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1. 4.1 Functions as Relations
SQU-Math2350
Dr. Yassir Dinar
Sultan Qaboos University
Department Of Mathematics
Math2350: Foundation of Mathematics
Fall 2019
SQU-Math2350 4.1 Functions as Relations Fall 2019 1 / 12

2. Functions as Relations
Definition 1.1
A function (or mapping ) from A to B is a relation from A to B such
that
(i) the domain of f is A and
(ii) if (x, y) ∈ f and (x, y) ∈ f, then y = z.
f : A → B reads“f is a function from A to B” or “f maps A to B”.
The set B is called the codomain of f and if A = B, we say f is a
function on A.
When (x, y) ∈ f, we write y = f(x). We read it as y is the image of f
at x (or value of f at x) and that x is a pre-image of y..
Functions whose domains and codomains are subsets of R are referred
to as real functions
SQU-Math2350 4.1 Functions as Relations Fall 2019 2 / 12

3. Relations: Examples
Example 1.2
Let A = {1, 2, 3} and B = {4, 5, 6} which of the following relations is a function
from A to B:
1 P = {(1, 4), (2, 5), (3, 6), (2, 6)}.
2 S = {(1, 5), (2, 5), (3, 4)}.
3 T = {(1, 4), (3, 6)}.
Solution.
SQU-Math2350 4.1 Functions as Relations Fall 2019 3 / 12

4. Functions: Examples
Example 1.3
Show that the relations g = {(a, b) : b + 2a − 7 = 0} from N to Z is a function.
Solution.
(i) Let a ∈ N. Then b = −2a + 7 ∈ Z and (a, b) ∈ g. Hence,
Dom(g) = N.
(ii) Assume (a, n) ∈ g and (a, m) ∈ g. Then by definition of g,
m = −2a + 7 = n.
Frome (i) and (ii) we conclude that g is a function from N to Z.
Note that since g is a function we can write b = g(a) = −2a + 7.
Example 1.4
Define F = {(x, y) : x2
+ 4y2
= 16}. Show that F is a function from [0, 4] to
[0, ∞).
Since it is a function, write y =
q
16−x2
4 . Find 3 different codomains!!!
SQU-Math2350 4.1 Functions as Relations Fall 2019 4 / 12

5. Functions: Non Examples
Example 1.5
Define F = {(x, y) : x2
+ 4y2
= 16}. Show that F is not a function when
1 F is a relation from R to R.
2 F is a relation from [−4, 4] to R.
3 F is a relation from [0, 4] to [0, 1].
Solution.
1 Note that 10 ∈ R but 10 /
∈ Dom(F) since 102 + 4y2 ≥ 100, ∀y ∈ R.
Hence F is not a function.
2 Dom(F) = [−4, 4] Prove it!. For (x, y) and (x, z) in F we have
4y2 = 4z2. Hence y = ±z. So F is not a function. Counterexample
both (2,
√
3) and (2, −
√
3) are in F.
3 Dom(F) 6= [0, 4]. Counterexample 2 /
∈ Dom(F) since otherwise
(2,
√
3) ∈ F but
√
3 /
∈ codomain(F).
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6. Domain and Range
Example 1.6
Assume that the domain of each of the following functions is the larges possible
subset of R, find the domain and range.
1 f(x) = x2
+5x+6
x+2 .
2 f(x) =
√
x − 3.
Solution.
1 Dom(f) = R − {−2}. Let b ∈ Rng(f). Then there exists
a ∈ R − {−2} such that f(a) = b.This implies that b = a + 3. But
a 6= −2. Hence, b 6= −1. Therefore, Rng(f) = R − {1}.
2
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7. Equal functions
Definition 1.7
Two functions f and g are equal if they are equal as sets.
Theorem 1.8
Two function f and g are equal if and only if
(i) Dom(f) = Dom(g) and
(ii) for all x ∈ Dom(f), f(x) = g(x).
Prove it!
Example 1.9
The function f(x) = x2
+5x+6
x+2 and g(x) = x + 3 are not equal as real values
functions since Dom(f) = R − {−2} while Dom(g) = R. If we redefine both of
them with domain [0, ∞), then they are equal.
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8. Section 4.2 Constructions of Functions
Let f : A → B and g : B → C. Then f−1 = {(a, b) : (b, a) ∈ f} ,
g ◦ f = {(x, y) : (∃z ∈ B), (x, z) ∈ f and (z, y) ∈ g}.
Example 2.1
Let f and g be real valued functions given by f(x) = 2x + 1 and g(x) = x2
.
1 Find f−1
and g−1
and determine if they are functions.
2 Find g ◦ f and f ◦ g and determine if they are functions.
Solution.
Inverse of function is not necessary a function.
SQU-Math2350 4.2 Constructions of Functions Fall 2019 8 / 12

9. Composition of functions
Theorem 2.2
Let f : A → B and g : B → C. Then g ◦ f is a function from A to C and
Dom(g ◦ f) = Dom(f) = A.
Proof.
(i) It is known that Dom(g ◦ f) ⊆ A. Let x ∈ A. Since f is a function
there exist b ∈ B such that (a, b) ∈ f.Also, since g is a function there
is c ∈ C such that (b, c) ∈ g. But then (a, c) ∈ g ◦ f and
a ∈ Dom(g ◦ f). Thus Dom(g ◦ f) = A.
(ii) Assume that (a, y) and (a, z) are in g ◦ f. Then by definition
SQU-Math2350 4.2 Constructions of Functions Fall 2019 9 / 12

10. Composition of functions: Examples
Example 2.3
Let f and g be real valued functions given by f(x) = sin x and g(x) = ex
. Then
domain of f and g are R. Thus
1 g ◦ f = . . . and Dom(g ◦ f) = . . .
2 f ◦ g = . . . and Dom(f ◦ g) = . . .
Example 2.4
Let f, g and h be real valued functions defined by f(x) = x + 2 and g(x) = 1
x−9
and h(x) =
√
x + 5. Then the domain of f, g and h are different. Thus
1 g ◦ f = . . . and Dom(g ◦ f) = . . .. Since Dom(g) = R − {9}.
2 h ◦ f = . . . and Dom(h ◦ f) = . . .. Since Dom(h) = .....
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11. Theorems on Composition of Functions
Theorem 2.5
Let f : A → B, g : B → C and h : C → D. Then h ◦ (g ◦ f) = (h ◦ g) ◦ f.
Theorem 2.6
Let f : A → B with Rng(f) = C. If f−1 is a function then f−1 ◦ f = IA
and f ◦ f−1 = IC.
Proof.
Suppose f : A → B and f−1 is a function. Then
(i) Dom(f−1 ◦ f) = Dom(f) = A = Dom(IA).
(ii) For x ∈ A, there is y ∈ B such that (x, y) ∈ f. Then (y, x) ∈ f−1.
Thus f−1 ◦ f(x) = x = IA(x).
From (i) and (ii), f−1 ◦f = IA. The proof of f ◦f−1 = IC left as Exercise
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12. Theorems on Composition of Functions
Theorem 2.7
Let f : A → B. Then f ◦ IA = f and IB ◦ f = f.
Proof.
SQU-Math2350 4.2 Constructions of Functions Fall 2019 12 / 12