional part of last product, 0.250008 by 2 gives 0.500016
The integer part of 0.500016 = 0
Multiplying the fractional part of last product, 0.500016 by 2 gives 1.000032
The integer part of 1.000032 = 1
Multiplying the fractional part of last product, 0.000032 by 2 gives 0.000064
The integer part of 0.000064= 0
Multiplying the fractional part of last product, 0.000064 by 2 gives 0.000128
The integer part of 0.000128= 0
Multiplying the fractional part of last product, 0.000128 by 2 gives 0.000256
The integer part of 0.000256= 0
Multiplying the fractional part of last product, 0.000256 by 2 gives 0.000512 ...
and this stream of (integer part =0) continues for next ten such steps.
Therefore, the binary representation of decimal number 0.4062510 can be
written as
(0.011010000000000000)2 ≈ (0.01101)2
Part(ii) The two binary values that sandwich (0.01101)2 are (0.0110)2 and
(0.0111)2