260 lines
6.8 KiB
Markdown
260 lines
6.8 KiB
Markdown
Page 449
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**Definition**
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A **function $f$ from a set $X$ to a set $Y$**, denoted: $f: X \to Y$, is a
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relation from $X$, the **domain** of $f$, to $Y$, the **co-domain** of $f$, that
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satisfies two properties: (1) every element in $X$ is related to some element in
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$Y$, and (2) no element in $X$ is related to more than one element in $Y$. Thus,
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given any element $x$ in $X$, there is a unique element in $Y$ that is related
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to $x$ by $f$. If we call this element $y$, then we say that "$f$ sends $x$ to
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$y$" or "$f$ maps $x$ to $y$" and write $x \xrightarrow{f} y$ or $f: x \to y$.
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The unique element to which $f$ sends $x$ is denoted
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$f(x)$ and is called $f$ of $x$, or the output of $f$ for the input $x$, or the
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value of $f$ at $x$, or the image of $x$ under $f$.
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The set of all values of $f$ taken together is called the _range of $f$_ or the
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_image of $X$ under $f$_. Symbolically:
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$$ \text{range of } f = \text{ image of } X \text{ under } f = \{y \in Y | y = f(x), \text{ for some } x \text{ in } X\} $$
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Given an element $y$ in $Y$, there may exist elements in $X$ with $y$ as their
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image. When $x$ is an element such that $f(x) = y$, then $x$ is called **a
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preimage of $y$** or **an inverse image of $y$**. The set of all inverse images
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of $y$ is called _the inverse image of $y$_. Symbolically:
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$$ \text{ the inverse image of } y = \{x \in X | f(x) = y\} $$
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---
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Page 451
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**Theorem 7.1.1 A Test for Function Equality**
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If $F: X \to Y$ and $G: X \to Y$ are functions, then $F = G$ if, and only if,
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$F(x) = G(x)$ for every $x \in X$.
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**Proof:**
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Suppose $F: X \to Y$ and $G: X \to Y$ are functions; that is, $F$ and $G$ are
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relations from $X$ to $Y$ that satisfy the two additional function properties.
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Then $F$ and $G$ are subsets of $X \times Y$, and for $(x, y)$ to be in $F$
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means that $y$ is the unique element related to $x$ by $F$, which we denote as
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$F(x)$. Similarly, for $(x, y)$ to be in $G$ means that $y$ is the unique
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element related to $x$ by $G$, which we denote as $G(x)$.
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Now suppose that $F(x) = G(x)$ for every $x \in X$. Then if $x$ is any element
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of $X$,
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$$ (x, y) \in F \Leftrightarrow y = F(x) \Leftrightarrow y = G(x) \Leftrightarrow (x, y) \in G $$
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because $F(x) = G(x)$.
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So $F$ and $G$ consist of exactly the same elements and hence $F = G$.
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Conversely, if $F = G$, then for every $x \in X$,
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$$ y = F(x) \Leftrightarrow (x, y) \in F \Leftrightarrow (x, y) \in G \Leftrightarrow y = G(x) $$
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because $F$ and $G$ consist of exactly the same elements.
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Thus, since both $F(x)$ and $G(x)$ equal $y$, we have that
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$$ F(x) = G(x) $$
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---
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Page 453
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**Definition Logarithms and Logarithmic Functions**
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Let $b$ be a positive real number with $b \neq 1$. For each positive real number
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$x$, the **logarithm with base $b$ of $x$**, written $\log_bx$, is the exponent
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to which $b$ must be raised to obtain $x$. Symbolically:
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$$ \log_bx = y \Leftrightarrow b^y = x $$
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The **logarithmic function with base $b$** is the function from $\mathbb{R}^+$
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to $\mathbb{R}$ that takes each positive real number $x$ to $\log_bx$.
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---
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Page 455
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**Definition**
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An **($n$-place) Boolean function** $f$ is a function whose domain is the set of
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all ordered $n$-tuples of $0$'s and $1$'s and whose co-domain is the set
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$\{0, 1\}$. More formally, the domain of a Boolean function can be described as
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the Cartesian product of $n$ copies of the set $\{0, 1\}$, which is denoted
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$\{0, 1\^n}$. Thus $f: \{0, 1\}^n \to \{0, 1\}$.
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---
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Page 457
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**Definition**
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If $f: X \to Y$ is a function and $A \subseteq X$ and $C \subseteq Y$, then
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$$ f(A) = \{y \in Y | y = f(x) \text{ for some } x \text{ in } A\} $$
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and
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$$ f^{-1}(C) = \{x \in X | f(x) \in C\} $$
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$f(A)$ is called the **image of $A$**, and $f^{-1}(C)$ is called the **inverse
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image of $C$**.
