Computer Fundamental

This document provides solutions to various questions related to Computer Fundamental. It covers topics such as function composition, sets and relations, number systems, and arrays. The document also includes a bibliography for further reading.

Running head: COMPUTER FUNDEMENTAL
Computer Fundamental
Name of the Student:
Name of the University:
Author Note

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COMPUTER FUNDEMENTAL
Question 1
(i)
f (g (h (10))) = f (g (10)) = f (4) = 12
f –1(h–1(6)) = f –1(7) = 11
(ii)
X 0 1 2 3 4 5 6 7 8 9 1
0
11 12
f(x) 0 1 6 9 10 5 2 1
1
8 3 4 7 12
f-1(x) 0 1 2 3 4 5 6 7 8 9 1
0
11 12
g(f-1(x)) 0 1 5 1 12 5 5 8 8 1 1
2
8 12

2
f’(x))
COMPUTER FUNDEMENTAL
(iii)
X 0 1 2 3 4 5 6 7 8 9 10 11 12
h(x) 0 1 11 3 4 8 7 6 5 9 10 2 12
h(h(x)) 0 1 2 3 4 5 6 7 8 9 10 11 12
The inverse of h(x) is h(h(x)).
(iv)
X 0 1 2 3 4 5 6 7 8 9 10 11 12
f(x) 0 1 5 1 12 5 5 8 8 1 12 8 12
f-1(x) 0 1 6 9 10 5 2 11 8 3 4 7 12

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COMPUTER FUNDEMENTAL
Question 2
The grower set G = {a,b,c,d}
The set of retailers R = {e,f,g}
The set of customer = {m,n,p,q,r}
Hence, G x R have relation A = {aAe, bAg, cAf, dAe, dAg}
R x C Having relation B = {eBn, eBr, fBm, fBq, gBp}
i)
M(A) =
E f g
A 1 0 0
B 0 0 1
C 0 1 0
D 1 0 1
M(B) =
m N p q r
e 0 1 0 0 1
f 1 0 0 1 0
g 0 0 1 0 0
ii)
Inverse of A = M (A) Transpose =

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COMPUTER FUNDEMENTAL
a B C d
e 1 0 0 1
f 0 0 1 0
g 0 1 0 1
Inverse of B = M (B) Transpose =
e F G
m 0 1 0
n 1 0 0
p 0 0 1
q 0 1 0
r 1 0 0
Question 3
a)
Decimal Octal Binary Hexadecimal
119 167 1110111 77
96 140 1100000 60
215 327 11010111 D7
b)
i)
0.53125 in decimal = 0.10001 in binary

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COMPUTER FUNDEMENTAL
ii)
0.09375 in decimal = 0.00011 in binary
iii)
(0.10001) in binary + (0.00011) in binary = 0.10100 in binary
iv)
0.10100 in binary = 0.6250 in decimal
c) i)
11011 + 1001111 = 10111010 in binary
ii)
2756 + 5724 = 10702 in octal
iii)
2AE5C + B52F7 = E0153 in hexadecimal
d)
i)
i) 29578 = 0010 1001 0101 0111 1000
+31495 = 0011 0001 0100 1001 0101
Hence,
0010 1001 0101 0111 1000 +
0011 0001 0100 1001 0101 =
0101 1010 1010 0000 1101 =

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COMPUTER FUNDEMENTAL
ii)
138526 + 490615
138526 = 0001 0011 1000 0101 0010 0110
490615 = 0100 1001 0000 0101 0001 0101
Hence,
0001 0011 1000 0101 0010 0110 +
0100 1001 0000 0101 0001 0101 =
0101 1100 1000 1010 0011 1011
Question 4
i)
a)
The provided number = 0010 1010
Since C = 1 therefore the number is 0000 0001 0010 1010 = 298 in decimal
b)
The provided number = 1100 1001
Since C = 0 therefore the number is 1100 1001 = 201 in decimal
c)
The provided number = 1011 1100
Since C = 1 therefore the number is 0000 0001 1011 1100 = 444 in decimal
d)

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COMPUTER FUNDEMENTAL
The provided number = 0110 1011
Since C = 0 therefore the number is 0110 1011 = 107 in decimal
e)
The provided number = 0111 0011
Since C = 1 therefore the number is 0000 0001 0111 0011 = 371 in decimal
ii)
a)
The provided number = 0010 1010
Since N= 0, V = 0 therefore the number is 0010 1010 = 42 in decimal
b)
The provided number = 1100 1001
Since N = 1 and V= 0 therefore the number is 0011 0110 + 1 = 00110111 = -55 in decimal
c)
The provided number = 1011 1100
Since N = 1 and V = 0 therefore the number is 0100 0011 + 1 = -68 in decimal
d)
The provided number = 0110 1011
Since N = 0, V = 0 therefore the number is 0110 1011 = 107 in decimal
e)
The provided number = 0111 0011

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COMPUTER FUNDEMENTAL
Since N = 0, V = 1 therefore the number is 1000 1100 + 1 = 10001101 = -141 in decimal
Question 5
Number of
columns in
array
Row number Column number Sequence
position
(i) 14 5 11 67
(ii) 36 1 27 27
(iii) 9 4 8 35
(iv) 28 17 11 459
(v) 30 4 8 98
(vi) 45 6 30 300
(vii) 24 5 9 85

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COMPUTER FUNDEMENTAL
Bibliography
Marschner, S., & Shirley, P. (2015). Fundamentals of computer graphics. CRC Press.
Mayer, J., Borges, P. V., & Simske, S. J. (2018). Introduction. In Fundamentals and
Applications of Hardcopy Communication (pp. 1-5). Springer, Cham.