Exam (elaborations)
COS ASSIGNMENT 3 SEMESTER 2
- Institution
- University Of South Africa (Unisa)
COS ASSIGNMENT 3 SEMESTER 2 FULL MEMO WITH WORKING OUT
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By: lebogangmafoko • 4 year ago
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, Discussion of assignment 03 semester 2.Suppose U = {2, 4, 6, a, b, c, {b, c}} is a universal set with the following subsets:
A = {6, b, c, {b, c}} and B = {2, 6, b, c}.
Answer questions 1 and 2 by using the given sets.
Question 1
Which one of the following relations from A to B is functional?
1. {(2, 6), (6, b), (b, c), (c, {b, c}}
2. {(6, 6), (c, c), (b, 2), ({b, c}, 6)}
3. {(6, 2), (b, 6), (c, b), (6, {b, c})}
4. {(b, 2), (b, 6), (b, b), (b, c)}
Answer: Alternative 2
Discussion
First we look at the definition for functionality:
Suppose R B × C is a binary relation from a set B to a set C. We may call R functional if the
elements of B that appear as first co-ordinates of ordered pairs in R do not appear in more than
one ordered pair of R.
We consider the relations provided in the different alternatives:
1. Let L = {(2, 6), (6, b), (b, c), (c, {b, c}} (say). L is not functional since 2 ∉ A.
Therefore (2, 6) cannot be in L, since L is not defined from A to B.
2. Let M = {(6, 6), (c, c), (b, 2), ({b, c}, 6)} (say). M is functional because each first co-ordinate
is in A, and each first co-ordinate is only used once.
3. Let N = {(6, 2), (b, 6), (c, b), (6, {b, c})} (say). N is a not functional because the element 6
appears twice as first co-ordinate in the relation.
4. Let S = {(b, 2), (b, 6), (b, b), (b, c)} (say). N is a not functional because b appears as first co-
ordinate in every ordered pair in the relation.
From the arguments provided we can deduce that alternative 2 should be selected.
Refer to study guide, p 98.
3
, Question 2
Let F = {(6, {b, c}), (b, 2), (c, c), ({b, c}, 6)} be a function from A to U. Which one of the following
alternatives is true for function F?
1. F is injective and surjective.
2. F is surjective, but not injective.
3. F is neither injective nor surjective.
4. F is injective, but not surjective.
Answer: Alternative 4
Discussion
First we look at the definition of a function:
Suppose R B × C is a binary relation from a set B to a set C. We may call R a function from B to
C if every element of B appears exactly once as the first co-ordinate of an ordered pair in R (i.e.
f is functional), and the domain of R is exactly the set B, i.e. dom(R) = B. We also look at the
definition of an injective and surjective function.
Injective function:
A function f: B C is injective iff f has the property that
whenever f(a1) = f(a2) then a1 = a2 (or whenever a1 a2 then f(a1) f(a2)) .
Surjective function:
Given a function f: B C we say that f: B C is surjective iff the range of f is equal to the
codomain of f, ie f(B) = C.
Each first co-ordinate is paired with only one unique second co-ordinate, therefore F is an
injective function. But F is not surjective. Although ran(A) = {2, 6, c, {b, c})} U, it is not true that
ran(A) = U.
From the arguments provided we can deduce that alternative 4 should be selected.
Refer to study guide, pp 105 - 107.
Question 3
Which one of the following relations is a bijective function on the set B = {t | (−4 t 4, t Z}? Hint:
First list the elements of set B.
1. {(-3, -3), (-1, -1), (1, 1), (3, 3)}
2. {(0, -3), (0, -2), (0, -1), (0, 0), (0, 1), (0, 2), (0, 3)}
3. {(-3, 0), (-2, -1), (-1, -1), (0, 0), (1, -3), (2, -3), (3, 3)}
4. {(-3, 0), (-2, -3), (-1, 3), (0, -1), (1, -2), (2, 2), (3, 1)}
Answer: Alternative 4
Discussion
Let us first look at the definition of a bijective function. A function is bijective iff it is both a
surjective and an injective function. A function f: A → B is injective iff it has the property
that 4
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