Last modified: 2026-10-27 20:53:54 (PDT)
Example 1 (A transitive relation and one that is not)
Theorem 1 (Equalities are transitive) If \(a=b\) and \(b=c\), then \(a=c\). That is, \(=\) is a transitive relation.
Theorem 2 (Substituting equivalent expressions) If \(a = b\), then for any function \(f(x)\), \(f(a) = f(b)\)
Definition 2 (Identity) An identity [2] is an equation between two expressions that holds for every value of its variables in a stated set, such as the real numbers.
Example 2 (An identity, and an equation that is not one)
Definition 3 (Strict inequalities) For real numbers \(a\) and \(b\), \(a < b\) (“\(a\) is less than \(b\)”) and \(a > b\) (“\(a\) is greater than \(b\)”) are strict inequalities [3].
Definition 4 (Non-strict inequalities) For real numbers \(a\) and \(b\), the non-strict inequalities also allow equality, unlike the strict inequalities (Definition 3):
Example 3 (Strict and non-strict inequalities)
Theorem 3 (Adding to both sides of an inequality) If \(a<b\), then \(a+c < b+c\)
Theorem 4 (Negating both sides of an inequality) If \(a < b\), then: \(-a > -b\)
Theorem 5 (Multiplying both sides of an inequality by a positive number) If \(a < b\) and \(c > 0\), then \(ca < cb\).
Theorem 6 (Negation is multiplication by \(-1\)) \[-a = (-1)*a\]
Definition 5 (Power) Let \(a\) be a real number and \(n\) a natural number. The \(n\)th power of \(a\) is the product of \(n\) copies of \(a\) [4]:
\[a^n \stackrel{\text{def}}{=}\underbrace{a \cdot a \cdot\cdots \cdot a}_{n \text{ factors}}\]
Definition 6 (Base of a power) In \(a^n\) (Definition 5), \(a\) is the base.
Definition 7 (Exponent of a power) In \(a^n\) (Definition 5), \(n\) is the exponent.
Definition 8 (Square) The second power \(a^2\) (Definition 5) is the square of \(a\).
Example 4 (Powers)
\[ \begin{aligned} 2^3 &= 2 \cdot 2 \cdot 2 \\ &= 8, \end{aligned} \]
with base \(2\) and exponent \(3\).
\[ \begin{aligned} (-3)^2 &= (-3) \cdot(-3) \\ &= 9, \end{aligned} \]
but
\[ \begin{aligned} -3^2 &= -(3^2) \\ &= -9: \end{aligned} \]
the exponent applies only to the \(3\).
\(5^0 = 1\).
\[ \begin{aligned} 2^{-2} &= \frac{1}{2^2} \\ &= \frac{1}{4}. \end{aligned} \]
Definition 9 (Powers with exponent zero or a negative integer) For the exponent \(0\) and negative integer exponents:
Remark 1 (Powers with real exponents). Definition 26 in Polynomials, Exponentials and Logarithms extends powers to real exponents when the base is positive.
Definition 10 (Square root) For a real number \(a \ge 0\), the square root of \(a\), written \(\sqrt{a}\) [5], is the non-negative real number whose square (Definition 5) is \(a\).
Example 5 (Square roots)
Definition 11 (Absolute value of a real number) The absolute value of a real number \(a\) [6] is
\[ \mathopen{}\left|a\right|\mathclose{} \stackrel{\text{def}}{=}\begin{cases} a, & a \ge 0 \\ -a, & a < 0 \end{cases} \]
Example 6 (Absolute values)
\(\mathopen{}\left|2.5\right|\mathclose{} = 2.5\), because \(2.5 \ge 0\).
\[ \begin{aligned} \mathopen{}\left|-3\right|\mathclose{} &= -(-3) \\ &= 3, \end{aligned} \]
because \(-3 < 0\).
\(\mathopen{}\left|0\right|\mathclose{} = 0\).
\[ \begin{aligned} \mathopen{}\left|-3\right|\mathclose{} &= 3 \\ &= \sqrt{9} \\ &= \sqrt{(-3)^2}: \end{aligned} \]
for every real \(a\), \(\mathopen{}\left|a\right|\mathclose{} = \sqrt{a^2}\) (Definition 10).
\(\mathopen{}\left|x - 1\right|\mathclose{} < 1\) says that \(x\) is less than \(1\) away from \(1\), that is, \(0 < x < 2\).
Remark 2 (Absolute value as distance). For real numbers \(a\) and \(b\), \(\mathopen{}\left|a - b\right|\mathclose{}\) is the distance between \(a\) and \(b\) on the number line.
The word “percent” comes from the Latin “per centum”; “centum” is Latin for 100, so “percent” means “per hundred” [7].
Definition 12 (Percent) For a real number \(x\), “\(x\) percent”, written \(x\%\), means \(x / 100\) [7]:
\[ x\% \stackrel{\text{def}}{=}\frac{x}{100} \]
The percent sign “%” is a shorthand for “\(/100\)”.
