a) Falseb) False c) True d) True

[XodoX]

Homework Statement

Mark as true or false.
(a) A function is injective if a 6$$\neq$$ b yields f(a) 6$$\neq$$ f(b).
(b) A function is injective if f(a) = f(b) in case that a = b.
(c) A function is injective if f(a) = f(b) only if a = b.
(d) A function is injective only when f(a) 6$$\neq$$ f(b) yields a 6$$\neq$$ b. 3. The Attempt at a Solution

a) False
b) True
c) True
d) True I know that f is said to be injective, if and only if f(a) = f(b) implies that a = b.

Homework Statement

Mark as true or false. If f : E ! E is a map on a finite set then
(a) If f is injective then f is surjective.
(b) If f is surjective then f is injective.
(c) There is an injective map f from the set of natural numbers N to itself which is not surjective.
(d) There is a surjective map f from the set of natural numbers N to itself which is not injective.

I don't really get this. Probably easy, but badly explained in the book. Can someone clarifiy this one for me, please?

Thank you

 

[mighty2000]

for your response! Here is an explanation for each statement:

(a) False - This statement is known as the "pigeonhole principle" and it states that if a function is injective (meaning each input has a unique output), then it must also be surjective (meaning every output has at least one input). This statement is false because there are functions that are injective but not surjective, such as f(x) = x^2 on the set of real numbers.

(b) False - This statement is the converse of statement (a) and it is also false for the same reason. Just because a function is surjective does not mean it is injective. An example of this is the function f(x) = 2x on the set of real numbers.

(c) True - This statement is saying that there exists a function from the set of natural numbers to itself that is injective but not surjective. An example of this is the function f(x) = x+1, which maps each natural number to the next consecutive natural number. This function is injective because each input has a unique output, but it is not surjective because there is no natural number that maps to 1.

(d) True - This statement is saying that there exists a function from the set of natural numbers to itself that is surjective but not injective. An example of this is the function f(x) = x^2, which maps each natural number to its square. This function is surjective because every natural number has a square root, but it is not injective because multiple inputs can map to the same output (for example, both 2 and -2 map to 4).