One-one Functions

Let A and B be sets and f is a function from A to B. The function f is called a one-one function or an injective function if the following holds:

f(a) = f(b)    ⇒   a = b       ………………………………………………………………………………………………………………………………… (1)

Loosely speaking, the function is one-one if no two different elements of A have the same image in B.

 

Example 1: (not an injective function)

Let f: \mathbb{R} \rightarrow \mathbb{R} be defined by f(x) = x2.

Note that f(-5) = (-5)2 = 25 and f(5) = 52 = 25.

In this case, f(-5) = f(5), but -5 ≠ 5 .

In this example, the condition (1) is not satisfied, thus we conclude that f is not an injective function.

 

Example 2: (injective function)

Let g: \mathbb{R} \rightarrow \mathbb{R} be defined by g(x) = x + 10.

We claim that g is an injective function.

To prove this, let g(a) = g(b).

As a consequence, a + 10 = b + 10.

Subtract 10 from both sides of the equation. Then we have a = b.

We have proved that g(a) = g(b) ⇒ a = b.

By (1), we conclude that g is an injective function.

 

Example 3: (not a one-one function)

Let h: \mathbb{R} \rightarrow \mathbb{R} be defined by h(x) = |x|.

Note that h(3) = |3| = 3 and h(-3) = |-3| = 3.

In this case, h(3) = h(-3) but -3 ≠ 3.

The condition (1) is violated, so we conclude that h is not a one-one function.

 

Example 4 (one-one function)

Let k: \mathbb{R} \backslash \{ 0 \} \rightarrow \mathbb{R} be defined by k(x) = \frac{1}{x}

We claim that k is a one-one function.

To prove this, let k(a) = k(b).

Consequently, \frac{1}{a} = \frac{1}{b}

By multiplying both sides of the equation by ab, we have a = b.

We have proved that k(a) = k(b) ⇒ a = b.

By (1), it follows that k is a one-one function.

 

Surjective Functions

Let A and B be sets and f is a function from A to B. The function f is called a surjective function or an onto function if the following holds:

∀b ∈ B ∃a ∈ A ∋ f(a) = b ………………………………………………………………………………………………………………………………………………………… (2)

Stated another way, a function f is said to be surjective if for every b ∈ B there is some a ∈ A such that f(a) = b. Loosely speaking, condition (2) means: “to each element in B there is assigned an element of A”.

 

Example 5 (not a surjective function)

Let f: \mathbb{R} \rightarrow \mathbb{R} be defined by f(x) = x2.

f is not a surjective function because -3 ∈ \mathbb{R} but there is no a \in \mathbb{R} such that f(a) = a2 = -3.

In this case, condition (2) is not satisfied. So we conclude that f is not a surjective function.

 

Example 6 (surjective function)

Let g: \mathbb{R} \rightarrow \mathbb{R} be defined by g(x) = x + 10

We claim that g is a surjective function

To prove it, let b \in \mathbb{R}.

Choose a = (b – 10) ∈ \mathbb{R}.

Consequently, g(a) = (b – 10) + 10 = b.

We have proved that for every b \in \mathbb{R} there is some a \in \mathbb{R} such that g(a) = b. By (2), we can conclude that g is a surjective function.

 

Example 7 (not a surjective function)

Let k : \mathbb{R} \backslash \{ 0 \} \rightarrow \mathbb{R} be defined by k(x) = \frac{1}{x}.

We claim that k is not a surjective function.

To prove it, choose 0 ∈ \mathbb{R}. For any a \in \mathbb{R} \backslash \{ 0 \} it is impossible that \frac{1}{a} = 0.

We have proved that there is a b \in \mathbb{R} such that for every a \in \mathbb{R} \backslash \{ 0 \} k(a) ≠ b. Hence, k is not surjective.

 

Example 8 (surjective function)

Let v : \mathbb{R} \backslash \{ 0 \} \rightarrow \mathbb{R} \backslash \{ 0 \} be defined by v(x) = \frac{1}{x}.

We claim that v is a surjective function.

To prove it, let b \in \mathbb{R} \backslash \{ 0 \}. It is obvious that b ≠ 0. Therefore, we can select a = \frac{1}{b} \in \mathbb{R} \backslash \{ 0 \}. This choice implies v(a) = \frac{1}{1/b} = b.

We have proved that for every b \in \mathbb{R} \backslash \{ 0 \} there is some a \in \mathbb{R} \backslash \{ 0 \} such that v(a) = b. By (2), we can conclude that v is a surjective function.

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