Example 7 (Ten percent is one tenth) Contrary to what you may have learned previously, the equality \(10\% = 0.1\) is true and correct, just as \(10 \text{kg} = 10,000 \text{g}\) is true and correct:
\[ \begin{aligned} 10\% &= \frac{10}{100} && \text{(definition of percent)} \\ &= 0.1 && \text{(divide)} \end{aligned} \]
You are welcome to switch between decimal and percent notation freely; just make sure you execute it correctly.
Exercise 1 (Smallest value, and where it occurs)
Solution 1.
Solution.
Evaluate \(g\) at each of its four inputs:
| \(x\) | \(0\) | \(1\) | \(3\) | \(4\) |
|---|---|---|---|---|
| \(g(x)\) | \(5\) | \(2\) | \(2\) | \(5\) |
The smallest value of \(g\) is \(2\), and \(g\) takes it at two inputs, \(x = 1\) and \(x = 3\). The answer to “where?” is the set \(\mathopen{}\left\{1, 3\right\}\mathclose{}\), not a single number.
Definition 13 (Minimum) Let \(A \subseteq \mathbb{R}\). A number \(m\) is the minimum of \(A\) [8], written \(\min A\), if \(m \in A\) and \(m \le a\) for all \(a \in A\).
Remark 3 (Not every set has a minimum). The set \(\mathopen{}\left\{3, 1, 4\right\}\mathclose{}\) has minimum \(1\), and the interval \([0, 1]\) has minimum \(0\). The interval \((0, 1]\) has no minimum: every element \(a\) of it has a smaller element, \(a/2\), also in it.
Definition 14 (Maximum) Let \(A \subseteq \mathbb{R}\). A number \(M\) is the maximum of \(A\) [8], written \(\max A\), if \(M \in A\) and \(a \le M\) for all \(a \in A\).
Example 8 (A maximum, and a set without one)
Definition 15 (Argmin) Let \(f : A \to \mathbb{R}\) be a function. The argmin of \(f\) over \(A\) [9] is the set of inputs where \(f\) takes its smallest value:
\[\arg \min_{x \in A} f(x) \stackrel{\text{def}}{=}\mathopen{}\left\{x \in A : \forall a \in A,\ f(x) \le f(a)\right\}\mathclose{}\]
When this set has exactly one element \(\hat{x}\), we write \(\hat{x} = \arg \min_{x \in A} f(x)\).
Remark 4 (Smallest value versus where it occurs). The smallest value itself is \(\min f(A)\), the minimum (Definition 13) of the image of \(f\), and the argmin is where that value is attained. For example, let \(f(x) = x^2\) on \(A = \mathopen{}\left\{-1, 0, 2\right\}\mathclose{}\). The image is \(f(A) = \mathopen{}\left\{1, 0, 4\right\}\mathclose{}\), so the smallest value is \(\min f(A) = 0\), and \(\arg \min_{x \in A} f(x) = \mathopen{}\left\{0\right\}\mathclose{}\).
The argmin is empty when \(f\) has no smallest value, for example \(f(x) = x\) on \((0, 1]\).
Definition 16 (Argmax) Let \(f : A \to \mathbb{R}\) be a function. The argmax of \(f\) over \(A\) [9] is the set of inputs where \(f\) takes its largest value:
\[\arg \max_{x \in A} f(x) \stackrel{\text{def}}{=}\mathopen{}\left\{x \in A : \forall a \in A,\ f(a) \le f(x)\right\}\mathclose{}\]
When this set has exactly one element \(\hat{x}\), we write \(\hat{x} = \arg \max_{x \in A} f(x)\).
Example 9 (An argmax with two points, and an empty one)
Example 10 (The positive part of the identity function) For \(f(x) = x\) on \(\mathbb{R}\), the positive part is \(f^{+}(x) = \max\mathopen{}\left\{x, 0\right\}\mathclose{}\). At \(x = 3\), \(f^{+}(3) = \max\mathopen{}\left\{3, 0\right\}\mathclose{} = 3\), and at \(x = -2\), \(f^{+}(-2) = \max\mathopen{}\left\{-2, 0\right\}\mathclose{} = 0\).
Definition 18 (Negative part of a function) For \(f\) as in Definition 17, the negative part of \(f\) is the non-negative function
\[ f^{-}(s) \stackrel{\text{def}}{=}\max\mathopen{}\left\{-f(s), 0\right\}\mathclose{}. \]
Example 11 (The negative part of the identity function) For \(f(x) = x\) on \(\mathbb{R}\), the negative part is \(f^{-}(x) = \max\mathopen{}\left\{-x, 0\right\}\mathclose{}\). At \(x = 3\), \(f^{-}(3) = \max\mathopen{}\left\{-3, 0\right\}\mathclose{} = 0\), and at \(x = -2\), \(f^{-}(-2) = \max\mathopen{}\left\{2, 0\right\}\mathclose{} = 2\).
Exercise 2 (A dip that is not the bottom) Let \(f(x) = x^3 - 3x\) for \(x \in \mathbb{R}\).
Solution 2.
