## one-to-one function

**one-to-one function or injective function is one of the most common functions used. **

One-to-One functions define that each element of one set say Set (A) is mapped with a unique element of another set, say, Set (B).

To understand this, let us consider ‘f’ is a function whose domain is set A. The function is said to be one to one if for all x and y in A,

x=y if whenever f(x)=f(y)

In the same manner if x ≠ y, then f(x) ≠ f(y)

Formally, it is stated as, if f(x)=f(y) implies x=y, then f is one-to-one mapped or f is 1-1.

**if f(x) = y then only f ^{-1}(y) = x **

**Example 1: Show that f: R → R defined as f(a) = 3a ^{3} – 4 is one to one function?**

**Solution:**

Let f (a1) = f ( a_{2} ) for all a_{1} , a_{2} ∈ R

So 3a_{1}^{3} – 4 = 3a_{2}^{3} – 4

a_{1}^{3} = a_{2}^{3}

a_{1}^{3} – a_{2}^{3} = 0

(a_{1} – a_{2}) (a_{1} + a_{1}a_{2} + a_{2}^{2}) = 0

a_{1} = a_{2} and (a_{1}^{2} + a_{1}a_{2} + a_{2}^{2}) = 0

(a_{1}^{2} + a_{1}a_{2} + a_{2}^{2}) = 0 is not considered because there are no real values of a_{1} and a_{2}.

Therefore, the given function f is one-one.

## Onto Function

To explain Onto function consider two sets, Set A and Set B which consist of elements. Ifthere is at least one or more than one element matching with A for every element of B, then the function is said to be **onto function** or surjective function.

You can see in the figure that for each element of B there is a pre-image or a matching element in Set A. Hence it is called an onto function. In the second figure it can be seen that one element in Set B is not mapped with any of the element of set A. Because of this it’s not an onto function

- Every onto function has a right inverse and every function with a right inverse is an onto function. When we compose onto functions, the result will be onto function only.

**Example: Let A={1,5,8,9) and B{2,4} And f={(1,2),(5,4),(8,2),(9,4)}. Then prove f is a onto function.**

**Solution:** From the question itself we get,

A={1,5,8,9)

B{2,4}

& f={(1,2),(5,4),(8,2),(9,4)}

Here, the entire element on B has a domain element on A or we can say element 1 and 8 & 5 and 9 has same range 2 & 4 respectively.

Therefore, f: A → B is ansurjective function.

Hence, the onto function proof is explained.