ONE TO ONE FUNCTION

Example 1 :

Check whether the following function is one-to-one 

f : N → N defined by f(n) = n + 2

Solution :

To check if the given function is one to one, let us apply the rule

f (x) = f (y) ----> x = y

f(x) = x + 2 and f(y) = y + 2

x + 2 = y + 2

x = y

For every element if set N has images in the set N.

Hence it is one to one function.

Example 2 :

Check whether the following function is one-to-one 

f : R → R defined by f(n) = n²

Solution :

To check if the given function is one to one, let us apply the rule

f(x) = f(y) ----> x = y

f(x) = x² and f(y) = y²

By equating f(x) and f(y), we get

x² = y²

x = y

From this we cannot decide that the function is one to one. Because every two different elements in the domain has same images is co-domain. That is,

If x = 1 then y = 1. If x = -1 also, then y is 1.

Hence the given function is not one to one.

Example 3 :

Check whether the following function is one-to-one 

f : R - {0} → R defined by f(x) = 1/x

Solution :

To check if the given function is one to one, let us apply the rule

f(x) = f(y) ----> x = y

f(x) = 1/x and f(y) = 1/y

By applying the above rule, we get

1/x = 1/y 

x = y

Every element in domain has different images in co-domain. 

Hence it is one to one function.

Example 4 :

Check whether the following functions are one-to-one and onto.

f : N ∪ {−1, 0} → N defined by f(n) = n + 2.

Solution :

As above, this function is one-to-one. If m is in the co-domain, then m − 2 is in the domain and

f(m − 2) = (m − 2) + 2 = m

thus m has a pre-image and hence this function is onto.

Example 5 :

If f : R - {-1, 1} --> R is defined by f(x) = x/(x² - 1) verify whether f is one to one or not.

Solution :

Let us start with the assumption, f(x) = f(y)

f(x) = x/(x² - 1)

f(y) = y/(y² - 1)

x/(x² - 1) = y/(y² - 1)

x(y² - 1) = y(x² - 1)

xy² - x = yx² - y

xy² - x - yx² + y = 0

xy² - yx² + y - x = 0

xy(y - x) + (y - x) = 0

(y - x) (xy + 1) = 0

y - x = 0 and xy + 1 = 0

y = x and xy = -1

Since the domain is all real numbers excluding -1 and 1, let us apply some random values. Here we dont recieve only y = x, we recieve xy = -1 as well. So, it is not one to one function.

Example 6 :

Check whether the function f(x) = x|x| defined on [−2, 2] is one-to-one or not. If it is one-to-one, find a suitable co-domain so that the function becomes a bijection.

Solution :

f(x) = f(y)

f(x) = x|x|

f(y) = y|y|

x|x| = y|y}

x/y = |y|/|x|

x/y = |y/x|

f(y), we must have x² = y². Also x and y are either both negative or both positive. This is possible only if x = y. Thus f is one-to-one. When x < 0, f(x) = −x² and when x ≥ 0, f(x) = x²

So the range is [−4, 4]. So f becomes a bijection from

[−2, 2] to [−4, 4].

Example 7 :

Let A = {1, 2, 3, 4} and B = {a, b, c, d}. Give a function from A → B for each of the following:

(i) neither one-to-one nor onto.

(ii) not one-to-one but onto.

(iii) one-to-one but not onto.

(iv) one-to-one and onto.

Solution :

A = {1, 2, 3, 4} and B = {a, b, c, d}.

i) {1, a) (2, a) (3, c) (4, d)

ii) If it is not to be one to one, one output may have more than one preimages. To be onto, each output should have preimage. Since we have equal number of elements in both sets, we cannot create the function which is not one to one but onto. Then, it is not possible.

iii) Again, it has same number of elements, so it is not possible to create a function which is one to one but not onto.

iv) 

{1, a) (2, b) (3, c) (4, d)

(or)

{1, c) (2, b) (3, a) (4, d)

We can create many functions like this.

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