= 4x + 3
PRACTICE
1. Consider the sets A={1,2,3,4,5 },
B = { 1, 4, 9, 16, 25 } and a function
f: A _B defined by f(1) = 1,f(2)= 4,
f(3) = 9, f(4) = 16 and f(5) = 25
(a) Show that f is bijective
(b) Find f-1
Answers
The rule f that assigns the square of an integer to this integer
is a function. Indeed, every integer has an image: its square. Also whenever
two squares are different, it must be that their square roots were different.
We write
f : Z → Z, f(x) = x
2
.
Its domain is Z, its codomain is Z as well, but its range is {0, 1, 4, 9, 16, . . .},
that is the set of squares in Z.
Definition 64. Let f be a function from X to Y , X, Y two sets, and consider
the subset S ⊂ X. The image of the subset S is the subset of Y that consists
of the images of the elements of S: f(S) = {f(s), s ∈ S}
We next move to our first important definition, that of one-to-one.
Definition 65. A function f is one-to-one or injective if and only if f(x) =
f(y) implies x = y for all x, y in the domain X of f. Formally:
∀x, y ∈ X(f(x) = f(y) → x = y).
In words, this says that all elements in the domain of f have different
images.
Example 98. Consider the function f : R → R, f(x) = 4x − 1. We want
to know whether each element of R has a different image. Yes, this is the
case, why? well, visually, this function is a line, so one may ”see” that two
distinct elements have distinct images, but let us try a proof of this. We have
to show that f(x) = f(y) implies x = y. Ok, let us take f(x) = f(y), that is
two images that are the same. Then f(x) = 4x − 1, f(y) = 4y − 1, and thus
we must have 4x − 1 = 4y − 1. But then 4x = 4y and it must be that x = y,
as we wanted. Therefore f is injective.
Example 99. Consider the function g : R → R, g(x) = x
2
. Do we also have
that two distinct reals have distinct images? Well no... because 1 and −1
are both sent to 1...so this function is not injective! If g(x) = g(y) = 1, we
cannot conclude that x = y, in fact this is wrong, it could be that x = −y.