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ASSIGNMENT 03
Total Marks: 100
Memorandum
ONLY FOR YEAR MODULE
All questions will be marked.
DO NOT USE A CALCULATOR TO OBTAIN YOUR ANSWERS-WHERE APPLICABLE, LEAVE
YOUR ANSWERS IN TERMS OF FACTORIALS, n Cr AND n Pr .
Question 1: 8 Marks
Suppose A = {a, b, c, d, e, f } and R is the relation on A defined by
R = {(a, e), (b, d), (c, c), (d, a), (e, b)}.
State, with reason, whether the relation is:
(1.1) A function from A to A. (2)
Solution
Since we do not have (x, y ) and (y , z) in R for y ̸= z, R is a function from A to A. Informally, no
element of the domain maps to more than one element of the range.
(1.2) An everywhere defined function. (2)
Solution
Since, for each x ∈ A, there exists a y ∈ A such that (x, y ) ∈ R, R is everywhere defined. Informally,
every element of A maps to something.
(1.3) An onto function. (2)
Solution
Since, for each y ∈ A, there exists an x ∈ A such that (x, y) ∈ R, R is onto. Informally, every element
of the range is reached.
(1.4) A one-to-one function. (2)
Solution
Since we do not have (x, z) and (y, z) in R for x ̸= y , R is a one-to-one function from A to A,
Informally, different elements of the domain map to different elements in the range.
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, Question 2: 10 Marks
Let f : A → B and g : B → C be functions. Show that if:
(2.1) f and g are onto, then g ◦ f is onto. (4)
Solution
We note that g ◦ f maps A to C, i.e.
f g
A B C
g◦f
In order to show that g ◦ f is onto, we need to show that for each c ∈ C, there exists an a ∈ A such
that (g ◦ f )(a) = c.
Suppose c ∈ C. Then, since g : B → C is onto, there exists a b ∈ B such that g(b) = c.
Also, since f : A → B is onto, there exists an a ∈ A such that f (a) = b. Hence g(f (a)) = c, i.e.
(g ◦ f )(a) = c. Hence, (g ◦ f ) is onto.
(2.2) f and g are one-to-one, then g ◦ f is one-to-one. (6)
Solution
In order to show that (g ◦ f ) is one-to-one, we need to show that
(g ◦ f )(a) = (g ◦ f )(a′ ) =⇒ a = a′ .
Suppose
(g ◦ f )(a) = (g ◦ f )(a′ ).
Then,
g(f (a)) = g(f (a′ )).
Since g is one-to-one, it follows that
f (a) = f (a′ ).
Now, since f is one-to-one, it follows that
a = a′ .
Thus we have proved that g ◦ f is one-to-one.
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ASSIGNMENT 03
Total Marks: 100
Memorandum
ONLY FOR YEAR MODULE
All questions will be marked.
DO NOT USE A CALCULATOR TO OBTAIN YOUR ANSWERS-WHERE APPLICABLE, LEAVE
YOUR ANSWERS IN TERMS OF FACTORIALS, n Cr AND n Pr .
Question 1: 8 Marks
Suppose A = {a, b, c, d, e, f } and R is the relation on A defined by
R = {(a, e), (b, d), (c, c), (d, a), (e, b)}.
State, with reason, whether the relation is:
(1.1) A function from A to A. (2)
Solution
Since we do not have (x, y ) and (y , z) in R for y ̸= z, R is a function from A to A. Informally, no
element of the domain maps to more than one element of the range.
(1.2) An everywhere defined function. (2)
Solution
Since, for each x ∈ A, there exists a y ∈ A such that (x, y ) ∈ R, R is everywhere defined. Informally,
every element of A maps to something.
(1.3) An onto function. (2)
Solution
Since, for each y ∈ A, there exists an x ∈ A such that (x, y) ∈ R, R is onto. Informally, every element
of the range is reached.
(1.4) A one-to-one function. (2)
Solution
Since we do not have (x, z) and (y, z) in R for x ̸= y , R is a one-to-one function from A to A,
Informally, different elements of the domain map to different elements in the range.
19
, Question 2: 10 Marks
Let f : A → B and g : B → C be functions. Show that if:
(2.1) f and g are onto, then g ◦ f is onto. (4)
Solution
We note that g ◦ f maps A to C, i.e.
f g
A B C
g◦f
In order to show that g ◦ f is onto, we need to show that for each c ∈ C, there exists an a ∈ A such
that (g ◦ f )(a) = c.
Suppose c ∈ C. Then, since g : B → C is onto, there exists a b ∈ B such that g(b) = c.
Also, since f : A → B is onto, there exists an a ∈ A such that f (a) = b. Hence g(f (a)) = c, i.e.
(g ◦ f )(a) = c. Hence, (g ◦ f ) is onto.
(2.2) f and g are one-to-one, then g ◦ f is one-to-one. (6)
Solution
In order to show that (g ◦ f ) is one-to-one, we need to show that
(g ◦ f )(a) = (g ◦ f )(a′ ) =⇒ a = a′ .
Suppose
(g ◦ f )(a) = (g ◦ f )(a′ ).
Then,
g(f (a)) = g(f (a′ )).
Since g is one-to-one, it follows that
f (a) = f (a′ ).
Now, since f is one-to-one, it follows that
a = a′ .
Thus we have proved that g ◦ f is one-to-one.
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