Exam (elaborations)
COS ASSIGNMENT 3 SEMESTER 2
- Course
- Institution
COS ASSIGNMENT 3 SEMESTER 2 FULL MEMO WITH WORKING OUT
[Show more]
1 review
By: lebogangmafoko • 4 year ago
Seller
Follow
bob_ross
Reviews received
Content preview
, 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
The benefits of buying summaries with Stuvia:
Guaranteed quality through customer reviews
Stuvia customers have reviewed more than 700,000 summaries. This how you know that you are buying the best documents.
Quick and easy check-out
You can quickly pay through credit card or Stuvia-credit for the summaries. There is no membership needed.
Focus on what matters
Your fellow students write the study notes themselves, which is why the documents are always reliable and up-to-date. This ensures you quickly get to the core!
Frequently asked questions
What do I get when I buy this document?
You get a PDF, available immediately after your purchase. The purchased document is accessible anytime, anywhere and indefinitely through your profile.
Satisfaction guarantee: how does it work?
Our satisfaction guarantee ensures that you always find a study document that suits you well. You fill out a form, and our customer service team takes care of the rest.
Who am I buying these notes from?
Stuvia is a marketplace, so you are not buying this document from us, but from seller bob_ross. Stuvia facilitates payment to the seller.
Will I be stuck with a subscription?
No, you only buy these notes for $4.55. You're not tied to anything after your purchase.
Can Stuvia be trusted?
4.6 stars on Google & Trustpilot (+1000 reviews)
67866 documents were sold in the last 30 days
Founded in 2010, the go-to place to buy study notes for 14 years now