2.5 KiB
Lecture Topic: Functions & Relations
When a function is onto, it means that the entire co-domain of a function is the range of the function
Proving a function is not onto:
Just prove for any element b in the co-domain B and show that b cannot be equal to f(a) for any a in the domain.
Example: g: \mathbb{R} \rightarrow \mathbb{R}, g(x) = x^2 + 10 is not onto
Proof:
For example, choose the element -1 from the co-domain \mathbb{R}
Consider any domain element $a$
The function will square a, and a^2 \geq 0 (because no squares are negative reals)
Then a^2 + 10 \geq 10, so this means that -1 is not in the co-domain
\therefore The function is not onto
You can also use a proof by contradiction by saying that: $g(a) = -1$ $\therefore a^2 + 10 = -1$ $\therefore a^2 = -11$ Squares of a real number cannot be negative, so the function is not onto
A function is not always numbers:
Example: Favourite UNB Course : S -> C Where S is all students and C is all courses
What would have to be true for this to be one to one or onto?:
- One to one:
- Onto:
Relations:
A binary relation R on two sets A and B is a subset of A \times B
- e.g. S = all students, C = all students, s is any given student, c is any given course
- Relation: HasTaken: for (s, c) s has taken c
- Relation Aplus: for (s,c) s got A+ in c
A relation can be between the same set, for example the set of A = {John, Jane, Bill, Sue, Mary, Betty}, we can define a relation that is the relationship between who loves who A \times A
- e.g. (John, Jane)
A binary relation R \subseteq A \times A is also called an equivalence relation if the following conditions hold:
- reflexive:
\forall a \in A, (a,a) \in R - symmetric:
\forall a,b \in A, (a,b) \in R \rightarrow (b,a) \in R - transitive:
\forall a,b,c \in A, [(a,b) \in R \wedge (b,c) \in R] \rightarrow (a,c) \in R
Example:
G = \{(x,y) | x \in \mathbb{R}, y \in \mathbb{R}, x > y\}
Is G reflexive? (Is is true that every real number is relation to itself)
No, because we can show that (5,5) \notin \mathbb{R} because 5 \ngtr 5
Is G symmetric? (Is it the case that whenever (a,b) \in G, (b,a) will also be in G)
No, for example (10,3) \in G because 10 > 3 but (3, 10) \notin G because 3 \ngtr 10
Is G transitive? (Is it true that whenever (a,b) and (b,c) are in G, (a,c) will also be in G)
Yes, let a,b,c be any real numbers where (a,b) \in G and $(b,c) \in G$
Since (a,b) \in G, we know that $a > b$
Since (b,c) \in G, we know that $b > c$
Then a > b > c, so that means that a > c, so also (a,c) \in G