course material

1. MATH 150 Discrete Mathematics
Answer:
Section I:
Two sets are disjoint if they have no element in common. Thus their intersection is an
empty set. Hence the statement is true: True
True
The big O is the limiting factor and its absolute value multiplied by a constant will still be
the upper limit of the function. True
Suppose x and y have the property that x-y is divisible by q; that is (x-y)/q is an integer then
x and y are congruent modulo m. That is x=y(mod)q. In this case x=40, y=3 and q=2.
(40-3)/2 Thus statement is False.
Octal expansion use digits {0,1,2,3,4,5,6,7. Hence the statement is
Section II
Cardinality of A=3
Cardinality of the power set=
Thus the answer d
is putting elements in an ascending order. Answer b
The worst case occur in liear search algorithm when item is the last element in the array or
is not there at all.
A = { 1,4,5, 8,10, 12,14,15}
B = { 1,2,3,10,13 }
C={8}

2. U = {1, 2, 3, 4, 5, 6, 7, 8, 10, 11, 12, 13, 14, 15}.
{ 1,4,5, 8,10, 12,14,15,8}
The necessary condition for the existence of an inverse is that the function has to be one-to-
one. That is, if f(a)=f(b) then a=b.
This means that every element in the domain has a corresponding value in the codomain..
There are 8 numbers so n=8, the following are the steps:
The process stops at i=7 because 9 has been found.1st Round
[5 1 4 2 8 ] [1 5 4 2 8 ] comparison is done between 1st 2 elements and swaps them.
[ 1 5 4 2 8 ) ( 1 4 5 2 8 ] swaps since
( 1 4 5 2 8 )( 1 4 2 5 8 ) swaps because 5
To find the GCD using Euclidean geometry, we first take the two numbers. Which are our
lengths, given to be our first pair. Thus (75, 55) is our first pair. Dividing 75 by 55, we get 1
and remain with 20. Our new pair becomes (55,20). Dividing 55 by 20, we get 2and a
remainder which is 15. We again form a new pair consisting of the previous remainder and
the just-got remainder: (20,15).
Dividing 20 by 15, we get 1 and a remainder of 5. We again a new pair, (15,5). When we
divide in this case, we get 5 and 0 as a remainder hence 5 is the greatest Common divisor.
A function is bijective if it is both Subjective And Injective. Thus we prove the two. For the
function to meet the surjectivity condition, any point say a, in the given range, there must be
another point, say b, in the domain that meets the condition: . That is, for any point a in the
codomain there is b in the domain
( a real number in the domain).
To prove injectivity, we show that
Hence a=b. Since both injectivity and surjectivity conditions have been met; the function is
bijective.