Functions as Relations

Functions as Relations
Dr. Yassir Dinar
Spring 2020
Dr. Yassir Dinar Functions as Relations Spring 2020 1 / 10

These slides cover the following topics
The deﬁnition of a function
Examples of inﬁnite functions
Relations which are not functions
Domain and range of real functions
Identical functions

Deﬁnition 1.1
A function f (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, z) ∈ f, then y = z.
In this case B is called the codomain of f. If A = B, we say f is a
function on A.
• The notation f : A → B reads“f is a function from A to B” or “f
maps A to B”.
• If (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

Relations and functions
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 S = {(1, 5), (2, 5), (3, 4)}.
2 T = {(1, 4), (3, 6)}.
3 P = {(1, 4), (2, 5), (3, 6), (2, 6)}.
Solution.
1 Dom(S) = A and the numbers 1,2 and 3 appears as ﬁrst coordinates
only once. Then S is a function.
2 2 ∈ A and 2 /∈ Dom(T), therefore T is not a function.
3 Dom(P) = A but (2, 5) ∈ P and (2, 6) ∈ P where 5 = 6. Therefore,
P is not a function.

Inﬁnite Functions: Example I
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.
From (i) and (ii) we conclude that g is a function from N to Z.
• Note that since g is a function we can write g : N → Z given by
b = g(a) = −2a + 7.

Inﬁnite Functions: Example II
Example 1.4
Deﬁne F = {(x, y) : x2
+ 4y2
= 16}. Show that F is a function from [0, 4] to
[0, ∞).
Solution.
(i) Assume x ∈ [0, 4]. Then x2 ≤ 16. Hence y = 16−x2
4 belongs to
[0, ∞). Then x2 + 4y2 = x2 + 4( 16−x2
4 )2 = 16. Thus (x, y) ∈ F
and Dom(F) = [0, 4].
(ii) Assume (t, z) ∈ F and (t, w) ∈ F. Then t2 + 4z2 = 16 = t2 + 4w2.
Hence z2 = w2. Since z and w nonnegative, it follows that z = w.
We conclude from (i) and (ii) that F is a function.
• Since F is a function, we can write F : [0, 4] → [0, ∞) given by
y = F(x) = 16−x2
4 .

Relations which are not functions
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) = [0, 4]. Counterexample 2 /∈ Dom(F) since otherwise
(2,
√
3) ∈ F but
√
3 /∈ codomain(F).

Domain and Range of Real Functions I
Example 1.6
Assume that the domain of the following function is the larges possible subset of
R, Find the domain and range.
f(x) =
x2
+ 5x + 6
x + 2
Solution.
From the deﬁnition Dom(f) ⊆ R − {−2}. Assume x ∈ R − {−2}.
Then y = x2+5x+6
x+2 = x + 3 ∈ R. Hence, x ∈ Dom(f) and
Dom(f) = R − {−2}.
Assume b ∈ Rng(f). Then there exists a ∈ R − {−2} such that
f(a) = b. This implies that b − 3 = a. But a = −2. Hence, b = 1.
Therefore, Rng(f) ⊆ R − {1}. Assume b ∈ R − {1}. set a = b − 3.
Hence, a = −2. Then a ∈ R − {−2} and f(a) = b − 3 + 3 = b.
Hence b ∈ Rng(f). Therefore, Rng(f) = R − {1}.

Domain and Range of Real Functions II
Example 1.7
Assume that the domain of the following function is the larges possible subset of
R, Find the domain and range.
g(x) =
√
x − 3
Solution.
From the deﬁnition, Dom(g) ⊆ [3, ∞). Assume x ∈ [3, ∞). Then
y =
√
x − 3 ∈ R and x ∈ Dom(g). Hence, Dom(g) = [3, ∞).
Assume y ∈ Rng(g). Then there exist x ≥ 3 such that y =
√
x − 3.
Hence y ≥ 0 and Rng(g) ⊆ [0, ∞). Assume b ∈ [0, ∞). Set
a = b2 + 3. Then a ∈ [3, ∞) and
g(a) =
√
a − 3 = b2 + 3 − 3 = b.
Hence b ∈ Rng(g). Therefore, Rng(g) = [0, ∞).

Equal functions
Deﬁnition 1.8
Two functions f and g are equal if they are equal as sets.
Theorem 1.9
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 1.10
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.

Functions as Relations

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