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