A relation f between two non-empty sets X and Y is called a function from X to Y if, for each x ∈ X there exists only one y ∈Y such that (x, y) ∈ f .
That is, f = {(x,y)| for all x ∈ X, y ∈ Y }.
This represents a function. Each input corresponds to a single output. | |
This represents a function. Each input corresponds to a single output. |
Question 1 :
Let f = {(x, y) | x, y ∈ N and y = 2x} be a relation on ℕ. Find the domain, co-domain and range. Is this relation a function?
Solution :
Since x and y ∈ N,
x = 1 y = 2(1) y = 2 |
x = 2 y = 2(2) y = 4 |
x = 3 y = 2(3) y = 6 |
x = 4 y = 2(4) y = 8 |
f = {(1, 2) (2, 4) (3, 6) (4, 8)................}
For each values of x, we get different values of y. So the given relation is a function.
Domain is the set of values of x
Domain = {1, 2, 3, 4, ............}
Co domain is the set of value of y. Since y ∈ N
Co domain = {1, 2, 3, 4, .............}
Range means the set of values of y, which are associated with x.
Range = {2, 4, 6, 8, .....}
Question 2 :
Let X = {3, 4, 6, 8}. Determine whether the relation ℝ = {(x, f (x)) | x ∈ X, f (x) = x^{2} + 1} is a function from X to ℕ ?
Solution :
Given that :
f (x) = x^{2} + 1
x ∈ X
if x = 3 f(3) = 3^{2}+1 f(3) = 10 |
if x = 4 f(4) = 4^{2}+1 f(4) = 17 |
if x = 6 f(6) = 6^{2}+1 f(6) = 37 |
if x = 8 f(8) = 8^{2}+1 f(8) = 65 |
R = { (3, 10) (4, 17) (6, 37) (8, 65) }
For each values of x, we get different values of f(x).
Hence it is a function
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