Answer to Question #221672 in Discrete Mathematics for Sabelo Zwelakhe

Question

Consider the following relations on the set A={1,2,3,4,5}.

a) R₁={(1,1), (1,2), (3,3), (3,2)}

b) R₂={(1,1), (2,2), (3,2), (4,5), (5,2)}

c) R₃={(1,4), (2,5), (3,3), (4,2), (5,2)}

State, with reasons, for a), b), c) whether the relation is:

I) a function from A to A

ii) an onto function

iii) a one-to-one function

iv) an everywhere defined function

Answer 1

(i). For (a),

R₁ will not be a function from A to A as there is more than one image of 1 and 3 in R₁ (i.e (1, 1), (1, 2), (3, 3), (3, 2)) which is not allowed in a function.

For (b),

R₂ will be a function as there is no unmatched element in the domain of R₂ and there is only one image of all the elements of domain.

For (c),

R₃ will be a function as there is no unmatched element in the domain of R₃ and there is only one image of all the elements of domain.

(ii). For (a),

R₁ will not be a function.

For (b),

R₂ will not be an onto function as 3 and 4 elements in the range are not matched with any element of the domain, which is not a property of an onto function.

For (c),

R₃ will not be an onto function as 1 element in the range is not matched with any element of the domain, which is not a property of an onto function.

(iii). For (a),

R₁ will not be a function.

For (b),

R₂ will not be a one-to-one function as 2, 3 and 5 have multiple images in the codomain which is not allowed in a one-to-one function.

For (c),

R₃ will not be a one-to-one function as 4 and 5 have multiple images in the codomain which is not allowed in a one-to-one function.

(iv). For (a),

R₁ will not be a function.

For (b),

R₂ will be an everywhere defined function, as the domain of function is the domain itself.

For (c),

R₃ will be an everywhere defined function, as the domain of function is the domain itself.