Since
\[ \begin{aligned} f(1) &= 1 - 3 \\ &= -2, \end{aligned} \]
so
\[ \begin{aligned} f(x) - f(1) &= x^3 - 3x + 2 \\ &= (x - 1)(x^2 + x - 2) \\ &= (x - 1)(x - 1)(x + 2) \\ &= (x - 1)^2 (x + 2). \end{aligned} \]
For \(x \in (0, 2)\), \((x - 1)^2 \ge 0\) and \(x + 2 > 0\), so their product is at least \(0\). By part 1, \(f(x) - f(1) \ge 0\), that is, \(f(x) \ge f(1)\). So \(f(x)\) is never smaller than \(f(1)\) on \((0, 2)\). It equals \(f(1)\) only when \((x - 1)^2 = 0\), that is, only at \(x = 1\), and it is larger than \(f(1)\) at every other \(x \in (0, 2)\).
Solution.
\[ \begin{aligned} f(-3) &= -27 + 9 \\ &= -18. \end{aligned} \]
Since \(-18 < -2 = f(1)\), \(f(1)\) is not the smallest value of \(f\) on \(\mathbb{R}\).
No. For each \(n \ge 2\),
\[ \begin{aligned} f(-n) &= -n^3 + 3n \\ &= -n(n^2 - 3) \\ &\le -n, \end{aligned} \]
so no value of \(f\) is smaller than all the others: for any \(x\), picking \(n \ge 2\) with \(-n < f(x)\) gives \(f(-n) < f(x)\).
Definition 19 (Global minimizer) Let \(A \subseteq \mathbb{R}^p\) and let \(f : A \to \mathbb{R}\) be a function. A point \(x^* \in A\) is a global minimizer of \(f\) over \(A\) if \(f(x^*) \le f(x)\) for all \(x \in A\).
Remark 5 (Global minimizers form the argmin). The global minimizers of \(f\) are exactly the elements of \(\arg \min_{x \in A} f(x)\) (Definition 15). A function can have more than one global minimizer. For example, \(f(x) = (x^2 - 1)^2\) on \(\mathbb{R}\) satisfies \(f(x) \ge 0\) for every \(x\), and \(f(x) = 0\) exactly when \(x = -1\) or \(x = 1\), so its global minimizers are \(-1\) and \(1\), and \(\arg \min_{x \in \mathbb{R}} f(x) = \mathopen{}\left\{-1, 1\right\}\mathclose{}\).
Definition 20 (Local minimizer) Let \(A \subseteq \mathbb{R}^p\) and let \(f : A \to \mathbb{R}\) be a function. A point \(x^* \in A\) is a local minimizer of \(f\) if there is a number \(\delta> 0\) such that \(f(x^*) \le f(x)\) for all \(x \in A\) with \(\mathopen{}\left\lVert x - x^*\right\rVert\mathclose{} < \delta\), where \(\mathopen{}\left\lVert\cdot\right\rVert\mathclose{}\) is the Euclidean norm.
Remark 6 (Local and global minimizers). For \(p = 1\), \(\mathopen{}\left\lVert x - x^*\right\rVert\mathclose{} = \mathopen{}\left|x - x^*\right|\mathclose{}\). For example, with \(x^* = 1\) and \(\delta= 1\), the condition \(\mathopen{}\left|x - 1\right|\mathclose{} < 1\) means \(0 < x < 2\).
Every global minimizer (Definition 19) is a local minimizer: take any \(\delta> 0\). For example, \(x^* = 0\) is a global minimizer of \(f(x) = x^2\) on \(\mathbb{R}\), so it is also a local minimizer.
The converse fails. In Exercise 2, \(x^* = 1\) is a local minimizer of \(f(x) = x^3 - 3x\) (take \(\delta= 1\)), but not a global one.
Example 12 (A point that is not a local minimizer) For \(f(x) = x^3 - 3x\), the point \(x = -1\) is not a local minimizer. For any \(t\),
\[ \begin{aligned} f(-1 + t) &= (-1 + t)^3 - 3(-1 + t) && \text{(substitute)} \\ &= (-1 + 3t - 3t^2 + t^3) - 3(-1 + t) && \text{(expand the cube)} \\ &= -1 + 3t - 3t^2 + t^3 + 3 - 3t && \text{(multiply } -3 \text{ into the parentheses)} \end{aligned} \]
\[ \begin{aligned} f(-1 + t) &= 2 - 3t^2 + t^3 && \text{(} 3t - 3t = 0 \text{ and } -1 + 3 = 2 \text{)} \\ &= 2 + t^2 (t - 3) && \text{(factor out } t^2 \text{)} \\ &= f(-1) + t^2 (t - 3), && \text{(} f(-1) = (-1)^3 - 3(-1) \text{, which equals } 2 \text{)} \end{aligned} \]
and \(t^2 (t - 3) < 0\) whenever \(t \ne 0\) and \(t < 3\). So whatever \(\delta> 0\) is, the point \(x = -1 + t\) with \(t = \min\mathopen{}\left\{\delta/ 2, 1\right\}\mathclose{}\) satisfies \(\mathopen{}\left|x - (-1)\right|\mathclose{} < \delta\) and \(f(x) < f(-1)\). For example,
\[ \begin{aligned} f(-0.9) &= -0.729 + 2.7 \\ &= 1.971 \\ &< 2. \end{aligned} \]