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---
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Page 463
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**Definition**
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Let $F$ be a function from a set $X$ to a set $Y$. $F$ is **one-to-one** (or
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**injective**) if, and only if, for all elements $x_1$ and $x_2$ in $X$,
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$$ \text{if } F(x_1) = F(x_2) \text{, then } x_1 = x_2 $$
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or, equivalently,
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$$ \text{if } x_1 \neq x_2 \text{, then } F(x_1) \neq F(x_2) $$
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Symbolically:
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$$ F: X \to Y \text{ is one-to-one } \Leftrightarrow \forall x_1, x_2 \in X \text{, if } F(x_1) = F(x_2) \text{ then } x_1 = x_2 $$
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---
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Page 466
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**Definition: Hash Function**
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A **hash function** is a function defined from a larger, possibly infinite, set
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of data to a smaller fixed-size set of integers.
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---
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Page 469
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**Definition**
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Let $F$ be a function from a set $X$ to a set $Y$. $F$ is **onto** (or
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**surjective**) if, and only if, given any element $y$ in $Y$, it is possible to
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find an element $x$ in $X$ with the property that $y = F(x)$.
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Symbolically:
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$$ F:X \to Y \text{ is onto } \Leftrightarrow \forall y \in Y, \exists x \in X \text{ such that } F(x) = y $$
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---
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Page 472
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**Laws of Exponents**
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If $b$ and $c$ are any positive real numbers and $u$ and $v$ are any real
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numbers, the following laws of exponents hold true:
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7.2.1
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$$ b^ub^v = b^{u + v} $$
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7.2.2
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$$ (b^u)^v = b^{uv} $$
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7.2.3
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$$ \frac{b^u}{b^v} = b^{u - v} $$
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7.2.4
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$$ (bc)^u = b^uc^u $$
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---
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Page 473
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**Theorem 7.2.1 Properties of Logarithms**
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For any positive real numbers $b$, $c$, $x$ and $y$ with $b \neq 1$ and
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$c \neq 1$ and for every real number $a$:
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a. $\log_b(xy) = \log_bx + \log_by$
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b. $\log_b\left(\dfrac{x}{y}\right) = \log_bx - \log_by$
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c. $\log_b(x^a) = a\log_bx$
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d. $\log_cx = \dfrac{\log_bx}{\log_bc}$
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---
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Page 475
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**Definition**
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A **one-to-one correspondence** (or **bijection**) from a set $X$ to a set $Y$
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is a function $F: X \to Y$ that is both one-to-one and onto.
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---
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Page 478
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**Theorem 7.2.2**
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Suppose $F: X \to Y$ is a one-to-one correspondence; in other words, suppose $F$
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is one-to-one and onto. Then there is a function $F^{-1}: Y \to X$ that is
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defined as follows:
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Given any element $y$ in $Y$,
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$$ F^{-1}(y) = \text{ that unique element } x \text{ in } X \text{ such that } F(x) \text{ equals } y $$
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Or, equivalently,
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$$ F^{-1}(y) = x \Leftrightarrow y = F(x) $$
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---
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Page 478
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**Definition**
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The function $F^{-1}$ of Theorem 7.2.2 is called the **inverse function** for
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$F$.
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---
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Page 479
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**Theorem 7.2.3**
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If $X$ and $Y$ are sets and $F: X \to Y$ is one-to-one and onto, then
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$F^{-1}:Y \to X$ is also one-to-one and onto.
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**Proof:**
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**$F^{-1}$ is one-to-one:**
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Suppose $y_1$ and $y_2$ are elements of $Y$ such that
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$F^{-1}(y_1) = F^{-1}(y_2)$. _[We must show that $y_1 = y_2$.]_ Let
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$x = F^{-1}(y_1) = F^{-1}(y_2)$. Then $x \in X$, and by definition of $F^{-1}$,
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$$ F(x) = y_1 \text{ since } x = F^{-1}(y_1) $$
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and
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$$ F(x) = y^2 \text{ since } x = F^{-1}(y_2) $$
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Consequently, $y_1 = y_2$ because each is equal to $F(x)$. _[This is what was to
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be shown.]_
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**$F^{-1}$ is onto:**
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Suppose $x \in X$. _[We must show that there exists an element $y$ in $Y$ such
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that $F^{-1}(y) = x$.]_ Let $y = F(x)$. Then $y \in Y$, and by definition of
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$F^{-1}$, $F^{-1}(y) = x$ _[as was to be shown.]_
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