Question

Let n ∈ N and f : [n] → [n] a function. Prove that f is a surjection if and only if f is an injection.

Answer #1

. Let f : Z → N be function.
a. Prove or disprove: f is not strictly increasing. b. Prove
or disprove: f is not strictly decreasing.

3. Let N denote the nonnegative integers, and Z denote the
integers. Define the function g : N→Z defined by g(k) = k/2 for
even k and g(k) = −(k + 1)/2 for odd k. Prove that g is a
bijection.
(a) Prove that g is a function.
(b) Prove that g is an injection
. (c) Prove that g is a surjection.

Let A be a finite set and let f be a surjection from A to
itself. Show that f is an injection.
Use Theorem 1, 2 and corollary 1.
Theorem 1 : Let B be a finite set and let f be a function on B.
Then f has a right inverse. In other words, there is a function g:
A->B, where A=f[B], such that for each x in A, we have f(g(x)) =
x.
Theorem 2: A right inverse...

Let f(n) be a negligible function and k a positive integer.
Prove the following:
(a) f(√n) is negligible.
(b) f(n/k) is negligible.
(c) f(n^(1/k)) is negligible.

Let f be a function from the set of students in a discrete
mathematics class to the set of all possible final grades.
(a) Under what conditions is f an injection?
(b) Under what conditions is f a surjection?
(please show all work)

Let A be a finite set and f a function from A to A.
Prove That f is one-to-one if and only if f is onto.

For each of the following, give an example of a function g and a
function f that satisfy the stated conditions. Or state that such
an example cannot exist. Be sure to clearly state the domain and
codomain for each function.
(a)The function g is a surjection, but the function fog is not a
surjection.
(b) The function g is not an injection, but the function fog is
an injection.
(c)The function g is an injection, but the function fog...

Suppose that f is a bijection and f ∘ g is defined. Prove:
(i). g is an injection iff f ∘ g is;
(ii). g is a surjection iff f ∘ g is.

Let A and B be nonempty sets. Prove that if f is an injection,
then f(A − B) = f(A) − f(B)

Let A and B be nonempty sets. Prove that if f is an injection,
then f(A − B) = f(A) − f(B)

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