Need help with proving onto functions

1) Let f: R→R be defined by f(x) = ax+b where a and b are real
numbers and a ≠0. Prove that f is onto.

2) Prove that there exists a set A such that f :R→A, defined by f(x) =
x^2, is NOT onto.

Any help with proving these 2 functions is appreciated. Thanks!

1) Let f: R→R be defined by f(x) = ax+b where a and b are real
numbers and a ≠0. Prove that f is onto.
for any x :
if y = f( x ) = ax + b
is equivalent to : x = (y - b)/a

this means that : for any real y, f( (y-b) / a ) y
so there exists
one and one only real x (id est : x = (y-b) / a )
so that f(x) = y
therefore f is a bijection and
hence f is "onto", so a surjection : f(IR) = IR
(every real from the target set has at least one antecedent)


2) Prove that there exists a set A such that f :R→A, defined by f(x) = x^2, is NOT onto.
let's take : A = IR
as negative real numbers have no antecedent,
f : IR ---> IR is not onto : f( IR ) =/= IR

et voilà !!

hope it' ll help !!

A function is onto if and only if for all y ∈ C, there exists x ∈ D such that f(x) = y. So let y ∈ ℝ, then we need to prove that there exists x ∈ ℝ such that ax + b = y. But if ax+b = y, then x = (y - b)/a, which is a real number since y ∈ ℝ and a ≠ 0. Therefore f(x) is onto.

2) Let A = [-1,1]. I'll leave the proof to you.