Discrete Mathematics Study Material

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DISCRETE MATHEMATIC
Student’s Name:
University Affiliation:

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Problem 1
The maximum value of the exponential x in 12-bits can be found as;
10x =22047 which tends to x=616.
As a result, the maximum positive number in 12 bits 2’s complement is
2047.
But 2047 is equivalent to 011111111111 in binary system.
You notice that the decimal component i.e 2047 is obtained by converting
011111111111 as illustrated below;
(1024 +512 +256 +128 + 64 + 32 + 16 +8 + 4 +2 + 1)~(0 1 1 1 1 1 1 1
1 1 1 1 1 )
2047.
Problem 2
To find the values of a 12-bits representation of a signed integer, the
following steps are involved;
i)Find the positive value.
ii)Switch to 1s and 0s.
iii)Add 1.
This can be expressed as;
(-128 + 64 + 32 +16 + 8 + 4 + 2 + 1) = -95
(-128 + 0+ 32 + 0 + 0 + 0 + 0 +0 + 1) = -95
(1 0 1 0 0 0 0 0 1)
Problem 3
Computers use a form of scientific notation for floating point representation.
The numbers themselves are written in scientific notation that have got
three components as a whole.
Therefore,
121.12 can also be expressed in scientific notation;

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121.12=1.2112 * 103
But;
121 can be expressed to be equivalent to (64 +32 + 16 + 8 + 0 + 0 + 1) in
a 12- bits notation.
This gives,
(1 1 1 1 0 0 1) ~ exponent
However, this is not the exact;
You can also express 1.2112 * 103 as 0.12112 * 104
Hence, you can use your 13 biased experiment and add 4 to give;
17 10=(16+ 0+ 0+0+1)
~(1 0 0 0 1 ) ~ mantissa
But;
The floating-point number is represented as sign| mantissa |Exponent;
Expressed as 0 |1 0 0 0 1 | 1 1 1 1 0 0 1|
The integer number is a positive hence expressed as o in the float.
Problem 4
The number stored in question 3 is not exact due to some errors.
The errors can be corrected by finding the quotient value;
1 1 1 0
=1 0 0 0 1 √ 1 1 1 1 0 0 1
- 1 0 0 0 1
1 1 1 1
1 1 1 1 0
-1 1 1 1 1
-1 0 0 0 1

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1 0 1 0
Hence the exact number to be stored is given as 1 1 1 0.
Problem 5
The bits in electronic form can only take two values i.e. 0 and 1.
A letter is usually stored using a byte (8 bits) which should be confirmed,
otherwise, a challenge.
“Bob” is a text which is stored on a computer by first converting each
character to an integer and then storing the integer.
For instance,
You are to store “B” as 66, and “O” as 79 as the integer numbers assigned.
But, “Bob” have got three characters each with independent integer values.
This is equivalent to;
66 + 66 + 79 =211.
211~(128 +64 +32+ 0 + 0 + 4 + 2 +1+ 1)
=(1 1 1 0 0 1 1 1 1)
Therefor, “Bob” would be stored as 9 bits in the computer system.
Problem 6
To get the actual pattern required to store “Bob”, You are required to assign
a different pattern of bits to each of the characters.
For instance, “Bob” has got only three characters.
Each is usually stored in 8 bits but with limitations, with one wasted bit. The
first bit in each of the 8 bits pattern is a 0.
To confirm the pattern;
Let character “B” and “b” be represented by 66 as the integer values and
character “o” as 79.