Answer :
In this case, the two incorrect statements in expressing that a function f is onto are:
- c. f is onto → ∀y € Y, 3x E X such that f(x) = y.
- d. f is onto → ∀ x E X, 3y EY such that f(x) = y.
The answer is option~C and D
To determine the two statements that are incorrect in expressing that a function f is onto:
a. This statement is correct. An onto function means that every element in the co-domain has a corresponding pre-image in the domain.
b. This statement is correct. An onto function implies that every element in the domain maps to a unique element in the co-domain.
c. This statement is incorrect. The quantifier "3x" should be replaced by "∀x" to correctly express the condition for an onto function.
d. This statement is incorrect. Similar to statement c, the quantifier "3y" should be "∀y" to accurately represent an onto function.
e. This statement is correct. For a function to be onto, its range (set of all possible outputs) should be equal to its co-domain.
Therefore, the two incorrect statements are:
c. f is onto → ∀y € Y, 3x E X such that f(x) = y.
d. f is onto → ∀ x E X, 3y EY such that f(x) = y.
These statements incorrectly use the existential quantifier "∃" instead of the universal quantifier "∀," which changes the intended meaning regarding the function being onto.
The answer is option~C and D
Your question is incomplete, but most probably the full question was:
All but two of the following statements are correct ways to express the fact that a function f is onto.
Find the two that are incorrect.
a. f is onto → every element in its co-domain is the image of some element in its domain.
b. f is onto → every element in its domain has a corresponding image in its co-domain.
c. f is onto → ∀y € Y, 3x E X such that f(x) = y.
d. f is onto → ∀ x E X, 3y EY such that f(x) = y.
e. f is onto → the range of f is the same as the co-domain off.
