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Let A = {1, 2} and B = {2, 4, 6}. Let f = {(x, y) : x ϵ A, y ϵ B and y > 2x + 1}. Write f as a set of ordered pairs. Show that f is a relation but not a function from A to B.
Given: A = {1, 2} and B = {2, 4, 6}
f = {(x, y): x ∈ A, y ∈ B and y > 2x + 1}
Putting x = 1 in y > 2x + 1, we get
y > 2(1) + 1
⇒ y > 3
and y ∈ B
this means y = 4, 6 if x = 1 because it satisfies the condition y > 3
Putting x = 2 in y > 2x + 1, we get
y > 2(2) + 1
⇒ y > 5
this means y = 6 if x = 2 because it satisfies the condition y > 5.
∴ f = {(1, 4), (1, 6), (2, 6)}
(1, 2), (2, 2), (2, 4) are not the members of ‘f’ because they do not satisfy the given condition y > 2x + 1
Firstly, we have to show that f is a relation from A to B.
First elements = 1, 2
All the first elements are in Set A
So, the first element is from set A
Second elements in F = 4, 6
All the second elements are in Set B
So, the second element is from set B
Since the first element is from set A and second element is from set B
Hence, F is a relation from A to B.
Function:
(i) all elements of the first set are associated with the elements of the second set.
(ii) An element of the first set has a unique image in the second set.
Now, we have to show that f is not a function from A to B
f = {(1, 4), (1, 6), (2, 6)}
Here, 1 is coming twice.
Hence, it does not have a unique (one) image.
So, it is not a function.