### Definition

Let A and B be two subsets of ℝ
A relation from A to B is a rule which assigns a member of B to each member of A.

### Definition

Let A and B are two subsets of ℝ
A function from A to B is a rule which assigns a unique member of B of each number of A.

In the example above, 5 being assigned to both 2 & 6. Hence the relation in that example is not a function.

Here is an example for a function;

### Definition

Let f: A → B a function.

If f(x1) = f(x2) ⇒ x1 = x2
We say that f is an one to one function

In the previous example fo(1) = fo(9) but 1 ≠ 9
∴ fo is not a one to one function.

### Definition

Let f: A → B be a function

If ∀y ∈ B, ∃x ∈ A s.t. f(x) = y
We say that f is an onto function

In the previous example ∃!x s.t. fo(x) = 4
∴ fo is not an onto function

### Definition

Let f: A → B be a function.

If f is both one to one and onto, we say that f is a bijection

In the previous example fo is neither one to one nor onto.
∴ fo is far from being bijection

### Definition

Let f: A → B be a function
1. A is called the Domain of A
2.   We write D(f) = A
3. The following set is called the range of f
4.   R(f) = {y | y ∈ B & y = f(x) for some x ∈ A}
When B = R(f), f is an onto function

### Definition

Let f: A → B be a function
1. If x1 < x2 ⇒ f(x1) ≤ f(x2), f is an increasing function
2. If x1 < x2 ⇒ f(x1) < f(x2), f is an strictly increasing function
3. If x1 < x2 ⇒ f(x1) ≥ f(x2), f is an decreasing
4. If x1 < x2 ⇒ f(x1) > f(x2), f is an strictly decreasing function
Eg: f: R → R
f(x) = 2x + 7

x1 = x2 ⇒ 2x1 = 2x2 ⇒ 2x1 + 7 = 2x2 + 7
⇒ f(x1) = f(x2)
∴ f is a function.

x1 < x2 ⇒ 2x1 < 2x2 ⇒ 2x1 + 7 < 2x2 + 7
∴ f is a strictly increasing function.

### Definition

Let x ∈ R

The greatest integer less than or equal to x is denoted by [x]
Eg: [7.3] = 7

### Definition

Let f: A → B be a function

The inverse relation f-1 R(f) → A is defined in the following manner.
f-1(y) = x where y = f(x) for some x ∈ A

in general f-1 is not a function

### Theorem

Let f: A → B be a function.
Then f-1 is a function ⇔ f is one to one

### Proof

Assume f-1 is a function. Let f(x1) = f(x2)

Write f(x1) = f(x2) = y

f(x1) = y ⇒ f-1(y) = x1
f(x2) = y ⇒ f-1(y) = x2
x1 = x2 because f-1 is a function
∴ f is one to one.

Assume f is one to one
Let y ∈ R(f) suppose f-1(y) = x1 & f-1(y) = x2
f-1(y) = x1 ⇒ f(x1) = y
f-1(y) = x2 ⇒ f(x2) = y
f(x1) = f(x2) & f is one to one
∴ x1 = x2

∴ f-1 is a function

### Definition

Let f: A → B & g: A → B be two functions
1. f + g: A → B
2.   (f + g)(x) = f(x) + g(x)
3. f - g: A → B
4.   (f - g)(x) = f(x) - g(x)
5. rf: A → B
6.   (rf)(x) = rf(x)
7. fg: A → B
8.   (fg)(x) = f(x).g(x)
9. f/g: A → B
10.   (f/g)(x) = f(x)/g(x)  (g(x) ≠ 0)

### Definition

Let f: A → B & g: B → C be two functions. The composite function (g o f) is defined as
g o f: A → C
(g o f)(x) = g(f(x))

eg:
 f: R → R g: R[-1, 1] f(x) = x2 – 3x + 2 g(x) = sin x

g o f: R → [-1, 1]
(g o f)(x) = g(f(x)) = g(x2 – 3x + 2) = sin(x2 – 3x + 2)

A function f: R → R is said to be an additive function if the following condition satisfied.

f(x + y) = f(x) + f(y) ∀x, y ∈ ℝ

Prove the following fact for an additive function.
1. f(0) = 0
2. f(rx) = rf(x) ∀r ∈ ℚ, ∀x ∈ ℝ
Proof of i.

0 + 0 = 0 ⇒ f(0 + 0) = f(0)
f(0) + f(0) = f(0) because f is additive.

∴ f(0) = 0

Proof of ii.

Let m ∈ ℤ+

Claim: f(mx) = mf(x) ∀x ∈ ℝ
It is clearly true for m = 1

Assume the claim is true for m = k
f{(k + 1)x} = f(kx + x) = f(kx) + f(x) [because f is additive]

= kf(x) + f(x)
=(k + 1)f(x)

∴ The claim is true for m = k + 1
∴ By induction, claim is true ∀m ∈ ℤ+

Let m ∈ ℤ-

Now m + n = 0 ⇒ (m + n)x = 0.x = 0

∴ f(mx + nx) = f(0) = 0 [by part I]
∴ f(mx) + f(nx) = 0 [because f is additive]

f(mx) + f(nx) = 0
f(mx) = -nf(x) = mf(x)

So we have proved that f(mx) = mf(x) ∀m ∈ ℤ

Let r ∈ ℚ
Write r = m/n with m ∈ ℤ, n ∈ ℤ+
Let x ∈ ℝ    x = n(1/nx)

f(x) = f[n(1/nx)] = nf(1/nx)
1/nf(x) = f(1/nx)
f(rx) = f(m/nx) = f[m(1/nx)] = mf(1/nx)
= m.1/nf(x) = m/nf(x) = rf(x)